Molecular genetics and gene expression sample question 1

5′-AUGCCAGGUUAA-3′; Met–Pro–Gly, then stop.

Molecular genetics and gene expression sample question 4

mRNA and protein of normal sequence are made at a much lower level, because the complex assembles poorly

Molecular genetics and gene expression sample question 5

translation initiation in bacteria depends on base pairing between the Shine–Dalgarno sequence of the mRNA and 16S rRNA.

MCAT · Biological and Biochemical Foundations · Biology

Molecular genetics and gene expression

Molecular genetics and gene expression is part of Biology in the MCAT Biological and Biochemical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include operon logic and mutant strains, mutation to protein outcome, rNA processing and transcript reading and transcription factor and reporter assay.

Common traps: treating operator mutations as trans-acting (they are cis-only); missing catabolite repression by glucose when lactose is present; reversing which operon type is repressed by its product.

Sample questions

Question 1
The template strand of a short gene segment reads 3′-TACGGTCCAATT-5′. Which choice gives the mRNA transcribed from it (written 5′ to 3′) and the amino acids it specifies, up to any stop codon?
- A.
  5′-AUGCCAGGUUAA-3′; Met–Pro–Gly, then stop.
- B.
  5′-UACGGUCCAAUU-3′; Tyr–Gly–Pro–Ile.
- C.
  5′-AAUUGGACCGUA-3′; Asn–Trp–Thr–Val.
- D.
  5′-UUAACCUGGCAU-3′; Leu–Thr–Trp–His.
Answer: A. 5′-AUGCCAGGUUAA-3′; Met–Pro–Gly, then stop.
RNA is synthesized 5′→3′ and is complementary and antiparallel to the template, giving 5′-AUG CCA GGU UAA-3′, Met–Pro–Gly and stop. B copies the template with U for T, C writes the complement but reads it backwards, and D reads the template itself as if it were the message.
Question 2
An in vitro bacterial replication system contains an origin-bearing template, helicase, primase, DNA polymerase III, DNA polymerase I and ATP, but one component is missing. The leading strand is complete, and the lagging strand is made as fragments of about 1000 nucleotides that have lost their RNA primers but are not joined to one another. Which component is missing?
- A.
  Helicase
- B.
  Primase
- C.
  DNA polymerase I
- D.
  DNA ligase
Answer: D. DNA ligase
D: primers were removed and replaced by DNA, so only the final nick sealing is missing, which is ligase's job. Without A the fork would not open at all, without B no fragments would be primed, and without C the primers would remain on the fragments.
Question 3
In a qPCR experiment, the target gene has Ct = 27 in control cells and Ct = 24 in treated cells. The reference gene has Ct = 20 in control cells and Ct = 21 in treated cells. Assuming 100% efficiency, what is the expression of the target in treated cells relative to control cells, normalized to the reference?
- A.
  1/16
- B.
  1/4
- C.
  4
- D.
  16
Answer: D. 16
ΔCt (control) = 27 − 20 = 7 and ΔCt (treated) = 24 − 21 = 3, so ΔΔCt = −4 and the fold change is 2⁴ = 16 (D). 1/16 reverses the direction (A); 1/4 reverses direction and treats ΔΔCt as linear (B); 4 uses ΔΔCt linearly rather than as an exponent of 2 (C).
Question 4
A point mutation destroys the TATA box of gene Z, about 25 base pairs upstream of the transcription start site; the coding sequence is unchanged. What is the expected result?
- A.
  A truncated protein is made, because transcription starts inside the coding sequence
- B.
  A protein with one amino acid replaced is made at the normal level, because the TATA box lies in the coding region
- C.
  mRNA and protein of normal sequence are made at a much lower level, because the complex assembles poorly
- D.
  Normal levels of mRNA are made, but no protein, because ribosomes cannot bind without the TATA box
Answer: C. mRNA and protein of normal sequence are made at a much lower level, because the complex assembles poorly
The TATA box binds the TATA-binding protein, which helps the polymerase II complex assemble at the promoter, so its loss mainly lowers initiation, not the sequence of the product. A and B treat the promoter as coding sequence; D confuses a DNA promoter element with the ribosome-binding step of translation.
Question 5
In E. coli, changing the Shine–Dalgarno sequence of one mRNA from 5′-AGGAGG-3′ to 5′-ACCACC-3′ reduces its translation tenfold. A mutation in the 16S rRNA gene that changes the anti–Shine–Dalgarno sequence from 3′-UCCUCC-5′ to 3′-UGGUGG-5′ restores translation of this mRNA but impairs translation of most other mRNAs. The best conclusion is that
- A.
  translation initiation in bacteria depends on base pairing between the Shine–Dalgarno sequence of the mRNA and 16S rRNA.
- B.
  bacterial ribosomes find the start codon by scanning from the 5′ cap, as eukaryotic ribosomes do from the 5′ end of the mRNA.
- C.
  the Shine–Dalgarno sequence is the start codon, which is recognized by the initiator tRNA.
- D.
  the 16S rRNA mutation repairs the DNA mutation in the gene so that the original mRNA is made.
Answer: A. translation initiation in bacteria depends on base pairing between the Shine–Dalgarno sequence of the mRNA and 16S rRNA.
Restoring complementarity (ACCACC pairs with UGGUGG) rescues initiation of that one mRNA, while the new anti-SD sequence no longer pairs with the usual AGGAGG-type sites, which explains the loss for other mRNAs; this identifies SD–16S pairing as the initiation signal. Bacterial mRNAs have no cap or scanning (B), the start codon is AUG (C), and rRNA mutations do not alter DNA (D).

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