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IIT JAM Biotechnology Mock Test 4 - IIT JAM MCQ


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30 Questions MCQ Test - IIT JAM Biotechnology Mock Test 4

IIT JAM Biotechnology Mock Test 4 for IIT JAM 2024 is part of IIT JAM preparation. The IIT JAM Biotechnology Mock Test 4 questions and answers have been prepared according to the IIT JAM exam syllabus.The IIT JAM Biotechnology Mock Test 4 MCQs are made for IIT JAM 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for IIT JAM Biotechnology Mock Test 4 below.
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IIT JAM Biotechnology Mock Test 4 - Question 1

What is the instrument that measures the amount of air inhaled and exhaled with each breath?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 1

A spirometer measures ventilation, the movement of air into and out of the lungs.

IIT JAM Biotechnology Mock Test 4 - Question 2

Which one of the following statements is incorrect regarding the pharynx?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 2

Pharynx does not contribute to the production of speech, it s larynx. 

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IIT JAM Biotechnology Mock Test 4 - Question 3

Emphysema, a chronic disorder is high in cigarette smokers. In such cases the ______ of the person is/are found damaged.

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 3

Emphysema is caused due to damage of bronchioles. 

IIT JAM Biotechnology Mock Test 4 - Question 4

Which of the following neurotransmitters is tryptophan derived

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 4

Tryptophan is also a precursor to the neurotransmitter serotonin, the hormone melatonin, and vitamin B3.

IIT JAM Biotechnology Mock Test 4 - Question 5

Which organ is mainly responsible for osmoregulation?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 5

Kidneys by selective reabsorption and secretion can maintain the osmolarity of blood.

IIT JAM Biotechnology Mock Test 4 - Question 6

Which part of the root is not involved in transport of water

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 6

Root cap is not involved in water absorption

IIT JAM Biotechnology Mock Test 4 - Question 7

Which of the following is correct about photosynthesis

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 7

Photosynthesis takes place only in the presence of photosynthetic pigment such as chlorophyll, carotenoids, phycocyanin, phycoerythrin. These pigments capture light necessary for photosynthesis.

IIT JAM Biotechnology Mock Test 4 - Question 8

In which of the following is chlorophyll present?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 8

Chlorophyll is the green pigment present in the thylakoid membrane. These thylakoids are stacked on top of one another and constitute grana. So, chlorophyll is present in the grana of chloroplasts.

IIT JAM Biotechnology Mock Test 4 - Question 9

Protein was purified to homogeneity. Determination of the molecular weight by size exclusion chromatography yields 60 kD. Chromatography in the presence of 6 M urea yields a 30-kd species. When the chromatography is repeated in the presence of 6 M urea and 10 mM b-mercaptoethanol, a single molecular species of 15 kd results. Describe the structure of the molecule?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 9

Size exclusion chromatography, also known as molecular sieve chromatography, is a chromatographic method in which molecules in solution are separated by their size and  molecular weight.
    Molecular weight of the protein was found to be 60 kDa by size exclusion chromatography. So, 60 kDa is the molecular weight of intact native protein.
    6 M  urea  will  denature the   protein,  that is protein will lose its secondary and tertiary structure, however its disulphide bonds will remain intact in the presence of urea. Urea does not reduce the disulphide bonds. Urea belongs to a class of compounds known as chaotropic denaturants, which unravel the tertiary structure of proteins by destabilizing internal, non-covalent bonds between atoms.
    So when size exclusion chromatography was carried out in the presence of 6 M urea, 30 kDa species was obtained, it’s because, urea denatures the proteins, so subunits of proteins (each subunit 30 kDa) were  separated from each other in the presence of urea.
    However, when size exclusion chromatography was carried out in the presence of 6 M urea and Beta mercaptoethanol, a single molecular species of 15 kDa was obtained. It’s because, beta mercaptoethanol is a reducing agent. It reduces the disulphide bonds (that  form between thiol groups of cysteine residues) present in the protein.
    Each single 30 kDa subunit of protein was further made up of 2 single 15 kDa subunits that were held together by disulphide bond, which was reduced in the presence of beta mercaptoethanol.
    Hence 60 kDa protein is a tetrameric protein held together by disulphide bridge.

IIT JAM Biotechnology Mock Test 4 - Question 10

Following figure describes the components of electron transport chain. 

Which of the following proteins is represented by middle cylindrical shaped protein (showing conversion of O2 to H2O)?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 10

Cytochrome oxidase also known as complex IV is the terminal component of the electron transport system. It receives an electron from  cytochrome c molecule and transfers it to oxygen  molecule converting the molecular oxygen to water. In this process it translocates protons across the membrane increasing the transmembrane difference of proton electrochemical potential which the ATP synthase then uses to synthesize ATP.

IIT JAM Biotechnology Mock Test 4 - Question 11

Phosphorylase kinase is activated by hormones that lead to the phosphorylation of the  subunit and by Ca2+ binding of the  subunit. Both types of stimulation are required for maximal enzyme activity as shown in the figure below:

Which of the following mutation at the active site of phosphorylase is likely to inhibit the regulation of above protein ?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 11

S123L means serine at 123 poition in the protein is replaced by leucine. In this case nature of amino acid changes as leucine does not allow phosphorylation, therefore option (a) is correct. Serine, Threonine and tyrosine are most commonly phosphorylated amino acids. Histidine is phosphorylated in prokaryotes and plants (now, also known to be phosphorylated in human also).

IIT JAM Biotechnology Mock Test 4 - Question 12

Which phenomenon is best explained by following graph?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 12

Radioactive decay is a first order reaction hence, the above graph is true for radioactive decays.

IIT JAM Biotechnology Mock Test 4 - Question 13

Which of the following protein conformations can be stretched to the highest distance?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 13

Pi helix is the most stretched and 2.27 helix is the least stretched helix. The α -helix is not the only helical structure in proteins. Other helical structures include the 310 helix, which is stabilized by hydrogen bonds o f the type ( i, i+ 3 ) and the π -helix, which is stabilized by hydrogen bonds of the type ( i, i + 5 ).
The 310 helix has a smaller radius compared to the α-helix while the π -helix has a Larger radius.

IIT JAM Biotechnology Mock Test 4 - Question 14

Shalini, a biochemistry student was performing an experiment on proteins. She isolated the protein and precipitated it with the help of tri-chloroacetic acid. While she was doing her experiment, a notorious boy Suraj came to the lab and added some surf excel grains to the solution. Which of the following statement is true about the protein?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 14

Change in the protein conformation is observed by detergents due to denaturation. During denaturation, tertiary and secondary structure of protein is lost whereas primary structure remains intact.

IIT JAM Biotechnology Mock Test 4 - Question 15

Autoantibodies to a lipid in the membrane of platelets are seen in the disease, systemic lupus erythrematous. Which of the following membrane lipids is most likely to be involved?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 15

Systemic lupus erythematosus (SLE) is an autoimmune disease in which the body’s immune system mistakenly attacks healthy tissue in many parts of the body including platelets.  
    People with SLE have low platelet and white blood cell counts because autoantibodies against cardiolipin present in membrane are made. Anti cardiolipin antibodies  are antibodies often directed against cardiolipin and found in diseases such as systemic lupus erythematosus (SLE).
    In mammalian cells, cardiolipin (CL) is found almost exclusively in the inner mitochondrial membrane where it is essential for the optimal function of numerous enzymes that are involved in mitochondrial energy metabolism. Cardiolipin is exclusively present in mitochondria

IIT JAM Biotechnology Mock Test 4 - Question 16

Hemoglobin and myoglobin are two proteins with high degree of similarity in the structure and function. Following are few statements regarding hemoglobin and myoglobin. Select the correct statement.

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 16

Hemoglobin’s oxygen binding affinity is relatively weak compared to myoglobin’s affinity for oxygen. Hemoglobin’s oxygen-binding curve is in the shape of a sigmoidal curve. In red blood cells, the oxygen-binding curve for hemoglobin displays an “S” shaped called a sigmoidal curve. A sigmoidal curve shows that oxygen binding is cooperative; that is, when one site binds oxygen, the probability that the remaining unoccupied sites that will bind to oxygen will increase.
    The sigmoid shape is a consequence of the four subunits of hemoglobin “cooperating” in the binding of oxygen.  
    The myoglobin saturation curve is a rectangular hyperbola, NOT sigmoid. This different shape is the result of having a single subunit.
    The single subunit character of myoglobin and the multiple subunit character of hemoglobin give each of these molecules their characteristic “behavior” with respect to oxygen binding and different physiological uses for which each is suited.

IIT JAM Biotechnology Mock Test 4 - Question 17

Among the following peptide which one is expected to fold in a alpha helix conformation?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 17

In this question. look for absence of proline or glycine, repeats of basic or acidic amino acids and terminals should have acidic amino acid at amino terminal and basic amino acid carboxy terminal. Different amino acid sequences have different propensities for forming α helical structure. Methionine. alanine. leucine. glutamate. and lysine ("MALEK" in the amino acid 1 letter codes) all have high helix forming propensities. whereas proline and glycine have poor helix forming propensities. Proline either breaks or kinks a helix. because it cannot donate an amide hydrogen bond (having no amide hydrogen) and also because its side chain interferes sterically with the backbone of the preceding turn inside a helix. this forces a bend of about 30° in the helix's axis. At the other extreme, glycine also tends to disrupt helices because its high conformational flexibility makes it emropically expensive to adopt the relatively constrained α -helical structure. 
Option (a) is correct. Peptide (a) has the highest propensity to form an alpha helix because it is the only peptide which has acidic amino acid (D) at the N terminal end and basic amino acid at the C terminal end (R) which is essential for stability of alpha helix. Also it has amino acids which have higher propensity to form alpha helix compared to amino acids present in other options. Although peptide (b) also has a acidic auto acid (D) at the N terminal end and basic amino acid at the C terminal end (R.), but still peptide (b) has the less propensity to form alpha helix because it consist of proline (P) and glycine (G) which disrupt the alpha helix. Peptide (c) has less probability to fonn alpha helix because it consist of repeats of acidic amino acids (DE). Peptide (d) lacks a basic amino acid at the C terminal end (Histidine is less basic than arginine). So, between peptide(a) and peptide (d). peptide (a) is more likely to take up an alpha helix, due to presence of a acidic amino acid (D) at the N terminal end and basic amino acid at the C terminal end (R). 

IIT JAM Biotechnology Mock Test 4 - Question 18

Which of the following represents a protein sequence?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 18

English alphabet letters  B, J, O, U X, Z do not encode for standard peptides.

IIT JAM Biotechnology Mock Test 4 - Question 19

Four proteins are in alpha helix, pi helix, beta sheet and random coil structures respectively. Which of the following will be most resistance to physical stress?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 19

Beta sheets are most resistant to physical stress as they are most stretched structures.

IIT JAM Biotechnology Mock Test 4 - Question 20

Motifs are super secondary structures obtained by characteristic organization of secondary structures in a protein. Which of the following motifs will be largest in length if the all motifs have same number of amino acids?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 20

Beta-alpha-beta motif will have highest length. The distance between adjacent amino acids in a beta strand is 3.5 A° which is longer in comparison to the 1.5 A° distance in alpha strand. Therefore beta-alpha-beta motif will be of longer length compared to alpha-beta-alpha motif, provided they are made up of same number of amino acids.

IIT JAM Biotechnology Mock Test 4 - Question 21

What kind of symmetry is observed in the ATP synthase protein at quaternary level of organization?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 21

ATP synthase  is a membrane protein that makes ATP from the energy of proton gradient across the mitochondrial membrane. Since the transmembrane part (cylinder part in the figure) of ATP synthase has cyclic symmetry, they are all oriented in the same direction perpendicular to the internal surface of the mitochondrial membrane. This makes sure that all the ATPs produced are on one side of the membrane.

IIT JAM Biotechnology Mock Test 4 - Question 22

A scientist working at enzymes manufacturing firm created a modified enzyme with application for blood brain barrier(BBB) crossing capabilities to deliver a conjugated drug. Although the composition of new formation was kept secret by company, someone from the company gave a clue that at least three successive amino acids at the external surface of proteins were replaced with one type of aromatic amino acid. Can you predict which of the following replacements were made in the protein?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 22

Abilities of phenyalanine is highest to cross BBB, hence substitution of few amino acids with F will help.

IIT JAM Biotechnology Mock Test 4 - Question 23

What is the molarity of a protein solution formed by dissolving a 276.8 grams of 25 amino acid long protein in 500 ml water?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 23

Average molecular weight of a amino acid residue =110 Dalton
Molecular weight of 25 amino acid long protein = 110 x 25 = 2750 Daltons
Moles of sohite = given weight molecular weight.
Moles of solute = 276.8 2750 = 0.1

IIT JAM Biotechnology Mock Test 4 - Question 24

For the hydrolysis of 1 mole of ATP at 37°C. the standard free enthalpy change ΔGº = -35 KJ mol-1. Calculate the free enthalpy change ΔG° at the ratio ATP ADP = 100:1
(Temperature 37°C. R = 8.3143 J K-1 mol-1. Concentrations of water and inorganic phosphate are to be omitted from the equilibrium equation, assuming that they do not change significantly)

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 24

IIT JAM Biotechnology Mock Test 4 - Question 25

How is it possible for ions to be at equilibrium across a plasma membrane and yet not be at the same concentration on both sides?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 25

It is possible for ions to be at equilibrium across a plasma membrane and yet not be at the same concentration on both sides because there is a charge difference across the plasma membrane.

IIT JAM Biotechnology Mock Test 4 - Question 26

An enzyme reaction obeys the Michaelis - Menten form
 E+S <------ > E.S → E+P
Which of the following curves best describes the amount of P as a function of time

 

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 26

Only graph D is hyperbolic in shape, while c graph is sigmoidal. As Michaelis Menten kinetics shows only hyperbolic curve, therefore option (b) is correct.

IIT JAM Biotechnology Mock Test 4 - Question 27

Match the enzyme from group A with the respective class in group B.

IIT JAM Biotechnology Mock Test 4 - Question 28

A.    Protein moved more slowly in an SDS- PAGE.
B.    Isoelectric focusing [IEF] showed that there was no change in the pI
C.    Mass spectrometric analysis showed that the modification was on serine
The modification that the protein undergoes is likely to be

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 28

Since, there is no change in the pI after post translational modification, it means the post translational modification cannot be phosphorylation, ADP ribosylation or ubiquitination as these three modifications will change the ionization state of protein and hence change the pI so on IEF their location will be different. Therefore the given protein has glycosylation modification. Also serine cannot undergo ubiquitination; ubiquitination occurs on lysine residue.

IIT JAM Biotechnology Mock Test 4 - Question 29

Which one of the following statements is NOT correct for the classification of carbohydrates?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 29

Gulose is a hexose and arabinose is pentose.

IIT JAM Biotechnology Mock Test 4 - Question 30

Lysozyme has optimal activity at a pH centered around pH 5.0. The active site of lysozyme contains a glutamic acid residue (pKa = 5.5) and an aspartic acid residue (pKa = 4.0). Which of the following statements is correct about the mechanism of lysozyme?

Detailed Solution for IIT JAM Biotechnology Mock Test 4 - Question 30

The catalytic mechanism of lysozyme involves both general acid and general base catalysis. Glu 35 participates in general acid catalysis (donates a proton) and Asp 52 participates in general base catalysis (stabilising the positive charge of the oxonium ion). During the entire catalytic mechanism, the aspartic acid residue remains unprotonated. Its because pH 5 is more than the pKa = 4.0 of aspartic acid.

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