Respiration in Plants Test

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Respiration in Plants Test - 86

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Question 1
1 / -0

Identify the sequence of steps involved during inspiration in rabbit.
(A) Volume of thoracic cavity increases
(B) Contraction of external intercostal muscles moves the rib cage forward
(C) Diaphragm becomes flat
(D) Thoracic cavity expands drawing air into lungs.

A
(C)$$\rightarrow$$(B)$$\rightarrow$$ (A)$$\rightarrow$$(D)

B
(C)$$\rightarrow$$(A)$$\rightarrow$$ (B)$$\rightarrow$$ (D)

C
(A)$$\rightarrow$$(B)$$\rightarrow $$(C)$$\rightarrow$$ (D)

D
(A)$$\rightarrow$$(D)$$\rightarrow$$ (B) $$\rightarrow$$ (C)

Solution

The urge to breathe comes from the respiratory center, located at the base of brain. It sends signals via the spinal cord to diaphragm and the muscles between ribs telling them when to contract or relax. During inspiration, the diaphragm -- the large muscle that divides chest and abdomen -- contracts and moves downward. Additionally, ribs move outward. This enlarges your chest and lungs expand. Lung expansion creates a vacuum. Air enters the nose and mouth and is pulled into windpipe -- the trachea. The trachea divides into smaller airways called bronchi. These continue to divide as they get farther from the trachea, like the branches of a tree. Finally, the tiny airways deliver the air to the smallest structures in your lung -- the alveoli -- where gas exchange takes place.
Question 2
1 / -0

The time taken from the fixation of $$CO_{2}$$ to the formation of one glucose molecule is about __________ seconds.

A
$$20$$

B
$$40$$

C
$$60$$

D
$$90$$

Solution

In anaerobic respiration, one glucose molecule produces a net gain of two ATP molecules (four ATP molecules are produced during glycolysis, but two are required by enzymes used during the process). In aerobic respiration, a molecule of glucose is much more profitable in that a maximum net production of 30 or 32 ATP molecules (depending on the organism) is generated. The time taken for the fixation is about 90 seconds
So, the correct answer is '90'90
Question 3
1 / -0

Enzymes of TCA cycle are located in the mitochondrial matrix except one which is located in inner mitochondrial membrane in eukaryotes and in cytosol in prokaryotes. This enzyme is

A
Succinate dehydrogenase

B
Lactate dehydrogenase

C
Isocitrate dehydrogenase

D
Malate dehydrogenase

Solution

All the oxidative enzymes of TCA are located in matrix except succinic dehydrogenase. This enzyme catalyzes the conversion of succinic acid into fumaric acid. The enzyme is an integral protein complex that is tightly bound to the inner mitochondrial membrane. In fact, this enzyme is the preferred marker enzyme for inner membranes while performing mitochondrial fractionations. In prokaryotes, it is located in the cytosol.
So, the correct answer is option A.
Question 4
1 / -0

A student has taken 6000 molecules of fructose 1, 6-bisphosphate and 2 molecules of aldolase into an enzyme reaction mixture to determine its activity. After 5 minutes, 50% substrate was found converted into products. Then, what is the TON of aldolase and the total number of trioses formed by the enzyme action at the end of 5 minutes?

A
300, 3000

B
300, 6000

C
600, 3000

D
30, 6000

Solution

Turnover number is equivalent to the number of substrate molecules converted to product in a given unit of time on a single enzyme molecule, when a enzyme is saturated by substrate. Aldolase converted one substrate molecule to one product molecule.
If the total number of product after 5 minutes is (3000*2) = 6000.
So, the TON = 3000/(5*2) = 300
The total number of trioses formed by the enzyme action at the end of 5 minutes is 6000 (as one aldolase converts fructose bis phosphate into 2 triose phosphate).
Question 5
1 / -0

Net yield of aerobic respiration during Krebs cycle per glucose molecule is

A
2 ATP molecules

B
8 ATP molecules

C
36 ATP molecules

D
38 ATP molecules

Solution

The tricarboxylic acid cycle is a series of chemical reactions to generate energy through the oxidation of acetyl CoA. During one cycle, one molecule of ATP is produced when succinyl-CoA gets converted into succinic acid. Since for 1 molecule of glucose 2 rounds of Krebs cycle occurs, there is the net yield of 2 ATP molecules.
Question 6
1 / -0

A scientist took 8 molecules of cytosoic aldolase into an enzyme reaction mixture to study its activity. After ten minutes of enzyme reaction, 60% of its substrate was found converted into 2400 molecules of products. Then what is the TON of aldolase and number of substrate molecules left over in the reaction mixture?

A
$$30, 1440$$

B
$$30, 960$$

C
$$15, 1280$$

D
$$15, 800$$

Solution

Turnover Number = number of molecules of substrate converted to product per time by one active site.
Fructose 1,6-bisphosphate is converted by the enzyme aldolase into the triose phosphates- dihydroxyacetone phosphate and glyceraldehyde 3-phosphate.
If the total number of products after 10 minutes is 2400, the number of converted substrates is 2400/2 = 1200.
Hence turnover number = 1200/10 x 8= 15.
Here, total number of substrate = 1200 x 100/60 = 2000
So, the remaining substrate molecule is 2000-1200 = 800
Thus the correct answer is option D.
Question 7
1 / -0

The puffed-up appearance of dough is due to fermentation by yeast. Identify the gas liberated during the process

A
Methane

B
Carbon dioxide

C
Hydrogen sulphide

D
Ammonia

Solution

Yeast respire anaerobically i.e., through fermentation and convert the carbohydrate in the dough to alcohol and releases carbon dioxide as a by-product. Due to the release of carbon dioxide dough becomes puffy.
Question 8
1 / -0

In which of the following reactions of glycolysis, a molecule of water is removed from the substrate?

A
Glucose $$\rightarrow$$ Glucose-6-phosphate

B
Fructose-6-phosphate $$\rightarrow$$ Fructose-1,6 bisphosphate

C
2-phosphoglycerate $$\rightarrow$$ Phosphoenol pyruvate

D
Phosphoenol pyruvate $$\rightarrow$$ Pyruvate

Solution

Options A and B represent the addition of phosphate group to glucose and fructose 6 phosphates respectively. Option D represents dephosphorylation of phosphoenolpyruvate into pyruvate. Dehydration of 2-phosphoglycerate to phosphoenolpyruvate includes reversible removal of a molecule of water from 2-phosphoglycerate.
So, the correct answer is option C.
Question 9
1 / -0

The last stage of cellular respiration which yields ATP by the oxidation of organic molecules, derived from glucose is

A
Chemiosmosis.

B
Citric acid cycle.

C
Electron transport chain.

D
Glycolysis.

E
Oxidation phosphorylation.

Solution

Cellular respiration is a set of metabolic reactions and processes that take place in the cells to convert biochemical energy from the nutrients to ATP and release waste materials. These catabolic reactions break large molecules into smaller ones, releasing energy in the process, and the weak so-called "high-energy" bonds are replaced by stronger bonds in the products.Through cellular respiration the cells harvest energy stored in food by a catabolic pathway for the production of adenosine triphosphate (ATP). Cellular respiration is again aerobic and anaerobic. Aerobic respiration takes place in the presence of oxygen, where ATP is is produced by the cells through oxidation of organic compounds. Aerobic respiration takes place in three stages glycolysis, TCA cycle and electron transport chain. As the end product of glycolysis, pyruvate enters the mitochondria and is fully oxidized by the Krebs cycle. During this process, carbon dioxide and water and energy takes the form of ATP. However, most of the ATP produced during aerobic respiration is through oxidative phosphorylation.
Question 10
1 / -0

Match the compounds given in Column I with the number of carbon atoms present in them which are listed under Column II. Choose the answer which gives the correct combination of alphabets of the two columns.
Column I
Column II
A. Oxaloacetate
(p) 6- C compound
B. Phosphoglyceraldehyde
(q) 5- C compound
C. Isocitrate
(r) 4- C compound
D. a-Ketoglutarate
(s) 3- C compound

(t) 2- C compound

A
A- s, B- t, C- q, D- r

B
A- r, B- s, C- p, D- q

C
A- r, B- t, C- p, D- q

D
A- q, B- s, C- p, D- t

Solution

The citric acid cycle begins with the transfer of a two-carbon acetyl group from acetyl-CoA to the four-carbon acceptor compound (oxaloacetate) to form a six-carbon compound (citrate).
Phosphoglyceraldehyde is the breakdown of one molecules of glucose which becomes two PGAL with 3 carbon atoms and 1 phosphate each. Isocitrate dehydrogenase (IDH) is an important enzyme in the tricarboxylic acid cycle, which occurs in the mitochondrial matrix.
IDH is responsible for catalyzing the reversible conversion of isocitrate to alpha-ketoglutarate and CO2 in a two-step reaction.
α-Ketoglutarate is a key intermediate in the Krebs cycle, coming after isocitrate and before succinyl CoA.

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Column I	Column II
A. Oxaloacetate	(p) 6- C compound
B. Phosphoglyceraldehyde	(q) 5- C compound
C. Isocitrate	(r) 4- C compound
D. a-Ketoglutarate	(s) 3- C compound
	(t) 2- C compound

Respiration in Plants Test - 86

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Respiration in Plants Test - 86

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Question 1

(C)$$\rightarrow$$(B)$$\rightarrow$$ (A)$$\rightarrow$$(D)

(C)$$\rightarrow$$(A)$$\rightarrow$$ (B)$$\rightarrow$$ (D)

(A)$$\rightarrow$$(B)$$\rightarrow $$(C)$$\rightarrow$$ (D)

(A)$$\rightarrow$$(D)$$\rightarrow$$ (B) $$\rightarrow$$ (C)

Solution

Question 2

$$20$$

$$40$$

$$60$$

$$90$$

Solution

Question 3

Succinate dehydrogenase

Lactate dehydrogenase

Isocitrate dehydrogenase

Malate dehydrogenase

Solution

Question 4

300, 3000

300, 6000

600, 3000

30, 6000

Solution

Question 5

2 ATP molecules

8 ATP molecules

36 ATP molecules

38 ATP molecules

Solution

Question 6

$$30, 1440$$

$$30, 960$$

$$15, 1280$$

$$15, 800$$

Solution

Question 7

Methane

Carbon dioxide

Hydrogen sulphide

Ammonia

Solution

Question 8

Glucose $$\rightarrow$$ Glucose-6-phosphate

Fructose-6-phosphate $$\rightarrow$$ Fructose-1,6 bisphosphate

2-phosphoglycerate $$\rightarrow$$ Phosphoenol pyruvate

Phosphoenol pyruvate $$\rightarrow$$ Pyruvate

Solution

Question 9

Chemiosmosis.

Citric acid cycle.

Electron transport chain.

Glycolysis.

Oxidation phosphorylation.

Solution

Question 10

A- s, B- t, C- q, D- r

A- r, B- s, C- p, D- q

A- r, B- t, C- p, D- q

A- q, B- s, C- p, D- t

Solution

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