Can You Score 5/5? NEET Biology High-Yield Genetics Challenge

Every Thursday is test day. NEET is not only about knowing the facts. It is about spotting the trap in a statement and the twist in the wording, under time pressure.

Set a timer for 4 minutes. Attempt all 5 questions without looking at the answers, and see if you can score a perfect 5/5.

Question 1: Single correct (genetics trap)

In a dihybrid test cross involving two genes that assort independently, what is the expected phenotypic ratio?

(A) 9 : 3 : 3 : 1
(B) 1 : 1 : 1 : 1
(C) 9 : 7
(D) 1 : 2 : 1

Question 2: Molecular basis (NCERT line hunter)

Which of the following statements about the lac operon is INCORRECT?

(A) The repressor of the operon is synthesised constitutively from the i gene.
(B) A very low level of expression of the lac operon has to be present in the cell all the time.
(C) Lactose acts as an inducer by binding directly to the operator region to switch on transcription.
(D) Regulation of the lac operon by a repressor is referred to as negative regulation.

Question 3: Inheritance

A colour-blind man marries a woman with normal vision whose father was colour-blind. What is the probability that their first child will be a colour-blind boy?

(A) 0%
(B) 25%
(C) 50%
(D) 75%

Question 4: Assertion and reason

Assertion (A): The lagging strand of DNA needs many RNA primers during semi-conservative replication.

Reason (R): DNA-dependent DNA polymerase can catalyse polymerisation in only one direction, 5′ → 3′.

(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.

Question 5: Match the enzyme to its role

Column I	Column II
p. Aminoacyl-tRNA synthetase	1. Joins Okazaki fragments
q. Peptidyl transferase	2. Charges tRNA with its amino acid
r. DNA ligase	3. Forms peptide bonds
s. Topoisomerase	4. Relieves supercoiling ahead of the fork

(A) p-2, q-3, r-1, s-4
(B) p-3, q-2, r-4, s-1
(C) p-2, q-3, r-4, s-1
(D) p-1, q-3, r-2, s-4

Write down your five answers before you check them.

1. Answer: (B) 1 : 1 : 1 : 1

A test cross crosses the F₁ dihybrid (AaBb) with the homozygous recessive parent (aabb). The four kinds of gamete from AaBb each meet ab, so the phenotypic and genotypic ratios are both 1 : 1 : 1 : 1. The 9 : 3 : 3 : 1 ratio comes from selfing (AaBb × AaBb), not from a test cross.

2. Answer: (C)

The inducer (allolactose, from lactose) binds to the repressor protein, not to the operator. The repressor changes shape and can no longer bind the operator, so transcription goes ahead.

3. Answer: (B) 25%

Father: XᶜY (colour-blind).
Mother: her father was colour-blind, so she carries one Xᶜ: XXᶜ.
Cross: XᶜY × XXᶜ gives four equally likely children: XXᶜ (carrier girl), XᶜXᶜ (colour-blind girl), XY (normal boy), XᶜY (colour-blind boy).

One in four children is a colour-blind boy: 25%. The trap: if the question had said "if the child is a boy", the answer would be 50%.

4. Answer: (A)

DNA polymerase builds only 5′ → 3′. On the lagging strand that direction points away from the replication fork, so the strand is made in short pieces (Okazaki fragments), and each piece needs its own RNA primer. R explains A.

5. Answer: (A) p-2, q-3, r-1, s-4

Aminoacyl-tRNA synthetase attaches each amino acid to its tRNA (charging).
Peptidyl transferase, the 23S rRNA ribozyme in bacteria, forms peptide bonds.
DNA ligase seals the gaps between Okazaki fragments.
Topoisomerase relieves the supercoiling that builds up ahead of the replication fork.

How did you score? 5/5 means your genetics and molecular basis are exam-ready. Anything less, and those are the cards to review this week.

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