IIT JAM Biotechnology Mock Test 4 - Biotechnology Engineering (BT) MCQ

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

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

Emphysema is caused due to damage of bronchioles. 

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

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

Root cap is not involved in water absorption

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

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.

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.

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.

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).

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

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.

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.

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

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.

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). 

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

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

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.

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.

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

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

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.

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.

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.

Gulose is a hexose and arabinose is pentose.

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.