Prasanna

Students can Download Chapter 4 Presentation of Data Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 4 Presentation of Data

Plus One Economics Presentation of Data One Mark Questions and Answers

Question 1.
Which of the following comes under geometric diagram?
(a) Histogram
(b) Bar diagram
(c) Ogives
(d) Frequency polygon
Answer:
(b) Bar diagram

Question 2.
Which of the following comes under frequency diagrams?
(a) Bar diagram
(b) Histogram
(c) Pie diagram
(d) All the above
Answer:
(b) Histogram

Question 3.
To draw time-series graph, time is presented on:
(a) X-axis
(b) Y-axis
(c) any of two
Answer:
(a) X-axis

Question 4.
Name the types of graphs.
Answer:

1. One dimensional graph
2. Two-dimensional graph
3. Three-dimensional graph
4. Pictograms

Question 5.
State whether true or false.

1. The width of bars in a bar diagram need not be equal.
2. The width of rectangles in a histogram should essentially be equal.
3. Histograms can only be formed with continuous classification of data.
4. Histogram and column diagram are the same method of presentation of data.
5. Mode of a frequency distribution can be drawn graphically with the help of histogram,
6. The median of a frequency distribution cannot be drawn from the Ogive.

Answer:

1. true
2. false
3. true
4. true
5. true
6. true

Plus One Economics Presentation of Data Two Mark Questions and Answers

Question 1.
Which of the following is a cumulative frequency curve?
Answer:
(a) Bar diagram
(b) Histogram
(c) Ogive
(d) Pie diagram
Answer:
(c) Ogive

Question 2.
Distinguish between captions and stubs.
Answer:
Captions refers to the column headings and stubs refers to the row heading.

Question 3.
Match the following.

Answer:

Plus One Economics Presentation of Data Three Mark Questions and Answers

Question 1.
What kind of diagrams are more effective in representing the following?

1. Monthly rainfall in a year
2. Composition of the population of Delhi by religion
3. Components of cost in a factory

Answer:

1. Simple bar diagram
2. Sub-divided or component bar diagram
3. Pie diagram

Question 2.
Name different types of diagrams.
Answer:
The different types of diagrams are:
1. Geometric diagram

• Bar diagrams
• Pie diagram

2. Frequency diagram

• Histogram
• Frequency polygon
• Frequency curve -Ogive

3. Arithmetic line graph

Question 3.
“Diagrams and graphs help us visualize the whole meaning of numerical complex data at a single glance”. Comment.
Answer:
One of the most convincing and appealing ways in which statistical results may be presented is through diagrams and graphs. The special feature of graphs and diagrams is that they do away with figures altogether. Diagrams and graph is a statistical method which can be used for simplifying the complexity of quantitative data and t make them easily intelligible.

It presents dry and uninteresting statistical facts in the shape of attractive and appealing pictures and charts. They are important methods of visual aids and are appealing t the eye and mind of the observer.

Question 4.
“There are generally three forms of diagrammatic presentation of data” explain.
Answer:
There are various methods to present data. But generally, three forms of presentation of data are there
which are noted below:

1. Geometric diagram
2. Frequency diagram
3. Arithmetic line graph

1. Geometric Diagram:
Bar diagram and pie diagram come in the category of geometric diagram for presentation of data. The bar diagrams are of three types-simple, multiple and component bar diagrams.

2. Frequency Diagram:
Data in the form of grouped frequency distributions are generally represented by frequency diagrams like histogram, frequency polygon, frequency curve, and ogive

3. Arithmetic Line Graph:
An arithmetic line graph is also called time-series graph and is a method of diagrammatic presentation of data. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph or time-series graph.

Question 5.
Explain Ogive?
Answer:
Cumulative frequency of any class is equal to the sum of the frequencies of all the classes preceding that class and its own frequency e.g., frequencies are 10, 7, 12, 17 and 22. Cumulative frequencies are 10, 10 + 7 = 17, 17 + 12 = 29, 29 + 17 = 46 and 46 + 22 = 68.
Cumulative frequency of the last class = Total frequency.

For drawing an Ogive, cumulative frequency (i.e. number of values) is taken on the Y-axis and limits of class intervals on the X-axis.
Ogive is of two types:

1. less than
2. more than

In a “less than” type Ogive, we plot the upper limit of each class along the X-axis and in a “more than” type Ogive, we plot the lower limit of each class along the X-axis. Along the Y-axis, we plot the cumulative frequencies at the end of each class. Ogive can be drawn even if the class interval are unequal or open end. Ogives are performed over frequency curves for comparative study.

Question 6.
Illustrate how classes can be formed while presenting the data?
Answer:
Classes can be formed in two ways:

1. Exclusive type
2. Inclusive type

1. Exclusive Type:
When the class intervals are so fixed that the upper limit of one class is the lower limit of the new class, it is known as exclusive method of classification.

In this method, higher value of the variable in the class is not included in that class i.e.,

2. Inclusive Type:
In this method, the students getting say 39% marks will be included in class 30 – 39 itself i.e.,

Plus One Economics Presentation of Data Four Mark Questions and Answers

Question 1.
Choose the correct answer
a. Bar diagram is a

1. one-dimensional diagram
2. two-dimensional diagram
3. diagram with no dimension
4. none of the above

b. Data represented through a histogram can help in finding graphically the

1. mean
2. mode
3. median
4. all the above

c. Ogives can be helpful in locating graphically the

1. mode
2. mean
3. median
4. none of the above

d. Data represented through arithmetic line graph help in understanding

1. long term trend
2. cyclicity in data
3. seasonality in data
4. all the above

Answer:
a. 1. one-dimensional diagram
b. 3. mode
c. 3. median
d. 1. long term trend

Question 2.
Point out major parts of a statistical table.
Answer:

1. Table number
2. Title
3. Headnote
4. Stub
5. Box head or caption
6. Body or field
7. Footnote
8. Source note

Question 3.
Give the rules for constructing tables.
Answer:
The rules of constructing diagrams are:

• Every diagram should be titled.
• It should suit the size of the paper
• It should be neat and attractive
• It should be neatly indexed
• It should contain footnotes
• The details in diagram should be self-explanatory

Plus One Economics Presentation of Data Five Mark Questions and Answers

Question 1.
Explain the advantages of diagrammatic presentation.
Answer:
The advantages of diagrammatic presentation are given below.

1. Diagram give a clear picture of data
2. Comparison can be made easy
3. Diagrams can be used university at any place
4. It saves time and energy
5. The data can be remembered easily

Question 2.
Show how pie diagram is drawn for the following data?

Answer:

Question 3.
Give steps in the preparation of pie diagram.
Answer:
A pie diagram is also a component diagram, but unlike a component bar diagram, a circle whose area is proportionally divided among the components it represents. It is also called a pie chart. The circle is divided into as many parts as there are components by drawing straight lines from the centre to the circumference.

The following steps in the preparation of pie diagram are given below:

• Convert each component as percentage of the total.
• Multiply the percentage by 360/100 = 3.6 to convert into degree.
• Starting with the twelve o’clock position on the circle draw the largest component circle
• Draw other components in clockwise succession in descending order of magnitude except for each all components

Like all others and miscellaneous which are shown last:

• Use different columns or shades to distinguish between different components
• Explain briefly the different components either within the components in the figure or outside by arrow.

Plus One Economics Presentation of Data Eight Mark Questions and Answers

Question 1.
Write short notes on the following

1. pie diagrams
2. frequency curves
3. frequency polygon
4. ogive
5. arithmetic line graph

Answer:
1. Pie Diagram:
A pie diagram is also a component diagram, but unlike a component bar diagram, a circle whose area is proportionally divided among the components it represents. It is also called a pie chart. The circle is divided into as many parts as there are components by drawing straight lines from the centre to the circumference. Pie charts usually are not drawn with absolute values of a category.

The values of each category are first expressed as percentage of the total value of all the categories. A circle in a pie chart, irrespective of its value of radius, is thought of having 100 equal parts of 3.6° (3607100) each. To find out the angle, the component shall subtend at the centre of the circle, each percentage figure of every component is multiplied by 3.6°.

2. Frequency Polygon:
A frequency polygon is a plane bounded by straight lines, usually four or more lines. Frequency polygon is an alternative to histogram and is also derived from histogram itself. A frequency polygon can be fitted to a histogram for studying the shape of the curve. The simplest method of drawing a frequency polygon is to join the midpoints of the topside of the consecutive rectangles of the histogram.

3. Frequency Curve:
The frequency curve is obtained by drawing a smooth freehand curve passing through the points of the frequency polygon as closely as possible. It may not necessarily pass through all the points of the frequency polygon but it passes through them as closely as possible

4. Ogive:
Ogive is also called cumulative frequency curve. As there are two types of cumulative frequencies, for example, less than type and more than type, accordingly there are two ogives for any grouped frequency distribution data. Here in place of simple frequencies as in the case of frequency polygon, cumulative frequencies are plotted along y-axis against class limits of the frequency distribution.

For less than give the cumulative frequencies are plotted against the respective upper limits of the class intervals whereas for more than ogives the cumulative frequencies are plotted against the respective lower limits of the class interval. An interesting feature of the two ogives together is that their intersection point gives the median

5. Arithmetic Line Graph:
An arithmetic line graph is also called time-series graph and is a method of diagrammatic presentation of data. Init, time (hour, day/date, week, month, year, etc.) is plotted along x-axis and the value of the variable (time series data) along y-axis. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph (time series graph). It helps in understanding the trend, periodicity, etc. in a long term time series data.

Question 2.
3 Forms of presentation of data

1. Textual
2. Tabular
3. Diagrams & graphs Prepare a flow chart.

Answer:

Students can Download Chapter 10 Respiration in Plants Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 10 Respiration in Plants

Plus One Botany Respiration in Plants One Mark Questions and Answers

Question 1.
Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under the aerobic condition, it forms
(a) Lactic acid
(b) CO₂+H₂O
(c) Acetyl CoA + CO₂
(d) Ethanol + CO₂
Answer:
(b) CO₂+ H₂O

Question 2.
Electron Transport System (ETS) is located in mitochondrial
(a) Outer membrane
(b) Inter membrane space
(c) Inner membrane
(d) Matrix
Answer:
(c) Inner membrane

Question 3.
Choose the correct statement:
(a) During the conversion of succinyl Co-A to succinic acid, a molecule of ATP is synthesized.
(b) Oxygen is vital in respiration for removal of hydrogen.
(c) Pyruvate is formed in the mitochondrial matrix.
(d) There is complete breakdown of glucose in fermentation.
Answer:
(a) During the conversion of succinyl Co-A to succinic acid, a molecule of ATP is synthesized.

Question 4.
Mitochondria are called powerhouses of the cell. Which of the following observations support this statement?
(a) Mitochondria have a double membrane
(b) The enzymes of the Krebs cycle and the cytochromes are found in mitochondria.
(c) Mitochondria synthesise ATP
(d) Mitochondria are found in almost all plants and animal cells.
Answer:
(c) Mitochondria synthesise ATP.

Question 5.
The end product of oxidative phosphorylation is
(a) NADH
(b) Oxygen
(c) ADP
(d) ATP+H₂O
Answer:
(d) ATP+H₂O

Question 6.
Fo-F1 particles participate in the synthesis of
Answer:
ATP molecules

Question 7.
The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively ……… and …….
Answer:
Lactic acid and Ethyl alcohol.

Question 8.
How many NADH₂ molecules are produced from molecule of acetyl co A in TCA cycle?
Answer:
Three.

Question 9.
Name the final acceptor of electrons in ETC.
Answer:
Oxygen.

Question 10.
Give the number of ATP molecules produced from 1 glucose molecule in aerobic respiration.
Answer:
38.

Question 11.
The process of oxidation in microorganisms are referred as
(a) Cellular respiration
(b) Aerobic respiration
(c) Fermentation
Answer:
(c) Fermentation

Question 12.
Calculate the RQ of tripalmitin with the help of equation given below.
2(C₅H₉₈O₆) + 145O₂ -»102 CO₂ + 98H₂O + energy
Answer:
\(\mathrm{RQ}=\frac{102 \mathrm{CO}_{2}}{145 \mathrm{O}_{2}}=0.7\)

Question 13.
In the respiration process both in aerobic and anaerobic, the first phase of reactions is the same. Write the name of reaction Answer:
Glycolysis.

Question 14.
A germinating seed released 120 CO₂ by utilizing 180 O₂ stage. Calculate the RQ. Name the respiratory substrate.
Answer:
The substrate is fat.

Question 15.
Respiratory pathway is an amphibhotic pathway. Comment.
Answer:
It involves both anabolic & catabolic pathways.

Question 16.
Name the unit of oxidative phosphorylation.
Answer:
Oxysomes or Elementary particles.

Question 17.
How many calories are produced by aerobic oxidation of 1 gm mole of glucose?
Answer:
686000 calories.

Question 18.
We commonly call ATP as the energy currency of the cell. Can you think of some other energy carriers present in a cell? Name any two.
Answer:
NADH₂ and FADH₂.

Question 19.
RQ value changes for wheat and castor oil seeds. Justify.
Answer:
R Q = 1 carbohydrate (wheat)
R Q < 1 fat(castor)

Plus One Botany Respiration in Plants Two Mark Questions and Answers

Question 1.
Pyruvate, which is formed by the glycolytic catabolism of carbohydrates undergoes oxidative phosphorylation.

1. Where does it occur?
2. Who first elucidate this?

Answer:

1. Cristae of mitochondria
2. Hans Kreb

Question 2.
Glycolysis is the first stage in the break down of glucose in all organisms. Do you agree with this statement? Where does it take place? Give the net. gain of ATP molecules in glycolysis?
Answer:
Yes, Cytoplasm, 8ATPs

Question 3
Observe the relation of the first pair and fill up the blanks

1. Symport: Molecules cross the membrance in the same direction
   …………..: Molecules move in the opposite direction
2. RQ = 1 : Carbohydrate
   RQ > 1: ………..

Answer:

1. Antiport
2. Organic acids

Question 4.
‘‘Photorespiration is called a wasteful process.” Comment on it.
Answer:
ATP is utilized. There is no production of ATP during this process. It does not produce any beneficial product.

Question 5
Respirometer is an apparatus used to measure R.Q.

1. What is R.Q?
2. R.Q. of glucose is equal to one. Give the reason.

Answer:

1. Amount of CO₂ released during respiration / Amount of O₂ absorbed during respiration,
2. Amount of CO₂ released= Amount O₂ absorbed

Question 6.
Plant physiologists observed a relationship between respiration and salt absorption. Is absorption of salt increased due to respiration? Explain.
Answer:
Yes. As a result of respiration, the energy is released in the form of ATP. This energy is used for the active absorption of salt.

Question 7.
Glycolysis is the first step of both aerobic and anaerobic respiration. During this process glucose undergoes partial oxidation.

1. What is the final product of glycolysis?
2. What is the total yield of ATP molecules during glycolysis?

Answer:

1. Pyruvic acid
2. 8 ATP

Question 8.
The RQ value of a respiratory substrate is 1. Find the type of substrate and comment on RQ.
Answer:
Carbohydrate RQ:
It is the ratio of volume of CO₂ evolved to the volume of O₂ consumed. The value of RQ is different for respiratory substrates eg fat(0.7), protein (0.9)etc.

Question 9.
Oxidative phosphorylation is an important process in cellular respiration. Explain it?
Answer:
Oxidative phosphorylation takes place in the inner membrane of mitochondria. In the F1 head the inorganic phosphate combined with ADP to form ATP. It occurs due to the entry of proton pair from intermembrane space to the matrix through proton channel.

Question 10.
Why does anaerobic respiration produce less energy than aerobic respiration?
Answer:

1. The regeneration of NAD fails to produce ATP as the electrons are not shifted to oxygen
2. The end product of anaerobic respiration can be further oxidized to release energy.

Question 11.
What is fermentation Name any two organic compounds produced in this process?
Answer:
Anaerobic respiration also called fermentation involves the production of energy from food nutrients in the absence of oxygen.

Question 12.
“Respiration is an amphibolic pathway”. Justify the statement with reason.
Answer:
Respiration involves both the catabolic and anabolic process. Fatty acid undergoes breackdown to form acetyl. co-A.
Acetyl Co-A is again used for the synthesis of Fat. In respiration cycle the two process are involved.i.e catabolism and anabolism. So it is called a Amphibolic Pathway”.

Question 13.
ADP is converted to ATP as a result of phosphorylation which takes place in photosynthesis and respiration during electron transport system. What is the difference between these two?
Answer:
photophosphorylation, inorganic phosphate combines with ADP to form ATP This process takes place in chloroplast in the presence light. But in oxidative phosphorylation inorganic phosphate combines with ADP to form ATP. It takes place in Mitochondria.

Question 14.
When does anaerobic respiration occur in man and yeast?
Answer:
In man during streneous exercise O_2, inedequecy is experience in Skeletal tissues. In such a case the respiration is anaerobic. In yeast, during fermentation the respiration takes place is anaerobic.

Question 15.
Mention the important series of events of aerobic respiration that occur in the matrix of the mitochondrion as well as ones that take place in inner membrane of the mitochondrion.
Answer:
Citric acid cycle,Terminal oxidation and electron transport system.

Question 16.
Name the end products aerobic and anaerobic glycolysis. List the two ways by which molecules of ATP are produced in glycolysis during aerobic respiration in a cell.
Answer:
2 Pyruvic acid, 2ATP and 2 NADH₂ Two ways of ATP formation in glycolysis are:

1. During dephosphorylation of 1, 3, DPGA called substrate-level phosphorylation
2. Reduced NADH₂ releases ATP molecules in electron transport chain

Question 17.
Define RQ What is its significance?
Answer:
It is the ratio of volume of CO₂ released to volume of O₂ taken during respiration. Value of R Q gives the indication of nature of substrate respired by a particular tissue.

Question 18.
Different substrates get oxidized during respiration. How does Respiratory Quotient (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidized?
R Q=A/B
What do A and B stand for?
What type of substrates have R.Q. of 1, < 1 or> 1?
Answer:
A-Volume CO₂ evolved B- Volume O₂ consumed RQ=1 (Carbohydrate) RQ=< 1 Fat/protein RQ=> 1 (Organic acids).

Question 19.
Which of the following will release more energy on oxidation? Arrange them in ascending order,
a. Moffat
b. 1 gm of protein
c. 1 gm of glucose
d. 0.5 g of protein + 0.5g glucose
Answer:
c → d → b → a

Question 20.
What is the importance of F0-F, particle in ATP production during aerobic respiration?
Answer:
F0 is an integral membrane protein complex and has a proton channel while F1 has enzyme ATP synthase to form ATP. One molecule of NADH produces 3 ATP molecules in ETS.

Question 21.
What is Respiratory Quotient? What is its value for fats?
Answer:
It is the ratio of the volume of CO₂ evolved to the volume of O₂consumed.
\(So, RQ =\frac{\text { Volume of } \mathrm{CO}_{2} \text { Evolved }}{\text { Volume of } \mathrm{O}_{2} \text { consumed }}\)

Question 22.
Give two example for anaerobic respiration occurs in organisms and products.
Answer:
Alcoholic fermentation and Lactic acid fermentation Ethyl alcohol and lactic acid.

Question 23.
The complete oxidation of pyruvic acid that occurs in mitochondria undergoes both oxidation and decarboxylation.

1. Which is the first step of oxidative-decarboxylation occurs?
2. Where does substrate-level phosphorylation occurs?

Answer:

1. Conversion of pyruvic acid to acetyl CoA
2. Conversion of succinyl CoA to Succinic acid.

Question 24.
Give the name of intermediate compounds having carbon atoms 6,5,4 and 2 of mitochondrial oxidation.
Answer:

• 6C – Citric acid, 5C- Alpha keto glutaric acid,
• 4C- Succinic acid/succinyl coAl malic acid/ OAA,
• 2C- Acetyl CoA.

Plus One Botany Respiration in Plants Three Mark Questions and Answers

Question 1.
Incomplete oxidation of food material in the absence of oxygen is anaerobic respiration.
(a) Give an example of a microorganism which carry out anaerobic respiration.
(b) Where does anaerobic respiration take place in human beings?
(c) What is the net gain of ATP during this process? Spa) Yeast
(b) Muscle cells
Answer:
(c) 2 ATP

Question 2.
ETS operated in the inner mitochondrial membrane, it involves terminal oxidation and oxidative phosphorylation

1. Who discovered chemiosmotic hypothesis in mitochondria
2. Name the elementary particle promotes ATP synthesis
3. What is chemiosmotic hypothesis

Answer:

1. Peter Mitchel
2. Fg-F, particle
3. Proton gradient leads to ATP production

Question 3.
Yeast cells can respire in the absence of O₂

1. Name the process
2. Write the end product of the process
3. Write the net production of ATP during process.

Answer:

1. Anaerobic respiration
2. CO₂ + C₂H₅OH (Carbon dioxide and ethyl alcohol)
3. 2 ATPs

Question 4.
Respiration is a breakdown process it involves various steps

1. Where does the common step of aerobic and anaerobic process occurs
2. Find out the number of carbon atoms of a compound as end product of the above reaction
3. Name the 6 C intermedite compound splits and forms another 3 C intermediates.

Answer:

1. Cytoplasm
2. 3 Carbon (Pyruvic acid)
3. Fructose 1, 6 biphosphate

Question 5.
ATP and NADPH₂ molecules synthesised in light reaction of photosynthesis are used for the synthesis of glucose in dark reaction.

1. Who proposed the dark reaction?
2. List out three phases in dark reaction.
3. Location of dark reaction in the chloroplast?
4. Expense of ATP and NADPH₂ for the synthesis of one molecule of glucose in dark reaction?

Answer:

1. Melvin Calvin
   • Carboxylation
   • Reduction
   • Regeneration of RUBP
2. Stroma
3. 12 NADPH₂ and 18 ATP molecules.

Question 6.
In the last step of aerobic respiration oxidation of reduced co-enzymes produced in glycolysis and Krebs cycle occur.

1. What are the important reduced coenzymes?
2. Why is this process called as terminal oxidation?
3. Where does terminal oxidation takes place?

Answer:

1. FADH₂, NADH₂.
2. It is the formation of metabolic water at the end of electron transport chain by combining protons, electrons, and O₂.
3. The inner mitochondrial membrane of mitochondria.

Question 7.
An athlete felt muscular pain after a race.

1. Explain this in terms of anaerobic respiration?
2. Name any two microorganisms in which anaerobic respiration occurs
3. Glycolysis is common for both aerobic and anaerobic respiration. In glycolysis, there is a net gain of 8 ATP. But in anaerobic repiration the net gain is only 2 ATP. Give reason.

Answer:
1. Athletes felt muscular pain is due to the inadequecy of 02. In the absence of 02 partial oxidation takes place. It involves the formation of pyruvic acid followed by lactic acid.

2. Yeast & Lactobacillus.

3. After the formation of pyruvic acid, 2 NADPH molecules are utilised for the formation of Lactic acid. So the net gain of ATP in Anaerobic respiration is 2 ATPs.

Question 8.
Mitochondria is called the “Power House’’ of the cell.

1. Is the statement correct?
2. Write down reasons.

Answer:

1. Yes
   • Synthesis of ATP takes place in mitochondria,
   • Inner membrance acts as electron transport chain.
   • Presence of ATP – synthetase enzyme.
   • Enzymes for aerobic respiration in mitochondrial matrix

Question 9.
Analyse the diagram and answer the questions.

1. What does the diagram represent?
2. Write the role of F0-F., unit in the process?
3. What is oxidative phosphorylation?
4. Where does it take place?

Answer:

1. ATP synthesis
2. Flow of proton
3. In the presenceC₂, oxidation takes place and ADP combines with inorganic phosphate to form ATP
4. Crystal

Question 10.
Respirometer is an apparatus used to measure R.Q.

1. What is R.Q?
2. R.Q. of glucose is equal to one. Give reason.
3. Name the respiratory substrate for which R.Q. is more than one?

Answer:

1. Amount of C0₂ released/Amount of O₂ absorbed
2. In glucose amount of CO₂ released = amount of O₂ absorbed
3. Organic acids

Question 11.
RuBP carboxylase, PEPcase, Pyruvate dehydrogenase, ATPase, cytochrome oxidase, Hexokinase, Lactate dehydrogenase. Select/choose enzymes from the list above which are involved in

1. Photosynthesis
2. Respiration
3. Both in photosynthesis and respiration

Answer:

1. RuBP carboxylase, PEPcase, ATPase,
2. Pyruvate dehydrogenase, ATPase, cytochrome oxidase, Hexokinase, Lactate dehydrogenase
3. ATPase

Question 12.
The following statements are about an important event of cellular respiration. Read them carefully and answer the questions.
It is the last step of cellular respiration It synthesise ATP Water is the end product It accepts oxygen.

1. Name the process
2. Where does it occur?
3. Name any 4 components of ‘it’

Answer:

1. Electron transport system/oxidative phosphorylation
2. Inner mitochondrial membrane
3. NADP, FAD, Ubiquinone (Co-Q)
   Cyt. b, cyt – a, cty a₃ etc.

Question 13.
Respiration is viewed most simply as the oxidative production of ATR Justify the statement.
Answer:
The energy-releasing process by oxidation of organic food materials in the living cell is respiration. During this process, the energy contained in the food is released and is trapped in the ATP molecules. NADH+H+ and FADH₂ formed during various steps of respiration are oxidised, and protons (H+) and electrons (e ) are released.

These electrons are transported to the oxygen through a series of electron carries in the electron transport system (ETS), and their energy is stored in ATP molecules. So, respiration is process of oxidative production of ATP. 45% of energy released during the oxidation of 1 glucose molecule is stored in 38 ATP molecules.

Question 14.
The second phase of aerobic respiration takes place within mitochondria. This phase is called TCA cycle. The different steps of this reaction were found out by a British Biochemist who was awarded Nobel Prize in 1953.

1. Identify the scientist and name the first product of the reaction.
2. Write the first step of this reaction, why it is called TCA cycle?
3. From where Acetyl Co. A comes into mitochondria?
4. In which step FADHJs formed.

Answer:

1. Hans Krebs, citric acid(Tricarboxylic acid)
2. oaa+acetylcoA citric acid .since it has three -COOH group
3. Cytoplasm
4. 5^th step (Succinic acid to malic acid)

Question 15.
Glycolysis is common and first phase of both aerobic and anaerobic respiration.

1. Who discovered this process?
2. What are the substrate and end products in this reaction?

Answer:

1. Embeden, Mayerhof & Parnas
2. Glucose – Substrate.
   The end product – Pyruvic acid.

Plus One Botany Respiration in Plants NCERT Mark Questions and Answers

Question 1.
Discuss “The respiratory pathway is an amphibolic pathway.”
Answer:
Glucose is the favoured substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired but then they do not enter the respiratory pathway at the first step.

Since respiration involves breakdown as well as synthesis of substrates, the respiratory process involves both catabolism. That is why respiratory pathway is considered to be an amphibolic pathway rather than as a catabolic one.

Question 2.
What is the significance of step wise release of energy in respiration?
Answer:
If energy will be released at one go then most of it will be most in the form of heat. Cells should be in a position to utilize all the energy to synthesize something. To facilitate proper usage of energy, it is released in a stepwise manner during respiration.

Question 3.
Define RQ. What is its value for fats?
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called the respiratory quotient (RQ) or respiratory ratio.

Question 4.
Distinguish between the following:

1. Aerobic respiration
2. Glycolysis and Fermentation
3. Glycolysis and Citric acid Cycle

Answer:
1. There is incomplete oxidation of glucose during anaerobic respiration, while there is complete oxi¬dation during aerobic respiration.

2. The pyruvic acid formed during glycolysis is first converted to Acetyl coenzyme A, which undergoes citric acid cycle to produce critic acid. At the end of citric acid cycle NADH+H+ is released.

3. In both (b) and (c) glycolysis is the first step cellular respiration. The product of glycolysis is further utilized by either fermentation or critic acid cycle.

Question 5.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised, but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:

1. There is a acquential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.

2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.

3. None of the intermediates in the pathway are utilised to synthesise any other compound.

4. Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

Plus One Botany Respiration in Plants Multiple Choice Questions and Answers

Question 1.
Common phase between aerobic and anaerobic modes of respiration is
(a) Krebs cycle
(b) EMP/glycolysis
(c) oxidative phosphorylation
(d) PPP
Answer:
(b) EMP/glycolysis

Question 2.
Complete oxidation of one gram mol. of Glucose give rise to
(a) 6860,000 cal
(b) 686,000 cal
(c) 68,600 cal
(d) 6860 cal
Answer:
(b) 686,000 cal

Question 3.
When a molecule of pyruvic acid is subjected to anaerobic oxidation and forms Lactic acid there is
(a) loss of 3 ATP molecules
(b) loss of 6ATP molecules
(c) loss of 2 ATP molecules
(d) loss of 4 ATP molecules
Answer:
(b) loss of 6ATP molecules

Question 4.
When one glucose molecule is completely oxidised, it changes
(a) 36 ADP mole into 36 ATP molecules
(b) 32 ADP mole into 32 ATP molecules
(c) 38 ADP mole into 38 ATP molecules
(d) 30 ADP mole into 30 ATP molecules
Answer:
(c) 38 ADP mole into 38 ATP molecules

Question 5.
Substrate phosphorylation occurs during
(a) Fumaric acid → malic acid
(b) Oxalosuccinic acid → alpha-ketoglutaric acid
(c) Succinic acid – fumaric acid
(d) a Ketoglutaric acid – Succinic acid
Answer:
(d) a Ketoglutaric acid – Succinic acid

Question 6.
Glycolysis is significant for energy production in
(a) RBC
(b) fungi
(c) plants
(d) none of the above
Answer:
(a) RBC

Question 7.
In aerobic respiration
(a) 2 PGAL are formed in Glycolysis and none in kerbs cycle
(b) 6 PGAL in glycolysis, 3 PGAL in kerbs cycle
(c) PGAL formation does not occur in respiration
(d) 8 PGAL in glycolysis, 3 PGAL in krebs cycle
Answer:
(a) 2 PGAL are formed in Glycolysis and none in kerbs cycle

Question 8.
Two enzymes common to EMP pathway and C₃ cycle are
(a) Aldolase & triosephosphate isomerase
(b) Aldolase & enolase
(c) Cytochrome oxidase & enolase
(d) phosphoglyceromutase & triosephosphate isomerase
Answer:
(a) Aldolase & triosephosphate isomerase

Question 9.
The RQ value of fat is 0.7, it indicates
(a) volume of CO₂ is greater than O₂
(b) volume of O₂ is greater than CO₂
(c) volume of CO₂ is greater than volume of O₂
(d) none of the above
Answer:
(b) volume of O₂ is greater than CO₂

Question 10.
The net gain of ATP in aerobic respiration is
(a) 18
(b) 36
(c) 38
(d) 8
Answer:
(b) 36

Question 11.
Three carboxylic group occurs in
(a) pyruvic acid
(b) citric acid
(c) malic acid
(d) Aspartic acid
Answer:
(b) citric acid

Question 12.
Which is the first 5C carbon compound formed in TCA cycle
(a) succinic acid
(b) oxaloacetic acid
(c) alpha-ketoglutaric acid
(d) fumaric acid
Answer:
(c) alpha-ketoglutaric acid

Question 13.
The concentration of alcohol in fermentation influence the
(a) death of cells
(b) growth of cells
(c) production of succinic acid
(d) production of lactic acid
Answer:
(a) death of cells

Question 14.
Acetyl coA is called
(a) 5C compound
(b) 2C compound
(c) 3C compound
(d) 4C compound
Answer:
(b) 2C compound

Question 15.
Wine and beer are produced directly by fermentation. Brandy and whisky require both fermentation and distillation because
(a) fermendation is inhibited at an alcohol level of 10-18%
(b) distillation prolongs storage
(c) distillation improves quality
(d) distillation purifies the beverage.
Answer:
(a) fermendation is inhibited at an alcohol level of 10-18%

Question 16.
The process by which ATP is produced in the inner membrane of a mitochondrion. The electron transport system transfers protons from the inner compartment to the outer, as the protons flow back to the inner compartment, the energy of their movement is used to add phosphate to ADP, forming ATP.
(a) Chemiosmosis
(b) Phosphorylation
(c) Glycolysis
(d) Fermentation
Answer:
(a) Chemiosmosis

Question 17.
In which of the following reactions of glycolysis, a molecule of water is removed from the substrate?
(a) Fructose – 6 – phosphate → Fructose -1, 6 – bisphosphate
(b) 3 – phosphate – glyceraldehyde → 1, 3 bisphosphoglyceric acid
(c) PEP → Pyruvic acid
(d) 2 – phosphoglycerate → PEP
Answer:
(d) 2 – phosphoglycerate → PEP

Students can Download Chapter 4 Anatomy of Flowering Plants Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 4 Anatomy of Flowering Plants

Plus One Botany Anatomy of Flowering Plants One Mark Questions and Answers

Question 1.
Both apical and intercalary meristems are meristems
(a) primary
(b) secondary
(c) lateral
(d) none of these
Answer:
(a) primary

Question 2.
Sclerieds present in
(a) fruit wall of nuts
(b) pulp of guava
(c) seed coat of legumes
(d) all of these
Answer:
(d) all of these

Question 3.
Which of the following condition of xylem is present in both monocot and dicot stems
(a) exarch
(b) endarch
(c) polyarch
(d) mesarch
Answer:
(b) endarch

Question 4.
Bark does not include
(a) secondary xylem
(b) secondary phloem
(c) periderm
(d) primary phloem
Answer:
(a) secondary xylem

Question 5.
…………… is the living mechanical tissue
(a) collenchyma
(b) parenchyma
(c) sclerenchyma
(d) none of these
Answer:
(a) collenchyma

Question 6.
What are the cells that make the leaves curl in plants during water stress?
Answer:
Bulliform cells/motor cells

Question 7.
What part of the plant would show the following:
(a) Radial vascular bundle
(b) Polyarch xylem
(c) Well developed pith
Answer:
(a) Radial vascular bundle

Question 8.
Apical meristem for elongation of Stem, Lateral meristem for ………..
Answer:
Increase in thickness.

Question 9.
While analysing the anatomy of a given slide, phloem, cambium & xylem are present. Identify the material.
Answer:
Dicot stem.

Question 10.
Tissues are specialised to perform mechanical function have thickening at corners, living and cellulosic. Name it.
Answer:
Collenchyma

Question 11.
A cross-section of plant materials show the following features under microscope there are many vascular bundles scattered in the parenchymatous tissue, xylem is endarch. What kind of plant part shows the above anatomy.
Answer:
Monocot stem

Question 12.
Name tissue which provides mechanical strength to the plant’s organs.
Answer:
Sclerenchyma and xylem

Question 13.
What does the fascicular cambium gives rise to?
Answer:
Secondary vascular tissues and medullary rays

Question 14.
From where do the secondary cambium appear?
Answer:
From permanent tissues

Question 15.
Which type of meristems can be classified on the basis of positions in the plant body.
Answer:
Apical, intercalary and lateral

Question 16.
Find the odd one and give reason:
Phellem, Phellogen, Phloem, Phelloderm
Answer:
Phloem. The other components are part of periderm

Question 17.
Fill in the blanks:
Stem: Endarch:: Root: ……….
interfascicular cambium: secondary xylem:: cork cambium: …………
Answer:
Exarch, phelloderm

Question 18.
Activity of cambium is different in two seasons and xylem vessels formed have narrow and wider lumens. Give the type of wood formed during the different seasons.
Answer:
Springwood and autumn wood.

Question 19.
What use are phloem fibres put to
Answer:
Phloem fibers of plants like jute are used for making ropes.

Question 20.
Identify the plant group from the data given below.
(a) Habit-Tree
(b) Xylem vessels present
(c) Absence of companion cells in phloem
(d) No fruits
Answer:
Gymnosperm

Question 21.
Ramu observed the growth of a plant in his garden. He realised that the plant grow both in length and width. Name the tissue responsible for the growth of the plant.
Answer:
Meristems

Question 22.
A cross-section of plant materials show the following anatomical features under microscope.

1. Vascular bundles are radially arranged
2. Four xylem strands with exarch condition of the protoxylem. To which organ should it be assigned?

Answer:
Dicot root

Question 23.
The transverse section of a plant material shows the following anatomical features.

1. vascular bundles are conjoint, scattered and surrounded by a sclerenchymatous bundle sheath.
2. Phloem parenchyma is absent. What will you identify as?

Answer:
The plant is a monocotyledon.

Question 24.
The cross-section of a plant material showed the following features when viewed under the microscope.

1. The vascular bundles were radially arranged.
2. Four xylem strands with exarch condition of protoxylem.

To which organ should it be assigned?
Answer:
Dicot root

Plus One Botany Anatomy of Flowering Plants Two Mark Questions and Answers

Question 1.
During the secondary thickening of a dicot stem, from the outer cotical region, a meristematic tissue is developed. Name this tissue and write the function of it.
Answer:
Phellogen/cork cambium – It produces phellum towards the outer sides & phelloderm towards the inner side. These 3 layers are known as periderm

Question 2.
Analyse the table and arrange the matter is an appropriate order.

Answer:

Question 3.
Differentiate the following characters of dicot stem and monocot stem.
Differentiated cortex, Protoxylem lacuna present, open bundle, bundle sheath, starch sheath, sclerenchymatous hypodermis, scattered bundles, V.Bs. arranged in a ring, closed bundles, phloem parenchyma absent, presence of pith, collenchymatous hypodermis.
Answer:

Question 4.
Xylem is the complex tissue that transports water in plants.

1. Name the main components of xylem.
2. Which of these is most likely suitable for conducting water?

Answer:

1. Xylem parenchyma, vessels, tracheid, xylem fibers.
2. Vessels

Question 5.
Examine the anatomy of two plants

1. Xylem vessels and tracheids are seen
2. Xylem vessels absent and tracheids present

Answer:

1. Angiosperm
2. Gymnosperm

Question 6.
Identify the figure ‘A’ and ‘B’ and differentiate between ‘A’ and ‘B’.

Answer:

Question 7.
Secondary growth occurs in dicots and no such growth in monocots. Justify.
Answer:
In dicots, cambium is present in their vascular bundle. In monocot, cambium is absent in the vascular bundles. So secondary growth occurs in dicotyledons and not in monocotyledons.

Question 8.
Phloem is called the complex tissue while parenchyma is a simple tissue. Make the differences
Answer:
Phloem is made up of more than one type of cells namely, sieve tube, companion cell, phloem parenchyma & phloem fiber. Parenchyma consists of same types of cells.

Question 9.
When the cross-section of a stem was examined, a student could notice collenchymatous hypodermis, Limited number of open vascular bundles, cortex, endodermis, and pith.

1. Identify the material.
2. Draw the diagram of the material showing these parts and label.

Answer:
1. Dicot stem
2.

Question 10.
Observe the diagrams and identify the tissue types.

Answer:
a. Collenchyma
b. Sclerenchyma

Question 11.
Analyse the table and arrange the matter in an appropriate order.

Answer:

Question 12.
While doing anatomy practical, students observe ten following characters:

• Undifferentiated ground tissue.
• Numerous vascular bundles and are scattered.
• Sclerenchymatous hypodermis
• Presence of bundle sheath

1. Identify the material.
2. Write any one example of a plant having such characters.
Answer:

1. Monocot stem.
2. Grass/Bamboo

Question 13.
What is the function of phloem parenchyma?
Answer:
The phloem parenchyma stores food material and other substances like resins, latex, and mucilage.

Question14.
How do you correlate the activity of cambium with changing seasons?
Answer:
During winter season the cambium stops its activity. In spring season cambium becomes more active and produce large quantity of secondary tissue. The secondary xylem produced during spring season is called springwood. During autumn season and summer season the cambium becomes less active and produces small quantity of secondary tissue. The secondary xylem produced during this period is called autumn wood.

Question 15.
Mention the characteristic features of sieve-tube members.
Answer:
Sieve tubes are cylindrical have endplates called sieve plates. They are dead elements of phloem.

Question 16.
It is difficult to open and close wooden doors and windows during the rainy season. What is the reason?
Answer:
Imbibition It is the absorption of water by water-loving particles without forming a solution

Question 17.
Heartwood is resistant to microorganisms. Justify the statement.
Answer:
Contains alkalloids , tannins & resins

Question 18.
The microscopic observation is given below.

1. Identify A and B.
2. Write any two distinguishing characters of the above diagram.

Answer:

1. A dicot stem B monocot stem
2. In dicot stem vascular budles open, limited no.&arranged in the form of a ring. In monocot stem, vascular bundles closed, numerous & scattered

Question 19.
Simple tissues that perform mechanical support in plants are usually dead tissues’
Name the living simple tissue that provides mechanical support to plant organs. In which category of plants (dicots/monocots) do they found?
Answer:
Sclerenchyma tissues Monocots

Question 20.
What is present on the surface of the leaves which helps the plant prevent loss of water but is absent in roots?
Answer:
The cuticle is the waxy substance prevents the excessive water loss.

Question 21.
What is the epidermal cell modification in plants which prevents water loss?
Answer:
The presence of bulliform cells in the leaves of monocot plants, which helps in rolling and unrolling of lamina and prevents water loss.

Question 22.
Arrange the following in the sequence you would find them in a plant starting from the periphery – phellem. phellogen. phelloderm.
Answer:
phellem, phellogen, phelloderm.

Question 23.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What is the significance of these rings?
Answer:
It is formed due to the activity of vascular ambium. The growth ring comprises springwood and autumn wood that is formed in spring season and winter season respectively. These are used to determine the age of tree by counting the number of concentric rings.

Question 24.
While eating peach or pear it is usually seen that some stone-like structures get entangled in the teeth, what are these stone-like structures called?
Answer:
USB Non-elongated sclerenchyma(sclereids)

Question 25.
When your teacher asked to collect different types of seeds. One of the students collected some seeds from a tree. These seeds have no covering or naked.

1. In which group of plants does this tree belong?
2. Which portion of the plant develops into the seed?

Answer:

1. Gymnosperms
2. Ovule

Question 26.
We cannot find the woody trunk in monocot plants. Why?
Answer:
Monocots do not possess cambium Hence, the secondary tissues are not formed.

Question 27.
Identify the following materials under microscope.

1. Xylem polyarch, Round vessels, homogenous cortex with xylem & phloem in radial manner.
2. Endarch xylem, conjoint bundle, surrounded by bundle sheath, palisade and spongy tissues distinguished and kidney-shaped guard cells

Answer:

1. Monocot root
2. Dicot leaf

Question 28.
The product of photosynthesis is transported from the leaves to various parts of the plants and stored in some cells before being utilised. What are the cells/ tissues that store them?
Answer:
parenchyma tissue

Question 29.
Name the kind of vascular bundle in Fig. a and b.

Answer:
a. Closed bundle
b. Open bundle

Question 30.
Heartwood is resistant to micro-organisms. Justify the statement.
Answer:
The deposition of substances like Tannin, resin, oil, gum, aromatic substances and essential oil in the central part of wood helps to resist the attack of microorganisms.

Question 31.
Cambial ring is responsible for the formation of secondary vascular tissues both in Dicot stem and Dicot root. Differentiate the origin of cambial ring in Dicot stem and Dicot root.
Answer:
1. In Dicot stem, the conjuctive tissues in between the vascular bundles become meristematic and form. interfascicular cambium, it joins with intrafascicular cambium to form a cambial ring.

2. In Dicot root, the tissue just below phloem and part of pericycle tissue above the protoxylem join together to form a wavy ring. Later it becomes circular.

Question 32.
Give reason

1. Grittiness nature of sapota fruit pulp
2. Increase thickness of stem intake and not in coconut

Answer:

1. Due to the presence of nonelogated sclerenchyma tissues (sclereids)
2. In Teak, cambium is present (dicot plant) In coconut, cambium is absent because it is a monocot plant.
   Cambial activity causes an increase in thickness

Question 33.
Match the following.

Answer:

Question 34.
You may have observed droplets of water at the tip of leaves of grasses during early morning. Your sister told it is a dewdrop.
As a science student do you agree with her? Explain. Compare this process with transpiration.
Answer:
No.

1. The droplets of water are found in the tip of grass leaves. This is called Guttation. It occurs in early morning.
2. Water drops come out from Hydathode seen in the vien endings of leaf.
3. Transpiration is the loss of water in the form of water vapour from stomata of leaves.

Question 35.
How can we correlate the activity of cambium with changing season?
Answer:
In spring season the activity of cambium is more and produces large no. of xylem vessels with wide lumen. But in Autumn season the activity of cambium is less and produces fewer no.of xylem vessels with narrow lumen.

Question 36.
Classify the following characters under dicot and monocot stem?
Bundle sheath is present, open bundle, bundle cap is present, closed bundle, arranged in a broken ring, scattered bundle, collenchymatous hypodermis, sclerenchymatous hypodermis.
Answer:

Question 37.
Distinguish between cork cambium and vascular cambium.
Answer:
1. Cork cambium is the meristematic layer that cut of the tissues outside called phellem and inside phelloderm.

2. Vascular cambium is the meristematic layer that cut of the tissues outside called secondary phloem and inside secondary xylem.

Question 38.
Identify the following diagram and write its function.

Answer:
Hydathode. It is the pore found in the vein ending of the grass leaf through which water loss occurs in the form of droplets.

Question 39.
Which one out of the root or stem shows endarch arrangement of xylem? What is meant by endarch arrangement?
Answer:
Stem, In edarch xylem protoxylem, is present towards the centre (pith) and metaxylem is faced towards outside.

Question 40.
Observe the diagram and identify the tissue types. Justify your answer.

Answer:

• The first figure is the Sclerenchymatus tissue.
• The second figure is the collenchymatous tissue.
• In sclerenchyma tissue, the cell walls are lignified and dead.
• In collenchymatous tissue, the cell walls are thick and living.

Question 41.
What are bulliform cells? Mention its significance
Answer:
These are the specialised cells found in the epidermis of monocot leaf. They are otherwise called as motor cells.
It helps in the rolling and unrollong of lamina

Question 42.
Where are the companion cells located in the flowering plants? What are their functions?
Answer:
Companion cells are located in the phloem cells of vascular tissue. They support the sieve tubes.

Question 43.
What category of permanent plant cell is a companion cell?
Answer:
Companion cells are specialised thin-walled parenchymatous cells found to be associated with sieve tubes.

Question 44.
Sieve tubes in angiosperms are associated with specialised parenchyma cells. Name those cells. How do they help sieve-tube members?
Answer:
Companion cells, help in physiological working of sieve-tube members.

Question 45.

1. Name the various components of xylem.
2. Which of them does not have a nucleus?

Answer:

1. Tracheids, vessels, xylem parenchyma, and xylem fibers.
2. Tracheids, vessels, and xylem fibers

Question 46.
How are exarch and endarch conditions different anatomically in stem and root?
Answer:
1. The anatomical condition is endarch in dicot as well as monocot stem.

2. In endarch condition protoxylem lie towards inner side of vascular budle However, in anatomy of root.the condition is exarch.i.e protoxylem is towards outer side and metaxylem is towards centre.

Question 47.
Cork cambium forms tissues that form the cork. Do you agree with this statement? Give reasons.
Answer:
Cork cambium develops in cortex region during secondary growth in stem. Cork cambium cut off cork (phellum) towards the outside and secondary cortex or phellodem towards the inner-side.

Question 48.
If one debarks a tree, what parts of the plant is being removed?
Answer:
Phloem is removed and food translocation becomes difficult.

Question 49.
What is periderm? How does periderm formation take place in the dicot stems?
Answer:
Phellogen, phellem, and phelloderm are together known as periderm. Due to secondary growth of the stem, the outer cortical and epidermis layer broken and replaced by another meristematic issue called phellogen. Phellogen or cork cambium cuts off cells on both sides the outer cells are called cork or phellem and the inner cells are phelloderm or secondary cortex.

Question 50.
Some anatomical details of material are given below vascular bumdles are conjoint and open, medullary rays are present, hypodermis collenchymatous, cortex heterogenous

1. Identify the material
2. Name the tissue given above from which interfascicular cambium develops during secondary growth.

Answer:

1. Dicot stem
2. Medullary ray

Question 51.
Three type of tissues formed during secondary growth together have protective functions.

1. Name the three types
2. Name the opening found in outer part of mature woody dicot stem and give its function

Answer:

1. phellem, phellogen, and phelloderm
2. lenticel, gaseous exchange

Question 52.

1. Distinguish between heartwood and sapwood.
2. Heartwood is durable why?

Answer:
1. Distinguish between heartwood and sapwood
Heartwood

• found in the central part
• nonfunctional
• dark coloured

Sapwood

• found in peripheral part
• functional
• light coloured

2. Heartwood is resistant to microbial attack and it is filled with resin, gum, essential oil, aromatic oil, and tannins. Hence heartwood is durable.

Plus One Botany Anatomy of Flowering Plants Three Mark Questions and Answers

Question 1.
A portion of the transverse section of maize stem is shown in the diagram.
Label a, b, c, d, e, f, g, h and i.

Answer:

1. a – Cuticle
2. b – Epidermis
3. c – Sclerenchymatous hypodermis
4. d – Sclerenchymatous sheath
5. e – Parentchymatous sheath
6. f – Phloem
7. g – Metaxylem
8. h – Protoxylem
9. i – Water cavity.

Question 2.
Differentiate the following

1. Bark & Periderm
2. Heartwood &Sapwood
3. Endarch & Exarch

Answer:
1. Bark – It is the tissue outside vascular cambium.
Periderm – It is the outer protective tissue It includes phellem, phellogen, and phelloderm.

2. Heartwood – It is the dark coloured, nonfunctional central part of wood resistant to the attack of microorganisms.
Sap wood – It is the light coloured, functional and peripheral part of wood.

3. Endarch – It is the arrangement of xylem in which protoxylem facing towards cente and metaxylem towards the periphery, eg. stem.
Exarch – It is the arrangement of xylem in which Metaxylem facing towards centre and protoxylem towards periphery.eg: Root.

Question 3.
Match column A with B and C.

Answer:

Question 4.
Match the Following.

Answer:

Question 5.
The following are the characters of Dicot stem and Monocot stem. Identify the character and write in appropriate column.

1. collenchymatous hypodermis
2. Sclerenchymatous hypodermis
3. Open bundle
4. Closed bundle
5. Bundle cap present
6. Bundle sheath present
7. Xylem polygonal
8. Xylem circular

Answer:

Question 6.
Plant growth is indeterminate in most plants it can go on indefinitely at the tips of roots and shoots. Special types of tissues are located at these regions

1. Name those tissues
2. Classify these tissues based on the position and origin.

Answer:

1. Meristematic tissues
2. Based on the position they are classified into three types,
   • apical meristem
   • intercalary meristem and
   • lateral meristem

Based on the origin, they are classified into 2.

1. primary meristem and
2. Secondary Meristem

Question 7.
1. Choose from among the following key terms and
complete the chart given below.
conjoint, closed, exarch open, …………. large, ……………. small endarch, radial ……….. few in a ring, ……….. many xylem and phloem bundles
few.

2. Comment on the utility of this chart.
Answer:

Question 8.
The following schematic representation shows the classification of complex tissues. Complete it.

Answer:

Plus One Botany Anatomy of Flowering Plants NCERT Mark Questions and Answers

Question 1.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Answer:
On the basis of their structure and location, there are three types of tissue systems:

1. Epidermal Tissue System. Epidermis, Stomata.
2. Ground or Fundamental Tissue System Parenchyma, Sclerenchyma and Collenchyma.
3. Vascular or conducting Tissue System. Phloem and Xylem.

Question 2.
The transverse section of a plant material shows the following anatomical features

1. the vascular bundles are conjoint, scattered and surrounded by a sclerenchymatous bundle sheath.
2. phloem parenchyma is absent. What will you identify as?

Answer:
As is clear from above TS of monocot stem vascular bundles are scattered in monocot stems and phloem parenchyma is absent.

Question 3.
Why are xylem and phloem called complex tissues?
Answer:
Xylem and phloem are composed of several types of cells and they work as a unit. Hence they are called complex tissues.

Question 4.
What is the stomatal apparatus? Explain the structure of stomata with a labelled diagram.
Answer:
Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stomata is composed of two bean-shaped cells known as guard cells. In grasses, the guard cells are dumbbell-shaped.

The outer walls of guard cells (away from the stomatal pore) are thin and the inner walls (towards the stomatal pore) are highly thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata.

Plus One Botany Anatomy of Flowering Plants Multiple Choice Questions and Answers

Question 1.
Interfascicular cambium is a
(a) primary meristematic tissue
(b) primordial meristem
(c) type of proto deny
(d) secondary meristematic tissue
Answer:
(d) secondary meristematic tissue

Question 2.
In an annual ring, the light coloured part is known as
(a) earlywood
(b) late wood
(c) heartwood
(d) sapwood
Answer:
(a) earlywood

Question 3.
The layer of cells outside the phloem meant for giving rise to the root branches is called
(a) cambium
(b) cork
(c) endodermis
(d) pericycle
Answer:
(d) pericycle

Question 4.
Cork cambium of dicot originates from
(a) epiblema
(b) pericycle
(c) the cambium of vascular bundles
(d) endodermis
Answer:
(b) pericycle

Question 5.
Alburnum is otherwise known as
(a) periderm
(b) sapwood
(c) heartwood
(d) cork
Answer:
(b) sapwood

Question 6.
Duramen is present in
(a) inner region of secondary wood
(b) part of sapwood
(c) outer region of secondary wood
(d) region of pericycle
Answer:
(a) inner region of secondary wood

Question 7.
Which one of the following is the correct sequence of tissues present in dicot stem during secondary growth
(a) Phellogen, cork, primary cortex, secondary cortex
(b) Cork, primary cortex, secondary cortex, phellogen
(c) Primary cortex, secondary cortex, phellogen, cork
(d) Secondary Cortex, phellogen, cork .primary cortex
Answer:
(d) Secondary Cortex, phellogen, cork .primary cortex

Question 8.
The meristem responsible for extra stelar secondary growth in dicot stem is
(a) interfascicular cambium
(b) intrafascicular cambium
(c) intercalary meristem
(d) phellogen
Answer:
(d) phellogen

Question 9.
Tracheids
(a) are the dominant cell types of xylem in angiosperms
(b) are primarily found in mosses and liverworts
(c) are responsible for water conduction and support in many land plants.
(d) first appeared during Palaeozoic era
Answer:
(c) are responsible for water conduction and support in many land plants.

Question 10.
In the following, how the sapwood is converted into heartwood?
(a) By degeneration of protoplast of living cells
(b) Tyloses formation
(c) By deposition of resins, oils, gums, etc
(d) All of the above
Answer:
(d) All of the above

Question 11.
As secondary growth proceeds in a dicot stem, the thickness of
(a) sapwood increases
(b) heartwood increases
(c) both sapwood and heartwood increase
(d) both sapwood and heartwood remains the same
Answer:
(c) both sapwood and heartwood increase

Question 12.
Which is not a characteristic of plant cell walls?
(a) Found only in the sporophyte phase of life cycle
(b) Among other compounds contains compounds built of simple sugars
(c) May contain enzymes that are biologically active
(d) Often contain strengthening polymers
Answer:
(a) Found only in the sporophyte phase of life cycle

Question 13.
Identify the correct order of the components with reference to their arrangement from outer side to inner side in a woody dicot stern.

1. Secondary cortex
2. Autumn wood
3. Secondary phloem
4. Phellem

(a) 2,3, 1,4
(b) 3,4,2,1
(c) 4, 1,3,2
(d) 1,2,4,3
Answer:
(c) 4, 1,3,2

Question 14.
Removal of ring wood of tissue outside the vascular cambium from the tree trunk kills it because
(a) water cannot move up
(b) food does not travel down and root become starved
(c) shoot become starved
(d) annual rings are not produced
Answer:
(b) food does not travel down and root become starved

Question 15.
Which tissue gives rise to secondary growth?
(a) Apical meristem
(b) Adventitious roots
(c) Germinating seed
(d) Vascular cambium
Answer:
(d) Vascular cambium

Question 16.
Growth rings are formed due to activity of
(a) extrastelar cambium
(b) intrastelar cambium
(c) interstelar cambium
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

Question 17.
A nail is driven into the trunk of a 30 years old tree at a point l in above the soil level. The tree grows in height at the rate of 0.5m a year. After three years, the nail will be
(a) 1m above the soil
(b) 1.5 m above the soil
(c) 2 in above the soil
(d) 2.5 m above the soil
Answer:
(a) 1m above the soil

Question 18.
In which of the following, there is no differentiation of bark, sapwood, and heartwood?
(a) Ashok
(b) Neem
(c) Mango
(d) Datepalm
Answer:
(d) Datepalm

Question 19.
A tree grows at a rate of 0.5 m per year. What will be the height of the board fixed at 1.5 m above the base five years ago?
(a) 4.0 m
(b) 3.5 m
(c) 1.5 m
(d) 4.5 m
Answer:
(c) 1.5 m

Question 20.
Secondary phloem remains functional generally
(a) for one year
(b) for less than one year
(c) for many years
(d) as long as the plant is alive
Answer:
(d) as long as the plant is alive

Students can Download Chapter 6 Cell Cycle and Cell Division Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 6 Cell Cycle and Cell Division

Plus One Botany Cell Cycle and Cell Division One Mark Questions and Answers

Question 1.
Meiosis results in
(a) Production of gametes
(b) Reduction in the number of chromosomes
(c) Introduction of variation
(d) all of the above
Answer:
(d) all of the above

Question 2.
At which stage of meiosis does the genetic constitution of gametes is finally decided
(a) Metaphase I
(b) Anaphase II
(c) Metaphase II
(d) Anaphase I
Answer:
(d) Anaphase I

Question 3.
Meiosis occurs in organisms during
(a) Sexual reproduction
(b) Vegetative reproduction
(c) Both sexual and vegetative reproduction
(d) None of the above
Answer:
(a) Sexual reproduction

Question 4.
During anaphase-l of meiosis
(a) Homologous chromosomes separate
(b) Non-homologous autosomes separate
(c) Sister chromatids separate
(d) non-sister chromatids separate
Answer:
(a) Homologous chromosomes separate

Question 5.
Mitosis is characterised by
(a) Reduction division
(b) Equal division
(c) Both reduction and equal division
(d) None of the above
Answer:
(b) Equal division

Question 6.
A bivalent of meiosis-l consists of
(a) Two chromatids and one centromere
(b) Two chromatids and two centromere
(c) Four chromatids and two centromere
(d) Four chromatids and four centromere
Answer:
(c) Four chromatids and two centromere

Question 7.
Cells which are not dividing are likely to be at
(a) G1
(b) G2
(c) Go
(d) S phase
Answer:
(c) Go

Question 8.
Which type cell divisions occur in meristematic cell of root apex?
Answer:
Mitosis

Question 9.
In which stage the actual reduction of chromosome number occurs in meiosis.
Answer:
Anaphase 1

Question 10.
Give the term for the failure of separation of homologous chromosomes.
Answer:
Non-disjunction

Question 11.
Name the cell divisions which help in growth and recombination of genes.
Answer:
Mitosis and meiosis

Question 12.
It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit ____ phase to enter an inactive stage called ____ of cell cycle. Fill in the blanks.
Answer:
G₁ and G₀

Question 13.
In the case of plant cells, the formation of new cell wall begins with a simple precursor. What is this precursor?
Answer:
Middle lamella

Question 14.
It is said that the one cycle of cell division in human cells (eukaryotic cells) takes 24 hours. Which phase of the cycle, do you think occupies the maximum part of cell cycle?
Answer:
Interphase.

Question 15.
At what stage of cell cycle does DNA synthesis take place?
Answer:
substage of Interphase

Question 16.
If the failure of division of cytoplasm occurs after nuclear division, What will happen to the cell?
Answer:
Free nucleii are formed

Question 17.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them?
Answer:
300 pollen mother cells

Question 18.
What is the peculiarity of zygotene?
Answer:
The pairing of the homologous chromosome called synapsis

Question 19.
It is the inactive stage of cell division but cell differentiation occurs. Name it.
Answer:
G₀ Phase /Quiscent stage.

Plus One Botany Cell Cycle and Cell Division Two Mark Questions and Answers

Question 1.
Meiosis is the type of cell division which maintain the race. Discuss.
Answer:
Reduces the chromosome number to half, so that chromosome number is maintained in the next generation.

Question 2.
Interphase in cell cycle is sometimes referred to as resting phase. Do you consider this statement true? Substantiate your answer.
Answer:
No.

1. Nucleus and cytoplasm are metabolically very active.
2. Amount of DNA becomes doubled.

Question 3.
A diagram of typical cell cycle of a higher plant is shown here. Identify each stage of the cycle and explain what happens during these stages.

Answer:

• G₁ – pre mitotic gap -synthesis of RNA &Proteins
• S – phase of synthesis – DNA replication
• G₂ – Post mitotic Gap phase synthesis of RNA &Proteins continues
• M – Mitotic phase
• G₀ – Inactive phase

Question 4.
Cytokinesis differs in plant and animal cell. Substantiate this statement.
Answer:

• Plant cell – formation of cell plate.
• Animal cell – Invagination of cell membrane.

Question 5.
Analyse the column A and B arrange the matter is an appropriate order.

Answer:

Question 6.
Identify the stages of Mitosis in which the following events take place:

1. Synapsis
2. Terminalisation of chiasmata.

Answer:

1. Zygotene
2. Diakinesis

Question 7.
Different stages of Prophase I of Meiosis are given in column A, arrange them in correct order and match them with the events in column B.

Answer:

1. (i) – d
2. (ii) – c
3. (iii) – a
4. (iv) – b

Question 8.
The pairing of homologous chromosomes is called synapsis.

1. Name each pair of homologous chromosomes.
2. Name the stage of prophase at which it takes place.

Answer:

1. Bivalent
2. Zygotene

Question 9.
Specific chromosome number of each species is conserved across generations in sexually reproducing organisms. What is the reason for this? Write the different steps of this process.
Answer:

1. Meiosis (meiosis I & II)
2. Prophase Meta phase Ana phase Telophase

Question 10.
The life cycle of a cell is called cell cycle. It consists of four stages such as Gv S, G2, and M.

1. Construct a pie diagram showing the different stages indicated above.
2. State the major events occurring in G, S, and G2 phases.

Answer:
1. Pie diagram of cell cycle.

2. G1 Phase -1, Cell grows in size and prepares the machinery needed for the DNA replication. RNA and proteins are synthesized. S phase – DNA replication. G2 phase – Synthesis of RNA and proteins.

Question 11.
Arrange the following stages of cell cycle in correct sequence S, G2, G1, M.
Answer:
S, M, G1, S, G2

Question 12.
X’ shaped structure called ‘chiasmata occurs during a particular stage of cell division.

1. Name the stage?
2. What is the significance of this type of cell division?

Answer:

1. Pachytene
2. The exchange of genes takes place between homologous chromosomes.

Question 13.
Given below are the five phases of prophase I of Meiosis I. Arrange them in correct order.
Zygotene, diakinesis, diplotene, leptotene, pachytene
Answer:
Leptotene, Zygotene, Pachytene, diplotene and diakinesis

Question 14.
Give the scientific term of the following.

1. Interchange of genetic material between non-sister chromatids of the homologues chromosomes
2. The plane of alignment of the chromosomes at metaphase

Answer:

1. Crossing over
2. Equatorial

Question 15.
Identify the diagram and label a, b,c and write the events during this.

Answer:
(a)G1
(b) S
(c)G2

• G1 – Interval between mitosis and initiation of DNA replication.
• S – DNA synthesis or replication of DNA occurs.
• G2 – In this phase proteins are synthesised for mitosis.

Question 16.
You have supplied I set of glass slides showing gametogenesis or garnets formation in an animal In one of the slide you observed the following features. Four cells with haploid number of chromosomes. Your friend told you that this is a meiotic division. (Hint: The diploid number of chromosome is 16.)
Are you agree with this statement. Justify your answer.
Answer:
Yes. Meiosis takes place in diploid cell or meiocyte to form four haploid cells. These cells contain 8 chromosomes each.

Question 17.
Differentiate reduction division from equational division.
Answer:
1. Reduction division:
It occurs in diploid cells to form 4 haploid cells, ie, the Chromosome number reduced to half in Meiosis I and results 2 daughter cells, which again divides to form 4 daughter cells. All cells formed in meiosis are haploid.

2. Equational division:
It is the mitotic division results 2 daughter cells carrying same set of chromosomes as that of parent cell. No genetic variation occurs.

Question 18.
Can there be mitosis without DNA replication in S phase?
Answer:
There cannot be mitosis without DNA replication in 5 phase of interphase because the trigger for mitosis is disturbance of nucleocytoplasmic ratio caused by DNA replication in S phase. Mitosis restores the quantity of genetic material to the species-specific level.

Question 19.
How does anaphase of mitosis and anaphase I of meiosis differ from each other?
Answer:
In anaphase of mitosis chromatids separate while in anaphase 1 of meiosis homologous chromosomes separate.

Question 20.
How does cytokinesis in plant cells differ from that in animal cells?
Answer:
1. In an animal cell a furrow in the plasma membrane.lt joins in the centre and dividing the cell cytoplasm into two.

2. In-plant cells, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. Then Cell division occurs.

Question 21.
How cytokinesis is different in an animal and a plant cell?
Answer:
In-plant cell cytokinesis occurs by cell plate formation while in the animal cell it occurs by cell furrow formation.

Question 22.
Why mitosis is called equational division ? Give the occurrence of mitosis.
Answer:
It keeps the chromosome number constant. It occurs in somatic cells.

Question 23.
What is the feature of a metacentric chromosome?
Answer:
The metacentric chromosome has a centromere in the middle region with two equal arms of the chromosome.

Question 24.
What is kinetochore? Give its function.
Answer:
It is a disc-like area in each chromatid and is site of attachment of spindle microtubule.

Question 25.
Why is meiosis essentially in sexually reproducing organisms?
Answer:
Meiosis reduces the chromosome number to half as it is followed by fertilization which restores diploidy.

Question 26.
Name the stage of cells cycle at which one of the following events occurs.

1. Chromosomes are moved to spindle equator
2. Centromere splits and chromatids separate
3. Pairing between homologous chromosomes. takes place
4. Crossing over between homologous chromosomes takes place.

Answer:

1. Metaphase
2. Anaphase
3. Zygotene
4. Pachytene

Question 27.
Downs syndrome and Klinefelter’s syndrome occurs due to mistake in cell division. What does it indicate?
Answer:
It is due to the failure of separation of homologous chromosomes during meiosis.

Question 28.
The events occur in prophase and telophase are one opposite to other,

1. Name the cell structures shown the above events
2. How many daughter nuclei are formed at the end of mitotic and meiotic prophase?

Answer:

1. Nuclear membrane, nucleolus and spindle fibres
2. Two

Question 29.
Mangolism or Trisomy is due to the failure of one event in cell division.

1. Which is the improperly functioning stage?
2. What is the event that not occurs?

Answer:

1. Anaphase I of meiosis
2. Separation of homologous chromosomes

Question 30.

1. What will be the ploidy level of dyad and tetrad of cells in meiosis?
2. How is it occurs?

Answer:

1. Dyad – haploid, Tetrad – Haploid
2. It occurs in meiosis due to separation of homologous chromosomes and chromosome number reduced to half.

Plus One Botany Cell Cycle and Cell Division Three Mark Questions and Answers

Question 1.
In which phase of meiosis is the following formed? Choose the answers from hint points given below.

1. Synaptonemal complex ______
2. Recombination nodules ______
3. Appearance/activation of enzyme recombinase _____
4. Termination of chiasmata _______
5. Interkinesis ________
6. Formation of dyad of cells ________

Answer:

1. Zygotene
2. pachytene
3. pachytene
4. diakinesis
5. After Meiosis-I before meiosis II
6. after first cytokinesis

Question 2.
The interphase stage is significant in mitotic and meiotic cell division

1. Give one specific event
2. Name the stage of interphase this event occurs
3. How will you differentiate interphase from interkinesis?

Answer:

1. DNA replication
2. S – phase
3. Interphase – cell prepares for cell division Interkinesis – short interval between meiosis I and meiosis II

Question 3.
The following events occur during the various phases of the cell cycle, Write the phase against each of the events.

1. Appearance of nucleolus
2. Division of centromere
3. Replication of DNA

Answer:

1. Telophase
2. Anaphase
3. Interphase

Question 4.
Life cycle of a cell is called cell cycle. ‘S’ phase is an important phase of cell cycle.

1. Justify your answer.
2. Name the stages of cell cycle at which the following events occur.
   • Crossing over of homologous chromosome.
   • Pairingof homologous chromosomes.
   • Chromosomes are arranged at the equatorial plane.

Answer:

1. phase of DNA synthesis
2. stages of the cell cycle
   • Pachytene
   • Zygotene
   • Metaphase

Question 5.
Name the stages of cell division in which the following events occur?

1. Chromosomes are moved to spindle equator.
2. Centromere splits and chromatids separate.
3. Crossing over between homologous chromosomes takes place.

Answer:

1. Metaphase
2. Anaphase
3. Pachytene

Question 6.
Match the words listed in column I with suitable words from column II.

Answer:

1. a) – Gametic meiosis
2. b) – Nuclear division
3. c) – Zygotic meiosis
4. d) – Cytoplasmic division
5. e) – Meiocytes
6. f) – Plant cells

Plus One Botany Cell Cycle and Cell Division NCERT Mark Questions and Answers

Question 1.
Distinguish cytokinesis from karyokinesis.
Answer:
The division of cytoplasm is called cytokinesis, while the division of the nucleus is called Karyokinesis.

Question 2.
What is G0 (quiescent phase) of cell cycle?
Answer:
Some cells in the adult animals do not appear to exhibit division (e.g., heart cells and many other cells divide only occasionally, as needed to replace cells that have been lost because of injury or cell death.

These cells do not divide further exit G₁ phase to enter an inactive stage called quiescent stage (G₀) of the cell cycle. Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Question 3.
Describe the event taking place during interphase.
Answer:
The interphase is divided into three further phases:
1. G₁ phase (Gap 1). G₁ phase corresponds to the interval between mitosis initiation of DNA replication. During G, phase the cell is metabolically active and continuously grows but does not replicate its DNA.

2. S phase (Synthesis). S or synthesis phase marks the period during which DNA synthesis or replication takes place.

3. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C then it increases to 4C. However, there is no increase in the chromosome number, if the cell had diploid or 2n number of chromosomes at G₁, even after S phase the number of chromosomes remains the same, i.e., 2n.

4. G₂ phase (Gap 2). In animal cells, during the S phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm during the G₂ phase, proteins are synthesised in preparation for mitosis White cell growth continues.

Question 4.
Why is mitosis called equational division?
Answer:
Since the number of chromosomes remains same in parent and daughter cells so mitosis is also called a equational division.

Question 5.
Name the stage of cell cycle at which one of the following events occur.

1. Chromosomes are moved to spindle equator.
2. Centromere splits and chromatids separate.
3. Pairing between homologous chromosomes takes place.
4. Crossing over between homologous chromosomes takes place.

Answer:

1. Metaphase
2. Anaphase
3. Metaphase I of meiosis
4. Prophase I of meiosis

Question 6.
What is the significance of meiosis?
Answer:
Significance of Meiosis:

1. Maintaining genetic identity by maintaining number of chromosomes.
2. Bringing variations to ensure better species.
3. Facilitates sexual reproduction.

Question 7.
Discuss with your teacher about.

1. haploid insects and lower plants where cell division occurs, and
2. Some haploid cells in higher plants where cell division does not occur.

Answer:

1. Male bees, wasps and ants are haploid organisms because they are produced from unfertilized eggs.
2. Synergids and antipodal cells in the ovule don’t undergo cell division.

Question 8.
Can there be mitosis without DNA replication in ‘S’ phase?
Answer:
DNA replication is necessary for cell division, and cell division cannot happen without DNA replication.

Question 9.
Can there be DNA replication without cell division?
Answer:
DNA replication takes place in order to prepare for cell division. Cell division is the next logical step after DNA replication.

Question 10.
Analyse the events during every stage of cell cycle and notice how the following two parameters change

1. number of chromosomes (N) per cell
2. amount of DNA content (C) per cell

Answer:

1. Number of chromosomes remains same after mitotic cell division and becomes half after meiotic cell division.
2. During S phase the DNA content doubles, but number of chromosomes remains the same.

Plus One Botany Cell Cycle and Cell Division Multiple Choice Questions and Answers

Question 1.
Cleavage is a unique form of mitotic cell division in which
(a) there is no growth of cells
(b) the nucleus does not participate
(c) no spindle develops to guide the cells
(d) the plasma membranes of daughter cells do not separate.
Answer:
(a) there is no growth of cells

Question 2.
In animal cells, cytokinesis involves
(a) the separation of sister chromatids
(b) contraction of the contractile ring of microfilament
(c) depolymerisation of kinetochore microtubules
(d) a protein kinase that phosphorylates other enzymes
Answer:
(b) contraction of the contractile ring of microfilament

Question 3.
During mitosis, the number of chromosomes gets
(a) change
(b) no change
(c) maybe change if cell is mature
(d) maybe change if cell is immature
Answer:
(b) no change

Question 4.
A diploid living organism develops from zygote by which type of the following repeated cell divisions?
(a) Meiosis
(b) Amitosis
(c) fragmentation
(d) Mitosis
Answer:
(d) Mitosis

Question 5.
If you are provided with root-tips of onion in your class and are asked to count the chromosomes, which of the following stages can you most conveniently look into?
(a) Metaphase
(b) Telophase
(c) Anaphase
(d) Prophase
Answer:
(a) Metaphase

Question 6.
At which stage of mitosis, chromatids separated and passes to different poles
(a) prophase
(b) Metaphase
(c) anaphase
(d) Telophase
Answer:
(c) anaphase

Question 7.
The two chromatids of a metaphase chromosome represent
(a) replicated chromosomes to be separated at anaphase
(b) homologous chromosomes of a diploid set
(c) non-homologous chromosomes joined at the centromere
(d) maternal and paternal chromosomes joined at the centromere
Answer:
(a) replicated chromosomes to be separated at anaphase

Question 8.
The process of cytokinesis refers to the division of
(a) nucleus
(b) chromosomes
(c) cytoplasm
(d) nucleus and cytoplasm
Answer:
(c) cytoplasm

Question 9.
Which of the following serves as mitotic spindle poison?
(a) Ca2
(b) azide
(c) Tubulin
(d) Colchicine
Answer:
(d) Colchicine

Question 10.
Pairing of homologous chromosomes occurs at which stage?
(a) Zygotene
(b) Leptotene
(c) Metaphase
(d) Pachytene
Answer:
(a) Zygotene

Question 11.
In meiosis, division is
(a) I reductional and II equational
(b) I equational and II reductional
(c) Both reductional
(d) Both equational
Answer:
(a) I reductional and II equational

Question 12.
Which type of chromosomes segregate when a cell undergoes meiosis?
(a) Homologous chromosomes
(b) Non-homologous chromosomes
(c) Both (a) and (b)
(d) centric and acentric chromosomes
Answer:
(a) Homologous chromosomes

Question 13.
Chiasmata are most appropriately observed in meiosis during
(a) diakinesis
(b) diplotene
(c) metaphase-ll
(d) pachytene
Answer:
(b) diplotene

Question 14.
During cell division, sometimes there will be failure of separation of homologous chromosomes. This event is called
(a) interference
(b) complementation
(c) non-disjunction
(d) coincidence
Answer:
(c) non-disjunction

Question 15.
The second meiotic division leads to
(a) separation of sex chromosomes
(b) fresh DNA synthesis
(c) separation of chromatids and centromere
(d) separation of homologous chromosomes.
Answer:
(c) separation of chromatids and centromere

Question 16.
Term meiosis was proposed by
(a) Farmer and Moore
(b) Flemming
(c) Strasburger
(d) Darlington
Answer:
(a) Farmer and Moore

Question 17.
Synapsis occurs in the phase of meiosis.
(a) zygotene
(b) diplotene
(c) pachytene
(d) leptotene
Answer:
(a) zygotene

Question 18.
When the number of chromosomes is already reduced to half in the first reductional division of meiosis, where is the necessity of second meiotic division
(a) The division is required for the formation of four gametes
(b) Division ensures equal distribution of haploid chromosomes
(c) Division ensures equal distribution of genes on chromosomes
(d) Division is required for segregation of replicated chromosomes
Answer:
(d) Division is required for segregation of replicated chromosomes

Students can Download Chapter 8 Excretory Products and their Elimination Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 8 Excretory Products and their Elimination

Plus One Excretory Products and their Elimination One Mark Questions and Answers

Question 1
Which one of the following is also known as an antidiuretic hormone?
(a) Oxytocin
(b) Vasopressin
(c) Adrenaline
(d) Calcitonin
Answer:
(b) Vasopressin

Question 2.
A large quantity of one of the following is removed from our body by lungs.
(a) CO₂ only
(b) H₂O only
(c) CO₂ and H₂O
(d) ammonia
Answer:
(c) CO₂ and H₂O

Question 3.
The pH of human urine is approximately
(a) 6.5
(b) 7
(c) 6
(d) 7.5
Answer:
(c) 6

Question 4.
Which of the following pairs is wrong?
(a) Uricotelic ________ Birds
(b) Ureotelic _________ Insects
(c) Ammonotelic ________ Tadpole
(d) Ureotelic _________ Elephant
Answer:
(c) Ammonotelic ________ Tadpole

Question 5.
Which one of the following statements is incorrect?
(a) The medullary zone of kidney is divided into a few conical masses called medullary pyramids projecting into the calyces.
(b) Inside the kidney the cortical region extends in between the medullary pyramids as renal pelvis.
(c) Glomerulus along with Bowman’s capsule is called the renal corpuscle.
(d) Renal corpuscle, proximal convoluted tabule (PCT) and distal convoluted tubule (DCT) of the nephron are situated in the cortical region of kidney.
Answer:
(b) Inside the kidney the cortical region extends in between the medullary pyramids as renal pelvis.

Question 6.
Take the odd one justify?
GFR, JGA, ANF, ADH, TSH
Answer:
TSH: all others are the hormones regulate urine formation.

Question 7.
Identify the ammonotelic animal from the following and give reason. (Man, Crow, Lizard, Tadpole)
Answer:
Tadpole

Question 8.
While studying the excretory system in man, a student noticed some words.
RAAS and JGA. Can you help him to expand these terms.
Answer:
RAAS – Renin – Angiotensin – Aldosterone System
JGA – Juxta glomerular apparatus

Question 9.
In ureotelic animals, urea is produced in
(a) Kidney
(b) Liver
(c) Flame cells
(d) Malpighian tubules
Answer;
(b) Liver

Question 10.
State whether true or false and correct the false statements if any:

1. Flame cells are the excretory structures in round worms.
2. Fat is absorbed in the form of chylomicrons through the intestinal villi.
3. Blood is absent in cockroach.
4. Joint between atlas and axis vertebrae is a pivot joint.

Answer:

1. False – Flame cells are the excretory structures in flat worms.
2. True
3. False – Blood is present in cockroach. It is colourless.
4. True

Plus One Excretory Products and their Elimination Two Mark Questions and Answers

Question 1.
Draw the schematic diagram of the urine formation. (Hint: Show all the three process involved)
Answer:

Question 2.
Complete the following sentences.

1. Reabsorption of water from DCT is facilitated by the hormone __________
2. Kidney failure can cause a bone disorder called _________
3. Angiotensin II activates the adrenal cortex to release __________
4. In cases of Kidney failure, urea can be removed by the process called __________

Answer:

1. ADH
2. Gout
3. Aldosterone
4. Hemodialysis

Question 3.
Pick out the correct word from list provided (Myasthenia gravis, Uremia, Angina, diuresis)

1. Autoimmune disorder affecting neuromuscular junction is __________
2. Reduce loss of water through urine is _________
3. A system of acute chest pain appears when no enough oxygen is reaching the heart muscle is _________
4. Accumulation of urea in blood is __________

Answer:

1. Myasthenia gravis
2. Diuresis
3. Angina
4. Uremia

Question 4.
Classify the following animals according to their type of excretion.
Frog, Tadpole, birds, man, Turtle, shark, insect, camel
Answer:

Question 5.
Tadpole is ammonotelic while the frog is ureotelic. Comment on this statement.
Answer:
Tadpole larvae’s habitat is aquatic whereas the frog lives in the terrestrial habitat. So in the aquatic habitat due to the presence of water, ammonia can be excreted easily.

But in frog the water must be conserved. So ammonia is changed into urea which needs only small quantity of water for excretion.

Question 6.
Observe the diagram and answer the following questions.

1. Label the parts (1) and (2).
2. Name the functional unit of the kidney.

Answer:

1. The parts are:
   • (1) – Renal cortex
   • (2) – Ureter
2. Nephron

Question 7.
Rearrange the following parts in their correct way.
Glomerulus → Capsular space → loop of Henle → PCT → DCT → Collecting duct → Collecting tubule → Ureter → Renal pelvis → Urethra → Urinary bladder.
Answer:
Glomerulus → Capsular space → PCT → Loop of Henle → DCT →Collecting tubule → Collecting duct → Renal pelvis → Ureter → Urinary bladder → Urethra.

Question 8.
Fill in the gaps:

1. Ascending limb of Henle’s loop is ____(1)____ to water. Whereas the descending limb is _____(2)_____ to it.
2. Reabsorption of water from distal parts of the tubules is facilitated by hormone __________
3. Dialysis fluid contain all the constituents as in plasma except ________
4. A healthy adult human excretes (on an average) _______ gm of urea/day.

Answer:

1. (1) – Impermeable, (2) -Permeable
2. ADH
3. Nitrogenous wastes
4. 25-30 gm of urea/day

Question 9.
Brain controls the kidney action.

1. Name the hormones involved in this.
2. How do they act on kidney?

Answer:

1. Antidiuretic Hormone (ADH)
2. The pituitary gland release the hypothalamic hormone ADH and its is transported to the kidney. The presence of ADH makes the tubules permeable to water. Thus the water is conserved or excreted according to the presence of fluid level in the body.

Question 10.
A patient approaches a doctor having symptoms like excretion of large amount of urine, excessive thirst and dehydration.

1. Identify the disease.
2. Which hormone deficiency causes this disease?

Answer:

1. Diabetes insipidus
2. ADH

Question 11.
A table showing the average quantity of urine in a person in two different seasons are given.

1. Compare it. Do you agree with this table? State the reason for producing different Quantity in different season.
2. Name the hormone responsible for that.

Answer:

1. Amount of water in the body is low in summer season and increase its quantity in winter season. In summer season water lose is very high due to various reasons like evaporation. So body eliminate less amount of waterthrough urine to conserve water in the body
2. ADH or vasopressin

Question 12.
Teacher asks one of the student to name the excretory organs in man. He names three organs beside kidney. Teacher appreciates him for the correct answer.

1. What was his answer?
2. Give the excretory role of the above organs

Answer:

1. Liver, skin, lungs
2. the excretory role of the above organs
   • Liver – Excretes cholesterol, bile pigments, etc.
   • Skin – Sweat gland- Excretes excess sodium chloride, small amount of urea lactic acid etc. Sebaceous gland excretes lipids such as waxes, steroles, fatty acid etc.
   • Lungs – Excretes CO₂ and water

Question 13.
Observe the diagram

1. Copy the diagram and fill the gaps.
2. Write the concentration of glomerular filtrate (Isotonic, Hypotonic, Hypertonic) in the given regions of the nephron.
   • Proximal convoluted tubule
   • Descending limb
   • Ascending limb
   • Distal convoluted tubule

Answer:
1.

2. Concentration of glomerular filtrate

• Proximal convoluted tubule – isotonic
• Descending limb – hypertonic
• Ascending limb – hypotonic
• Distal convoluted tubule – Hypertonic

Question 14.
A patient with renal failure in waiting for a kidney donor for a transplantation operation. The urea level is high and the patient is developing symptoms of uremia. What method you can suggest to keep the patient live till a donor can be found. Explain the principle behind it.
Answer:
Haemodialysis. The haemodialyser has a cellophane tube bathed in a fluid with ionic concentration similarto that of plasma. The process of separation of small solutes from macromolecular colloids is done hereby the process of diffusion. So the blood will be free from urea, uric acid and creatinine.

Question 15.
About 125ml of glomerular filtrate is produced per minute in our kidney. However 1ml of urine is produced each minute. What happens to the other 124ml of the filtrate.
Answer:
In man about 125ml of glomerular filtrate is formed per minute. But most of the substances are reabsorbed into tubular network of capillaries. This process is known as selective reabsorption. High threshold substances reabsorbed by active transport and low threshold substances reabsorbed by diffusion.

Question 16.
Consumption of alcohol tends to frequent urination.

1. Name the hormone that control it.
2. Draw a flow chart showing the action of ethanol present in alcohol and its consequences on urinary system.

Answer:
1. Antidiuretic hormone (ADH)

2.

Question 17.
Prepare a flow chart showing hormonal control by Juxta glomerular apparatus (JGA) in regulation of kidney function.
Answer:

Question 18.
Flow of blood in vasa recta is known as counter current system. Give reason.
Answer:
The vasa recta is thin walled capillaries lying parallel to the loop of Henle. The blood entering the descending limb of each vasarecta come in close contact with the outgoing blood in ascending limb. This system is called counter current system.

The two limbs of the loop of Henle form another counter current system. These two systems are concentrating the urine in mammalian kidney by diffusion and osmosis.

Question 19.
‘Micturition is an involuntary as well as a voluntary process.’ Justify the statement.
Answer:
The elimination of urine from the urinary bladder is called micturition. The gradual filling of the urinary bladder causes stretching. When the bladder is filled to its limit, the stretch receptors send impulses to the brain to excrete the urine.

The initiation and inhibition of micturition is voluntary in adult but involuntary in children. Due to lack of nervous control the micturition in some humans become involuntary causing bed wetting.

Question 20.
In alcoholic drinkers the urine is dilute. Why?
Answer:
The process of reabsorption of water by the distal parts ofthe kidney tubules become low efficient. This occurs due to the deficiency of ADH. lt leads to diabetes insipedus.

Question 21.
Distinguish between diabetis mellitus and diabetis insipedus.
Answer:
Insulin deficiency causes large amounts of blood sugar to be lost in the urine. It leads to diabetis melitus while the deficiency of ADH leads to diabetes insipedus i.e dilute urine disease.

Question 22.
The failure of removing toxic substances from the blood ultimately leads kidney failure.

1. How is it temporarily solved.
2. Name the machine is used.

Answer:

1. Hemodialysis is an artificial process of removing toxic substances from the blood
2. Artificial kidney or Hemodialysis machine.

Plus One Excretory Products and their Elimination Three Mark Questions and Answers

Question 1.
State the differences between the following.

1. Tubular reabsorption and Tubular secretion.
2. ADH and RAAS.

Answer:
1. Tubular reabsorption:
The process of selective reabsorption of most of the useful substances from the glomerular filtrate in the renal tubule.

Tubular secretion:
It is the process of active secretion of certain substances such as Hydrogen ions, ammonia etc. from the blood into the lumen of the urinary tubules by the tubular epithelium during urine formation.

2. ADH:
ADH is the hormone from posterior pituitary gland, makes the wall of the distal convoluted tube and collecting tubule permeable to water so that more water is reabsorbed.

RAAS:
Renin – Angiotensin – Aldosterone system increases the blood pressure by constricting arterioles and increases the blood volume by stimulating proximal convoluted tubule.

Question 2.

1. What does the diagram represent?
2. Name the parts A, B, C, D.
3. What is the physiological function taking place in part ‘B.

Answer:

1. Nephron
2. The parts are
   • A – Bowman’s capsule
   • B – Henle’s loop
   • C – Proximal convoluted tubule
   • D – Collecting duct

3. Henle’s loop is the site of osmoregulation in human kidney. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. The ascending limb is impermeable to water but allows transport of electrolyte actively or passively.

Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Question 3.
Observe the diagram and answer the questions.

1. Identify the labelled parts A, B, C, D.
2. What does the diagram represents?

Answer:

1. The labelled parts are
   • A – Afferent arteriole
   • B – Efferent arteriole
   • C – Glomerulus
   • D – Bowman’s capsule
2. Glomerular Filtration

Question 4.
Observe the diagram given below which represents the counter current system in nephron and vasa recta. This mechanism helps in maintaining concentration gradient in the medullary interstitium. Substantiate.

Answer:
The flow of Glomerular filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current. The flow of blood through the two limbs of vasa recta is also in a counter current pattern.

The proximity between the Henle’s loop and vasa recta, as well as the counter current in them help in maintaining an increasing osmolarity towards the inner medullary interstitium i.e, from 300 mOsmolL1 in the cortex to about 1200 mOsmolL1 in the inner medulla.

This gradient is mainly caused by NACI and urea. NaCl is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta.

Similarly, small amounts of urea enter the thin segment of the ascending limb of Henle’s loop which in transported back to the interstitium by the collecting tubule. This counter current mechanism helps to maintain a concentration gradient in the medullary interstitium.

Question 5.
The steps involved in the treatment of a uremic patient is given below.

1. Complete the missing steps (a) and (b)
2. Diagonise the treatment
3. Name the organ which is under failure

Answer:

1. The missing steps are
   • (a) – Anticoagulant like heparin
   • (b) – Anti – heparin
2. Hemodialysis
3. Kidney

Question 6.

1. Name the mechanism by which JGA plays a complex regulatory role.
2. Draw a flow chart showing the mechanism.
3. Mention the feed back which act as a check on the above mechanism,

Answer:
1. Renin-Angiotensin mechanism

2.

3. An increase in blood flow to the atria of the heart can cause the release of Atrial Natriuretic factor (ANF). ANF can cause vasodilation and thereby decrease the blood pressure. ANF mechanism act as a check on the renin-Angiotensin mechanism.

Question 7.
Note the relationship between the first two words and suggest a suitable word in the missing place.

1. Cockroach – Malpighian Tubule : Flatworm – _________
2. Man – Kidney : Earthworm – _________
3. Prawn – Greengland : Amphioxus – __________

Answer:

1. Flame cells
2. Nephridia
3. Protonephridia

Plus One Excretory Products and their Elimination NCERT Questions and Answers

Question 1.
Define Glomerular Filtration Rate(GFR)
Answer:
Glomerular Filtration Rate (GFR). The amount of the filtrate formed by the kidneys per minute is called glomercular filtration rate (GFR). GFR in a healthy individual is approximately 125ml/minute, i.e., 180 litres per day. On an average, 1100 -1200 ml of blood pumped out by each ventricle of the heart in a minute.

Question 2.
Explain the autoregulatory mechanism of GFR.
Answer:
Regulation of GFR. The kidney have built-in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus(JGA).

JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomercular blood flow and thereby the GFR back to normal.

Question 3.
Indicate whether the following statements are true or false:

1. Micturition is carried out by a reflex
2. ADH helps in water elimination, making the urine hypotonic.
3. Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
4. Henle’s loop plays an important role in concentrating the urine.
5. Glucose is actively reabsorbed in the proximal convoluted tublue.

Answer:

1. True
2. False
3. True
4. True
5. True

Question 4.
Give a brief account of the counter mechanism.
Answer:
Counter current Mechanism. The Henle’s loop and vasa recta play a significant role in this. The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a countercurrent.

The flow of blood through the two limbs of vasa recta is also in a countercurrent pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter current in them help in maintaining an increasing osmolarity towards the inner medullary in terstitium, i.e., from 300 mOsmolL^-1 in the cortex to about 1200 mOsmolL^-1 in the inner medulla.

This gradient is mainly caused by NaCI and urea. NaCI is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta. NaCI is returned to the intestitium by the ascending portion of vasa recta.

Question 5.
What is meant by the term osmoregulation?
Answer:
Osmoregulation is the regulation of blood volume, body fluid volume and ionic concentration

Question 6.
Match the items of column I with those of column II.
Column I Column II

Answer:
(a) – (iii)
(b) – (v)
(c) – (iv)
(d) – (i)
(e) – (ii)

Question 7.
Name the following.

1. A chordate animal having flame cells as excretory structures.
2. Cortical portions projecting between the medullary pyramids in the human kidney.
3. A loop of capillary running parallel to the Henle’s loop.

Answer:

1. Cephalochordata
2. Columns of Bertini
3. Vasa Recta

Question 8.
Fill in the gaps:

1. Ascending limb of Henle’s loop is ________ to water whereas the descending limb is __________ to it.
2. Reabsorption of water from distal parts of the tubules is facilitated by hormone ___________
3. Dialysis fluid contain all the constituents as in plasma except ___________
4. A healthy adult human excretes (on an average) _________gm of urea/day.

Answer:

1. impermeable, permeable
2. ADH
3. nitrogenous waste
4. 25 to 30

Plus One Excretory Products and their Elimination Multiple Choice Questions and Answers

Question 1.
What will happen if the stretch receptors of the urinary bladder wall are totally removed’?
(a) Urine will not collect in the bladder
(b) Micturition will continue
(c) Urine will continue to collect normally in the bladder
(d) There will be no micturition
Answer:
(c) Urine will continue to collect normally in the bladder

Question 2.
Green glands present in some arthropods help in
(a) respiration
(b) excretion
(c) digestion
(d) reproduction
Answer:
(b) excretion

Question 3.
Excretory product of spider is
(a) uric acid
(b) ammonia
(c) guanine
(d) None of these
Answer:
(c) guanine

Question 4.
Which one is the component of ornithine cycle
(a) Ornithine, citrulline and fumaric acid
(b) Ornithine, citrulline and arginine
(c) Ornithine, citrulline and alanine
(d) Amino acids are not used
Answer:
(b) Ornithine, citrulline and arginine

Question 5.
Urea synthesis occurs in
(a) kidney
(b) liver
(c) brain
(d) muscles
Answer:
(b) liver

Question 6.
Renin is secreted from
(a) juxtaglomerular cells
(b) podocytes
(c) nephridia
(d) stomach
Answer:
(a) juxtaglomerular cells

Question 7.
Loop of Henle is associated with
(a) excretory system
(b) respiratory system
(c) reproductive system
(d) digestive system
Answer:
(a) excretory system

Question 8.
If one litre of water is introduced in human blood, then
(a) BMR increases
(b) RBC collapses and urine production increases
(c) RBC collapses and urine production decreases
(d) BMR decreases
Answer:
(b) RBC collapses and urine production increases

Question 9.
The urine is
(a) hypotonic to blood and isotonic in medullary fluid
(b) hypotonic to bloodand isotonic to medullary fluid
(c) isotonic to blood and hypotonic to medullary fluid
(d) hypertonic to blood and isotonic to medullary fluid
Answer:
(d) hypertonic to blood and isotonic to medullary fluid

Question 10.
Which substance is in higher concentration in blood than in glomerular filtrate’?
(a) Water
(b) Glucose
(c) Urea
(d) Plasma proteins
Answer:
(d) Plasma proteins

Question 11.
A large quantity of fluid is filtered everyday by the nephrons in the kidneys. Only about 1% of it is excreted as urine. The remaining 99% of the filtrate
(a) gets coiected in the renal pelvis
(b) is lost as sweat
(c) is stored in the urinary bladder
(d) is reabsorbed into the blood
Answer:
(d) is reabsorbed into the blood

Question 12.
The characteristic that is shared by urea, uric acid and ammonia is/are
I. They are nitrogenous wastes.
II. They all need very large amount of water for excretion.
III. They are all equally toxic.
IV. They are produced in the kidneys.
(a) I and III
(b) I and IV
(c) I, III and IV
(d) I only
Answer:
(d) I only

Question 13.
A bird excretes nitrogenous waste materials in the form of?
(a) uric acid
(b) ammonia
(c) urea
(d) amino acids
Answer:
(a) uric acid

Question 14.
Which of the following is correct with reference to haemodialysis?
(a) Absorbs and resends excess of ions
(b) The dialysis unit has a coiled cellophane tube
(c) Blood is pumped back through a suitable artery after haemodialysis
(d) Anti-heparin is added priorto haemodialysis
Answer:
(b) The dialysis unit has a coiled cellophane tube

Question 15.
Find the incorrect statement regarding mechanism of urine formation in man.
(a) the glomerular filtration rate is about 125 mL/min
(b) the ultra filtration is opposed by the colloidal osmotic pressure of plasma
(c) tubular secretion takes place in the PCT
(d) aldosterone induces greater reabsorption of sodium
Answer:
(c) tubular secretion takes place in the PCT

Question 16.
The excretory material of bony fish is
(a) urea
(b) protein
(c) ammonia
(d) amino acid
Answer:
(c) ammonia

Question 17.
The yellow colour of urine is due to the presence of
(a) urea
(b) uric acid
(c) urochrome
(d) bilirubin
Answer:
(c) urochrome

Question 18.
Malpighian tubules are
(a) excretory organs of insects
(b) excretory organs of frog
(c) respiratory organs of insects
(d) endocrine glands of insects
Answer:
(a) excretory organs of insects

Question 19.
The size of filtration slits of glomerulus
(a) 10wn
(b) 15nm
(c) 20nm
(d) 25nm
Answer:
(d) 25nm

Question 20.
The conversion of dangerous nitrogenous waste into less toxic excretory matter is carried out in man in the
(a) blood
(b) liver
(c) kidney
(d) skin
Answer:
(b) liver

Students can Download Chapter 3 Trigonometric Functions Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Plus One Maths Trigonometric Functions Three Mark Questions and Answers

Question 1.
Prove the following

Answer:
i) LHS

ii) LHS

iii) LHS = sin 2x + 2 sin 4x + sin 6x
= 2 sin 4xcos2x + 2sin 4x
= 2 sin 4x(cos2x + 1) = 4 cos² x sin 4x

iv) LHS

v) LHS

vi) LHS

vii) LHS = sin² 6x – sin² 4x

= 2 sin 10x sin(-2x)
= 2 sin 10x sin2x

viii) LHS

Question 2.
Find the general solution of the following equations.

1. cos4x = cos2x
2. sin 2x +cosx = 0
3. cos3x + cosx – cos2x = 0

Answer:
1. Given; cos 4x = cos 2x
⇒ cos4x – cos 2x = 0
⇒ -2 sin 3x sin x = 0
General solution is
⇒ sin3x = 0; ⇒ 3x = nπ ⇒ x = \(\frac{n \pi}{3}\), ∈ Z
Again we have;
⇒ sinx = 0; ⇒ x = nπ; n ∈ Z

2. Given; sin 2x + cosx = 0
⇒ 2sin xcosx + cosx = 0
⇒ cosx(2sin x + 1) = 0
General solution is
⇒ cosx = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
Again we have; 2sin x + 1 = 0

3. Given; cos3x +cosx – cos2x = 0
⇒ 2 cos2x cosx – cos2x = 0
⇒ cos2x(2cosx – 1) = 0
General solution is

Again we have; 2cosx -1 = 0

Question 3.
In Triangle ABC, if a = 25, b = 52 and c = 63, find cos A and sin A.
Answer:

Question 4.
For any ΔABC, prove that a(b cosC – c cosB) = b² – c²
Answer:
LHS = ab cos C – ac cos B

Question 5.
For any ΔABC, prove that, \(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}}\).
Answer:

Question 6.

1. Convert \(\frac{2 \pi}{3}\) radian measure into degree measure. (1)
2. Prove that \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x\) (2)

Answer:
1. \(\frac{2 \pi}{3}=\frac{2 \pi}{3} \times \frac{180}{\pi}=120^{\circ}\)

2. LHS

Plus One Maths Trigonometric Functions Four Mark Questions and Answers

Question 1.
For any ΔABC, prove that

Answer:

Question 2.
For any ΔABC, prove that \(\sin \frac{B-C}{2}=\frac{b-c}{a} \cos \frac{A}{2}\).
Answer:

Question 3.
(i) Which of the following is not possible. (1)
(a) sin x = \(\frac{1}{2}\)
(b) cos x = \(\frac{2}{3}\)
(c) cosec x = \(\frac{1}{3}\)
(d) tan x = 8
(ii) Find the value of sin 15°. (2)
(iii) Hence write the value of cos 75° (1)
Answer:
(i) (c) cosec x = \(\frac{1}{3}\)

(ii) sin 15° = sin(45° – 30°)
= sin45°cos30°- cos45°sin30°

(iii) sin 15° = sin(90° – 75°) = cos 75°

Plus One Maths Trigonometric Functions Six Mark Questions and Answers

Question 1.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B, the angle of elevation is 60°, where B is a point at a distance d from the point A measured along the line AB which makes angle 30° with AQ. Prove that d = h(\(\sqrt{3}\) – 1).
Answer:

From the figure we have ∠PAQ = 45°, ∠BAQ = 30°and ∠PBH = 60°
in right ∆AQP
Clearly ∠APQ = 45°, ∠BPH = 30° , giving ∠APB = 15° ⇒ ∠PAB = 15°
In ∆APQ ,PQ = AQ = h
AP² = h² + h² = 2h² ⇒ AP = \(\sqrt{2}\)h
From ∆ABP,

Question 2.
A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.
Answer:

Let BC represent the tree, A be the point 35m down the hill from the base of the tree and h be the height of the tree.
Clearly in ∆ABC
∠BAC = 60°- 15° =45°;
∠ACB = 30°; ∠ABC = 105°

Question 3.
(i) If sin x = cos x, x ∈ [0, π] then is
(a) 0
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{3}\)
(d) π
(ii) Write the following in ascending order of tits values, sin 100°, sin 0°, sin 50°, sin 200°
(iii) Solve: sin2x – sin4x + sin6x = 0
Answer:
(i) (b) \(\frac{\pi}{4}\)

(ii) sin 100° = sin(l 80 – 80) = sin 80°
sin 200° = sin(l 80° + 20°) = -sin 20°
The ascending order is
sin 200°, sin 0°, sin 50°, sin 100°

(iii) sin2x + sin6x – sin4x = 0
⇒ 2sin 4x cos2x – sin 4x = 0
⇒ sin 4x(2 cos 2x – 1) = 0
⇒ sin4x = 0 or (2cos2x – 1) = 0
⇒ 4x = nπ or cos2x = \(\frac{1}{2}\)

Plus One Maths Trigonometric Functions Practice Problems Questions and Answers

Question 1.
Convert the following degree measure into radian measure.
i)  45°
ii) 25°
iii) 240°
iv) 40°20′
v) -47°30′
Answer:

Question 2.
Convert the following radian measure into degree measure,
i)   6
ii) -4
iii) \(\frac{5 \pi}{3}\)
iv) \(\frac{7 \pi}{6}\)
v) \(\frac{11}{16}\)
Answer:

Question 3.
The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14)
Answer:
60 minutes = 360 degrees.
1 minutes = 6 degrees.
40 minutes = 240 degrees.
240° = 240 × \(\frac{\pi}{180}=\frac{4 \pi}{3}\)
The required distance travelled = l = rθ
= 1.5 × \(\frac{4 \pi}{3}\) = 2 × 3.14 = 6.28 cm

Question 4.
In a circle of diameter 40 cm, the length of a cord is 20 cm. Find the length of minor arc of the chord.
Answer:
The radius and chord join to form a equilateral triangle. Therefore
l = rθ = 20 × \(\frac{\pi}{3}\)
= 20 × \(\frac{3.14}{3}\) = 20.933.

Question 5.
If the arcs of the same lengths in the two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.
Answer:
We have l = rθ, the radius and angle are inversely proportional. Therefore;

Question 6.
Find the values of the other five trigonometric functions in the following; (2 score each)

1. cos x = \(-\frac{3}{5}\), x lies in the third quadrant.
2. cot x = \(-\frac{5}{12}\), x lies in the second quadrant.
3. sin x = \(\frac{1}{4}\), x lies in the second quadrant.

Answer:
1. Given;
cos x = \(-\frac{3}{5}\)

2. Given;
cot x = \(-\frac{5}{12}\)

3. Given;
sin x = \(\frac{1}{4}\); cosecx = 4

Question 7.
Find the value of the trigonometric functions. (2 score each)

Answer:

Question 8.
Find the value of the following.

iv) sin 75°
v) tan 15°
Answer:

iv) sin 75° = sin(45° + 35°)
= sin 45° cos30° + cos45° sin 30°

v) tan 15° = tan(45° – 30°) = \(\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}\)

Question 9.
Find the principal and general solution of the following.

1. sin x = \(\frac{\sqrt{3}}{2}\)
2. cosx = \(\frac{1}{2}\)
3. tan x = \(\sqrt{3}\)
4. cos ecx = -2

Answer:
1. Given; sin x = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)
General solution is; x = nπ + (-1)ⁿ\(\frac{\pi}{3}\),
n ∈ Z
Put n = 0, 1 we get principal solution; x = \(\frac{\pi}{3} ; \frac{2 \pi}{3}\).

2. Given; cosx = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
General solution is; x = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z
Put n = 0, 1 we get principal solution;
n = 0 ⇒ x = \(\frac{\pi}{3}\); n = 1 ⇒ x = 2π – \(\frac{\pi}{3}\) = \(\frac{5\pi}{3}\).

3. Given; tan x = \(\sqrt{3}\) = tan\(\frac{\pi}{3}\)
General solution is; ⇒ x = nπ + \(\frac{\pi}{3}\), n ∈ Z
Put n = 0, 1 we get principal solution;
n = 0 ⇒ x = \(\frac{\pi}{3}\); n = 1 ⇒ x = π + \(\frac{\pi}{3}\) = \(4\frac{\pi}{3}\).

4. Given; cosecx = -2
⇒ sin x = \(-\frac{1}{2}\) = – sin \(\frac{\pi}{6}\) = sin(-\(\frac{\pi}{6}\) )
General solution is; x = nπ – (-1)ⁿ \(\frac{\pi}{6}\), n ∈ Z
Put n = 1, 2 we get principal solution;

Students can Download Chapter 8 Mineral Nutrition Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 8 Mineral Nutrition

Plus One Botany Mineral Nutrition One Mark Questions and Answers

Question 1.
Which one of the following roles is not characteristic of an essential element?
(a) being a component of biomolecules
(b) changing the chemistry of soil
(c) being a structural component of energy-related chemical compounds
(d) activation or inhibition of enzymes
Answer:
(b) changing the chemistry of soil

Question 2.
Which one of the following statements can best explain the term critical concentration of an essential element?
(a) essential element concentration below which plant growth is retarded.
(b) essential element concentration below which plant growth becomes stunted.
(c) essential element concentration below which plant remains in the vegetative phase.
(d) none of the above
Answer:
(b) essential element concentration below which plant growth becomes stunted.

Question 3.
During protein synthesis subunits of ribosomes are combined due to the presence of the particular element. Name it.
Answer:
Magnesium

Question 4.
Deficiency symptoms of an element tend to appear first in young leaves.
It indicates that the element is relatively immobile. Which one of the following elemental deficiency would show such symptoms?
(a) sulphur
(b) Magnesium
(c) Nitrogen
(d) Potassium
Answer:
(a) sulphur

Question 5.
With regard to the Biological Nitrogen Fixation by Rhizobium in association with soybean, which one of the following statement/statements does not hold true.
(a) Nitrogenase may require oxygen for its functioning
(b) Nitrogenase is MO- Fe protein
(c) Leg-hemoglobin is a pink coloured pigment.
(d) Nitrogenase helps to convert N₂ gas into two molecules of ammonia.
Answer:
(a) Nitrogenase may require oxygen for its functioning

Question 6.
Name the free-living microorganisms which can fix nitrogen.
Answer:
Azotobactor and Rhodospirillum

Question 7.
In what form do plants absorb molebdenum from the soil.
Answer:
Molybdate ions M_oO₂²⁺

Question 8.
Mineral requirements of plants is explained by a method in the nutrient solution is Called ………. and it was demonstrated by ……….. for the first time.
Answer:
Hydroponics,Julius von sachs

Question 9.
Find the odd one out.
Boron, Copper, Zinc, Phosphorous
Answer:
Phosphorous

Question 10.
A plant cell when kept in a certain solution got plasmolysed. What was the nature of the solution?
Answer:
Hypertonic Solution.

Question 11.
Plants can be cultivated in water. Name the type of cultivation.
Answer:
Hydroponics

Question 12.
Find the odd one among the following Carbon, Manganese, potassium, Nitrogen
Answer:
Manganese

Question 13.
A farmer adds Azotobacter culture to the soil before sowing maize. Which mineral element is being replenished?
Answer:
Nitrogen

Question 14.
In the diagram below, Label the cell Ain Nostoc

Answer:
Heterocyst

Question 15.
Which one of the following statements can best explain the term critical concentration of an essential element?
(a) essential element concentration below which plant growth is retarded.
(b) essential element concentration below which plant growth becomes stunted.
(c) essential element concentration below which plant remains in the vegetative phase.
(d) none of the above
Answer:
(a) essential element concentration below which plant growth is retarded.

Question 16.
Deficiency symptoms of an element tend to appear first in young leaves. It indicates that the element is relatively immobile. Which elemental deficiency would show such symptoms?
Answer:
Sulphur or calcium

Question 17.
From where do plants receive hydrogen?
Answer:
Water absorbed by plants.

Plus One Botany Mineral Nutrition Two Mark Questions and Answers

Question 1.
Give example for

1. Free-living N2 fixing bacteria.
2. Symbiotic N2 fixing bacteria.

Answer:

1. Acetobacter, Nitrococcus
2. Rhizobium

Question 2.
Crop plants cannot grow well in the nitrogen-deficient soil while plants like Drosera and Nepenthes show vigorous growth.

1. Justify the statement
2. Mention any two deficiency shown by such crop plants.

Answer:
1. These plants are insectivorous plants. They capture and feed on insects in order to maintain the N₂ content of their body.

2. Two deficiency shown by such crop plants

• Stunted growth
• Chlorosis

Question 3.
Study the relation of the given pair and fill up the blanks:

1. Potassium: Stomatal movement:………… Constituent of chlorophyll
2. ………………: pollen germination Zinc: biosynthesis of auxin.
3. Ion exchange: Passive absorption ………….: Active absorption
4. Protoplasmic streaming theory: De Vries: ………….Active absorption

Answer:

1. Mg
2. Boron
3. Carrier concept
4. Munch

Question 4.
Plants can be grown in defined nutrient solution in the absence of soil.

1. Name the technique.
2. Who demonstrated the technique for the first time?

Answer:

1. Hydroponics
2. Julius von sachs

Question 5.
How is nitrogenase enzyme protected?
Answer:
In leguminous plants, the enzyme nitrogenase is protected from Oxygen by the pigment leghaemoglobin.

Question 6.
The functional roles of certain essential elements are given below. Identify the element based on the function.

1. Essential for the formation of Chlorophyll.
2. Involved in the pollen germination.

Answer:

1. Mg
2. Boron

Question 7.
Explain the scientific reason for growing legume crops prior to cereal crops.
Answer:
Increase the fertility of the soil. Biological N₂ fixation.

Question 8.
Name the pink coloured pigment present in the root nodules? Write its role in N₂ fixation?
Answer:
Leghemoglobin is an oxygen scavenger. It creates anaerobic condition for the action of nitrogenase enzyme to fix N₂.

Question 9.
Give example for the following taking part in biological nitrogen fixation.

1. Free-living N₂ fixing bacteria
2. Free-living N₂ fixing cyanobacteria
3. Symbiotic N₂ fixing bacteria.
4. Symbiotic N₂ fixing cyanobacteria

Answer:

1. Azotobacter, Clostridium
2. Nostoc, Anabaena, Oscillatonia.
3. Rhizobium leguminosarum
4. Nostoc, Anabaena

Question 10.
Prepare a flow chart which shows the conversion of N₂ to ammonia in biological N₂ fixation.
Answer:

Question 11.
A pigment similar to hemoglobin is present in leguminous plants. Name the pigment and write its role.

OR

How is nitrogenase enzyme protected?
Answer:
Leghemoglobin which is the scavenger of 0₂. Nitrogenase enzyme needs oxygen-free condition.

Question 12.
Observe the figure given below and answer the following questions.

1. Name this technique.
2. Write two uses of this technique.

Answer:

1. Hydroponics
2. uses
   • Used as kitchen garden,
   • Used to study jdeficiency symptoms.

Question 13.
For the normal growth of plants it requires minerals. Write two examples of micro and macro elements.
Answer:

• Macro – N, P
• Micro-Boron, copper

Question 14.
In a survey done in an agricultural area the crops in the area shows various symptoms.

• Chlorosis
• Necrosis
• Delay flowering.

Name the elements whose deficiency causes these symptoms.
Answer:

• N, P, K, Mg
• Chlorosis – N, K, Mg, S, Fe, Mn, etc.
• Necrosis – K, Mg, Ca and Cu Delay flowering – N, S, and Mo

Question 15.
Iron is not a structural component of chlorophyll but its deficiency causes yellowing of leaves. Give reason.
Answer:
Iron activates catalise enzyme in the formation of chlorophyll.

Question 16.
Amides are derivatives of amino acids in which the hydroxyl group is replaced by the amino group (NH2)
a) How are asparagine and glutamine formed.
b) Amides contain more than amino acids.
Answer:

Question 17.
In a survey done in an agricultural area the crops in the area shows various symptoms.

• Chlorosis
• Necrosis
• Delay flowering

Name the elements whose deficiency causes these symptoms.
Answer:

• N, P, K, Mg
• Chlorosis – N, K, Mg, S, Fe, Mn, etc.
• Necrosis – K, Mg, Ca and Cu
• Delay flowering – N, S, and Mo

Question 18.
A farmer supplies Nitrogen fertilizer to pea plants.

1. Is there any necessity to supply Nitrogen fertilizer?
2. Justify your answer with reason.

Answer:

1. No
2. Symbiotic N2 fixation

Question 19.
Name the respecive mineral nutrient element that

1. Forms the core constituent of the ring structure of chlorophyll
2. Activates carboxylases
3. Forms the components of nitrogenase
4. Synthesises middle lamella of plant cell.

Answer:

1. Magnesium
2. Manganese
3. Molybdenum and iron
4. Calcium

Question 20.
Observe the relationship between the first pair and fill in the blanks.
Posassium: Opening and closing of stomata::
Boron: …………
Answer:
Boron: Pollen germination/ pollen tube growth

Question 21.
State whether True or False.

1. Potassium plays an important role in opening and closing of stomata.
2. The movement of ions is usually called flux.
3. In an LHC (Light-Harvesting complex), Chlorophyll b is the reaction center.
4. The first stable product in calvin cycle is 3-PGA.

Answer:

1. True
2. True
3. False
4. True

Question 22.
A few inorganic elements are given. Match them with their specific functions.

Answer:

Question 23.
Pick the suitable minerals from the list, for following processes?

1. Synthesis of auxin
2. Pollen germination
3. Photolysis of water
4. Nitrogen metabolism

Answer:

1. Zn
2. B
3. Mn
4. Mo

Question 24.
Rhizobium leguminosarum cannot do N₂ fixation outside root nodules. Give reasons.
Answer:
Rhizobium can fix atmospheric N₂ only if the symbiotic relationship with leguminose plant is established.

Question 25.
What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Answer:
It acts as 0₂ scavenger that means it helps in preventing the combining of 0₂ with nitrogenase.

Question 26.
Leguminous plants can be cultivated in between rice cultivation. Why?
Answer:
Because it increases the fertility of soil (enrich N₂ content) for next year rice cultivation.

Question 27.
Match the element with its associated functions/ roles.

Answer:

1. A – iv
2. B – i
3. C – iii
4. D – ii
5. E – v

Question 28.
To get maximum yeild a farmer added exess amount of manganese-containing fertiliser. The plants showed some deficiency symptoms. Can you give the reason for this?
Answer:
Excess of the micronutrient -Mn causes Toxicity because it prevent the uptake of Iron, Magnesium, and calcium. This results in deficiency symptom.

Question 29.

1. Identify the phenomenon?
2. A pink coloured protein protects nitrogenase from oxygen. Name that protein.

Answer:

1. Ammonification
2. Leghemoglobin

Question 30.

Answer:

Question 31.

1. Name the oxygen-binding pigment found in the root nodule of plants like pea and bean.
2. What is the normal colour of this pigment?

Answer:

1. Leghemoglobin
2. Pink-colored

Question 32.
How the presence of magnesium fulfills the requirements of the essentiality of elements from growth and development of plants?
Answer:

1. It is a constituent of chlorophyll molecule and is essential for photosynthesis.
2. It cannot be replaced by any other element for the same function.
3. It is also required as a co-factor by many enzymes involved in cellular respiration and other metabolic pathways.

Question 33.
Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment?
Answer:
In prokaryotes, photosynthetic autotrophs (cyanobacteria) like nostoc and Anabaena have special N₂ fixing cell which helps in N₂ fixation.

Question 34.
How is sulphur important for plants? Name the amino acids in which it is present.
Answer:
Sulphur, besides being present in some amino acids essential for protein synthesis, is also a constituent of several coenzymes, vitamins, and ferrodoxin which are involved in some biochemical pathways.

Question 35.
How do some bacteria carry out nitrification? What are such bacteria called?
Answer:
Such types of bacteria convert ammonia into nitrates and obtain energy for their activities. They are called chemosynthetic autotrophs.

Question 36.
Name the respective mineral nutrient elements of plants that:

1. Is needed in the synthesis of auxins
2. Is a constituent of ferredoxin
3. Forms the core constituent of the ring structure of chlorophyll
4. forms the components of nitrogenase and nitrate reductase

Answer:

1. Zinc
2. Sulphur
3. Magnesium
4. Molybdenum

Question 37.
Why do plants of legume family contain more protein than the other plants?
Answer:
Plants in legume family bears root nodules which bear symbiotic nitrogen-fixing bacteria i.e rhizobium

Question 38.
Nepenthes carries out photosynthesis and still traps insects. Why?
Answer:
Nepenthes grow in nitrogen-deficient soils so these trap, digest and absorb amino acids from the insects supplementing their nitrogen supply.

Question 39.
Mycorrohiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Answer:
It is an association of root and fungi in which surface area of root is increased and more minerals can be absorbed. Fungus get nutrients and sugar from the plant.

Question 40.
Carnivorous plants like Nepenthes and Venus flytrap have nutritional adaptations. Which nutrient do they especially obtain and from where?
Answer:
In these, insect body is digested and plant gets Nitrogen as nutrients.

Question 41.
How are organisms like Pseudomonas and Thiobacillus of significance in nitrogen cycle?
Answer:
Pseudomonas and Thiobacillus carry out denitrification process wherein the nitrate present in the soil is reduced to nitrogen thus contributing to the atmospheric nitrogen.

Question 42.
Think of a plant which lacks chlorophyll. From where will it obtain nutrition? Give an example of such a type of plant.
Answer:
This case is observed in total parasites. In such plants they absorb food, water, and minerals from host. Eg cuscuta.

Question 43.
Nitrogen is essential element for plants and is found in abundance as atmospheric nitrogen. But most plants unable to use it. Why is it so and in what form do plants utilize them?
Answer:
Most plants cannot use it because atmospheric nitrogen is inert in its nature. Nitrogen fixers in soil can convert nitrogen gas into nitrates or nitrite or ammonium forms. These compounds enter the plants as nutrients through the roots and are assimilated as organic nitrogen.

Question 44.
How insectivorous plants fulfill their nitrogen requirements?
Answer:
Insectivorous plants fulfill their nitrogen requirements from insects and small animals trapped by their leaves(a pitcher).

Question 45.
All elements that are present in a plant need not be essential to its survival. Give reasons.
Answer:
The Criteria for essentiality are

1. The element must be necessary for supporting normal growth and reproduction
2. In the absence of elements, the plants do not complete their life cycle or set the seeds.
3. The element must be directly involved in the metabolism of the plant.
4. The requirement of the element must be specific and not replaceable by another.

Question 46.
Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs?
Answer:

• If the elements are actively mobilized, the older part of plant show deficiency symptoms.
• If they are not mobilized, the deficiency symptoms are first observed in younger regions.

Question 47.
Mineral elements are re-exported from one place to other or some are immobile and deficiency symptoms observed

1. Which is the part of the plant show deficiency symptoms in the former?
2. Name the mineral element is immobile and which part of the plant shows deficiency symptoms?

Answer:

1. Older regions because minerals are remobilised from older regions into younger regions.
2. Ca, Deficiency symptoms are noticed in younger tissues because it forms the structural part.

Plus One Botany Mineral Nutrition Three Mark Questions and Answers

Question 1.
Give the technical term for the following.

1. Leaf yellowing
2. death of tissues
3. soilless cultivation

Answer:

1. chlorosis
2. Necrosis
3. Hydroponics

Question 2.
Plants pick up nitrogen from the soil in the form of ammonia ions or nitrate ions, ammonia being the main product of biological nitrogen fixation. From the above view answer the following.
a) Which soil bacteria convert ammonia to NO₂ – and NO_3.
b) Draw a schematic diagram of progressive reduction of one molecule of nitrogen in the presentee of enzyme nitrogenase to yield two molecule of ammonia.
c) Name the enzyme for nitrogen assimilation in plants.
Answer:

Question 3.
Macro and micronutrients are required in particular amount for plants.

1. What will be the moderate increase of micronutrient affect the plants
2. Name the micronutrient helps in pollen germination
3. What is the concentration of macro and micronutrients required for plants?

Answer:

1. It cause toxicity to plants and affect the uptake and utilisation of macronutrients
2. Boron
3. Macronutrients – Excess 10 millimole /kilogram of dry matter
   Micronutrients – Less than 10 millimole /kilogram of dry matter

Question 4.
During formation of nitrate in plants, different steps occur.

1. Name the steps leading to the conversion of atmospheric nitrogen to nitrate
2. Which is the reductive and oxidative steps?
3. Name the pigments and Enzymes that are required and the steps they act on.

Answer:

1. Nitrogen fixation, Nitrification
2. Nitrogen fixation—reductive process, Nitrification—Oxidative process
3. pigment—Leghaemoglobin, Enzyme— Nirogenase Pigments and enzymes acts on Nitrogen fixation step

Question 5.

1. Name the process
2. Which enzyme catalyses the process
3. How the enzyme in the process is protected from oxygen in root nodules?

Answer:

1. N₂ fixation
2. Nitrogenase
3. Due to the presence of pigment Leghemoglobin. It acts as the scavenger of oxygen.

Plus One Botany Mineral Nutrition NCERT Mark Questions and Answers

Question 1.
‘All elements that are present in a plant need not be essential to its survival’. Comment.
Answer:
The criteria for essentially of an element are given below:

1. The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element, the plants do not complete their life cycle or set the seeds.

2. The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.

3. The element must be directly involved in the metabolism of the plant. All elements that are present in a plant do not fulfill these criteria hence cannot be essential for plant survival.

Question 2.
Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Answer:
In 1860, Julius von Sachs, a prominent German botanist, demonstrated, for the first time, that plants could be grown to maturity in a defined nutrient solution in complete absence of soil. The essence of all these methods involves the culture of plants in a soil-free, defined mineral solution. These methods require purified water and mineral nutrient salts.

Purification of water and nurtient salt is important to rule out other influencing factors. The presence of pure nutrients will give clearcut scientific results. This will help in making a sound basis for the right prediction.

Question 3.
What are the steps involved in the formation of a root nodule?
Answer:
Steps in the development of root nodules:
1. Rhizobium bacteria contact a susceptible root hair, divide near it.

2. Upon successful infection of the root hair cause it to curl

3. Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteriods and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation.

4. A mature nodule is complete with vascular tissues continuous with those of the root.

Plus One Botany Mineral Nutrition Multiple Choice Questions and Answers

Question 1.
1. Farmers in a particular region were concerned that premature yellowing of leaves of a pulse crop might cause a decrease in the yield. Which treatment could be most beneficial to obtain maximum seed yield?
(a) Frequent irrigation of the crop
(b) Treatment of the plants with cytokinins along with a small dose of nitrogenous fertilizer
(c) Removal of all yellow leaves and spraying the remaining green leaves with 2, 4, 5- thchlorophenoxy acetic acid
(d) Application of iron and magnesium to promote synthesis of chlorophyll
Answer:
(d) Application of iron and magnesium to promote synthesis of chlorophyll

Question 2.
For nitrogen fixation, useful pigment is
(a) nitrogenase
(b) hemoglobin
(c) myoglobin
(d) leghaemoglobin
Answer:
(d) leghaemoglobin

Question 3.
Plants cultivated in nutrient solution without soil is called
(a) somatic hybridization
(b) tissue culture
(c) hydroponics
(d) suspension culture
Answer:
(c) hydroponics

Question 4.
The process of decay of dead organic matter is known as
(a) denitrification
(b) nitrification
(c) nitrogen fixation
(d) ammonification
Answer:
(d) ammonification

Question 5.
Zn, Mo, Fe, Cu are
(a) trace elements
(b) non-essentials
(c) macronutrients
(d) both a and b
Answer:
(a) trace elements

Question 6.
An essential element is that which
(a) improves health of the plant
(b) is irreplaceable and indispensable for the growth of plants
(c) is found in plant ash
(d) is available in the soil
Answer:
(b) is irreplaceable and indispensable for the growth of plants

Question 7.
N₂ +8e +8H +16ATP → 2NH₃+2H⁺ + 16ADP + 8e^– The above equation refers to
(a) amnionification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(c) nitrogen fixation

Question 8.
Micronutrients are
(a) as important as macronutrients but are required in small amount
(b) less important than macronutrients
(c) called micro as they play only a minor role in plant nutrition
(d) required greater than 10m mole/Kg of dry matter
Answer:
(a) as important as macronutrients but are required in small amount

Question 9.
Which element is located at the centre of the porphyrin ring in chlorophyll?
(a) Potassium
(b) Manganese
(c) Calcium
(d) Magnesium
Answer:
(d) Magnesium

Question 10.
Which element is required for the germination of pollen grains?
(a) Boron
(b) Calcium
(c) Chlorine
(d) Potassium
Answer:
(a) Boron

Question 11.
Select the correct statement.
(a) Legumes are incapable of fixing nitrogen
(b) Legumes fix nitrogen through bacteria living fruits
(c) Legumes fix nitrogen only by bacteria present in root nodules
(d) frankia forms symbiotic association with algae
Answer:
(c) Legumes fix nitrogen only by bacteria present in root nodules

Question 12.
Chlorosis is caused due to deficiency of
(a) magnesium
(b) calcium
(c) boron
(d) manganese
Answer:
(a) magnesium

Question 13.
Enzyme nitrogenase is responsible for
(a) nitrification
(b) nitrogen fixation
(c) nitrite reduction
(d) nitrate reduction
Answer:
(b) nitrogen fixation

Question 14.
Maximum percentage of which element occurs in plant ash?
(a) Magnesium
(b) Zinc
(c) Potassium
(d) Calcium
Answer:
(d) Calcium

Question 15.
Which of the following metals causes bone cancer
(a) Lead
(b) Cobalt
(c) Uranium
(d) strontium90
Answer:
(d) strontium90

Question 16.
Premature leaf fall is due to deficiency of
(a) phosphorus
(b) nitrogen
(c) calcium
(d) potassium
Answer:
(a) phosphorus

Question 17.
The function of leg haemoglobin during biological nitrogen fixation in root nodules of legumes is to
(a) convert atmospheric nitrogen to ammonia
(b) convert ammonia to nitrite
(c) transport oxygen for activity of nitrogenase
(d) protect nitrogenase from oxygen
Answer:
(d) protect nitrogenase from oxygen

Question 18.
Which of the following gene is responsible for biological nitrogen fixation?
(a) Nitrogenase
(b) N if gene
(c) Yeast alanine tRNA synthetase
(d) RNA synthetase
Answer:
(b) N if gene