GATE Biotechnology Mock Test - 1 - GATE Biotechnology MCQ

Professional is the person who does something as a part of his job. Amateur is a person who does something because he loves doing it.

A = 2( l + b)h
A = 2(3 + 2) × 1 = 10
New area = 2(6 + 1) × 1/2 = 7
Decrease = (3/10) × 100
= 30%

Selling price of goods sold by Nidhi = Rs. 100
Selling price of goods sold by Deepika = Rs. 90
Now, according to the question:
110% of selling price of goods sold by Gurpreet = Rs. 90
Selling price of goods sold by Gurpreet = (100/110) x 90 = Rs. 900/11
So, Gurpreet's goods are cheaper than Nidhi's goods by

Therefore, the customer would have saved 18.18%, if he bought the goods from Gurpreet instead of buying them from Nidhi.

Ratio of ages of Meena and Leena = 4 : 3
Let the present age of Meena be 4x.
Present age of Leena = 3x
According to the question:
4x + 3x = 28
7x = 28
x = 4
Therefore, the present age of Meena = 4 × 4 = 16 years
Age of Leena = 3 × 4 = 12 years
After 8 years,
Age of Meena = 16 + 8 = 24 years
Age of Leena = 12 + 8 = 20 years
Therefore, required ratio = 24 : 20 or 6 : 5

Ratio of milk and water

Eigen values of a skew-symmetric matrix are either zero or pure imaginary, occuring in conjugate pairs.

Bioaugmentation is an in-situ bioremediation technique in which selected standardized bacteria are added to an area that has been contaminated with an unwanted substance. These bacteria breakdown the contaminants. Biostimulation involves modification of the environment to stimulate the existing bacteria for bioremediation. In biosparging, air (oxygen) and nutrients are injected in a saturated zone to increase microbial biological activity. Phytoremediation is a technique that employs plants for bioremediation.

Hepatitis B is the only hepatitis virus to have a double stranded DNA (dsDNA) as its genome. All other hepatitis viruses have single stranded RNA as their genome.

PC is present in the plasma membranes of human cells such as the RBCs and neurons, but it is totally absent from that of the E. coli. The major phospholipids present in the membrane of E. coli are PE and PS that collectively account for 85% of its weight.

Restriction Fragment Length Polymorphism (RFLP) is a genetic marker with high genomic abundance. RFLPs have only two alleles. It is inherited as a codominant marker.

f(x) = x²e^-x
Or f'(x) = 2xe^-x - x²e^-x
= xe^-x (2 - x)
f''(x) = (x² - 4x+ 2) e^-x
Now for maxima and minima, f'(x) = 0
xe^-x (2 - x) = 0
or x = 0, 2
at x = 0   f''(0) = 2 (+ ve)
at x = 2   f''(2) = - 2e^-2(- ve)
Now f''(0) = 2 and f''(2) = - 2e^-2 < 0. Thus, x = 2 is the point of maxima. 

We have 
f(x) = sin x + 2cos x
f'(x) = cos x - 2 sin x ....(i)
f"(x) = - sin x - 2 cos x ....(ii)
For maxima and minima, put f'(x) = 0
cos x = 2 sin x
tan x= 1/2 
x = 26.56° 
At x = 26.56° 
f''(x) = - 2.236(-ve)
Thus f(x) is maximum at x = 26.56° 

Given: 2P (X = 5) = P (X = 4) + P (X = 6)

⇒ 2 ⁿC₅ = ⁿC₄ + ⁿC₆
⇒ n = 7 or 14.

Trypsin cleaves carboxyl side of basic residues, Lys and Arg, while chymotrypsin cleaves the carboxyl sides of aromatic amino acids Phe, Trp and Tyr.

Many enzymes need a cofactor for their activity. A specific enzyme is functional in association to a specific metal ion cofactor.
A carboxypeptidase is a protease enzyme that hydrolyzes (cleaves) a peptide bond at the carboxy-terminal (C-terminal) end of a protein or peptide. Carboxypeptidase needs Zn ions for its activity.
Superoxide dismutase is an enzyme that helps break down potentially harmful oxygen molecules in cells. Superoxide dismutase requires Cu ions as a co factor.
Enolase, also known as phosphopyruvate hydratase, is a metalloenzyme responsible for the catalysis of the conversion of 2-phosphoglycerate (2-PG) to phosphoenolpyruvate (PEP). Enolase requires Mg ions as a cofactor.
Urease is an enzyme that catalyzes the hydrolysis of urea, forming ammonia and carbon dioxide. Urease needs Ni ions for its activity and functionality.

The number of generations, n = Total time passed/doubling time
n = 2.5/0.5 = 5
N_o = N/2ⁿ = 32 × 10⁶/25 = 10 × 10⁵ cells

The map distance between the genes is equal to the recombination frequency of the genes.
Recombination frequency = Number of recombinant offsprings/Total number of offsprings.
Recombination frequency = (36 + 39)/(36 + 39 + 160 + 165) = 0.1875 or 18.75%
Hence, the genes A and B are 18.8 map units apart.

2.303 log (N/N0) = -k_dt
Or
k_d = - (2.303/t) log (N/N_o)
Putting the values, we get
kd = - (2.303/1) log (10⁴/10⁶) = 4.606 hr^-1

During the M phase, the number of chromosomes(48) in a cell is double that of the actual number so as to ensure equal distribution of chromosomes in the daughter cells. The cell enters the G1 phase after cell division in the M phase is over. So, the number of chromosomes in the G1 phase would be half of what was in the M phase i.e 24.