Gene Expression and Regulation: Unit Diagnostic (Biology Unit 6)

What this diagnostic is for

Use this page to find which Unit 6 topics to revisit. There is one question per topic, and two for the larger Topics 6.4, 6.5 and 6.7. Each answer names the study guide to read if you got it wrong.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your total does not predict an exam score.

Calculator and data. A calculator is allowed on both sections of the Biology exam. Here a basic one is enough, and every number and codon you need is given. All organisms, genes and data are fictional unless a real system is named.

How to work. Answer every question, with a short reason, before opening any answer. A right answer for the wrong reason is still a gap.

Question 1 (multiple choice · 6.1)

The genome of a fictional virus is double-stranded RNA. Adenine makes up 22% of its bases. What percentage of its bases is cytosine?

(A) 22%
(B) 28%
(C) 56%
(D) It cannot be found, because RNA has uracil, not thymine.

Answer and explanation

Answer: (B). In a double strand, A pairs with U and G with C. So U = 22%, A + U = 44%, and the other 56% is shared equally: G = C = 28%.

(A) pairs cytosine with adenine; A pairs with U.
(C) is G + C together, not cytosine alone.
(D) forgets that uracil simply takes thymine’s place in RNA.

If you missed this: “Purines and pyrimidines” in study guide 6.1.

Question 2 (multiple choice · 6.2)

Researchers mix purified DNA polymerase, a single-stranded DNA template and all four DNA nucleotides. No new DNA is made. When they add a short RNA strand that pairs with the template near its 3′ end, a new DNA strand grows from it. Which statement best explains this?

(A) DNA polymerase can only add nucleotides to the 3′ end of an existing strand.
(B) The RNA strand unwinds the template for DNA polymerase.
(C) DNA polymerase builds DNA 3′ → 5′, starting from the 5′ end of the RNA.
(D) The RNA strand joins DNA fragments, as ligase does.

Answer and explanation

Answer: (A). DNA polymerase cannot start a strand from nothing; it can only add nucleotides to the 3′ end of an existing one. In cells, RNA primers give it that start.

(B) gives the primer helicase’s job, and the template is already single-stranded.
(C) reverses the direction. New DNA is always built 5′ → 3′.
(D) describes ligase, which joins fragments.

If you missed this: “The direction rule: 5′ → 3′ only” and “At the replication fork” in study guide 6.2.

Question 3 (multiple choice · 6.3)

The transcribed region of a fictional human gene is 4,200 base pairs long. Its mature mRNA, not counting the cap and the poly-A tail, is only 1,500 nucleotides long. Which statement best explains the difference?

(A) RNA polymerase copies only the exons and skips the introns.
(B) The mRNA copies one strand only, so it is shorter than the double-stranded gene.
(C) About 2,700 nucleotides of introns were cut out and the exons joined.
(D) The poly-A tail replaces part of the transcript at its 3′ end.

Answer and explanation

Answer: (C). RNA polymerase copies the whole region, introns included, into pre-mRNA. In the nucleus the introns are removed and the exons joined: 4,200 − 1,500 = 2,700 nucleotides of intron.

(A) is a common mix-up: introns are transcribed, then removed.
(B) confuses strands with length: one strand of the region is still 4,200 nucleotides.
(D) is wrong: the tail is added, replacing nothing.

If you missed this: “RNA processing in eukaryotes” in study guide 6.3.

Question 4 (multiple choice · 6.4)

In a test-tube translation system, researchers change the amino acid on one kind of tRNA. Its anticodon, 3′-CGA-5′, still pairs with the codon GCU (alanine), but it now carries cysteine. They translate an mRNA with several GCU codons. What do they find in the new polypeptides?

(A) Alanine at every GCU codon, because the ribosome rejects the wrong amino acid
(B) Short polypeptides, because translation stops at each GCU codon
(C) Extra cysteine at every cysteine codon, and nothing at GCU codons
(D) Cysteine wherever the mRNA has a GCU codon

Answer and explanation

Answer: (D). The ribosome matches codon with anticodon; it does not check the amino acid. Whatever the tRNA carries is added wherever its anticodon pairs.

(A) is tempting, but accuracy in cells depends on each tRNA carrying the right amino acid.
(B) treats GCU as a stop codon. Only stop codons end translation.
(C) assumes the amino acid, not the anticodon, decides where the tRNA goes.

If you missed this: “The players” and “Three stages of translation” in study guide 6.4.

Question 5 (multiple choice · 6.5)

A mutant E. coli strain makes a trp repressor that cannot bind tryptophan but is otherwise normal. The cells grow in tryptophan-rich medium. What happens to the trp operon?

(A) It is transcribed, because the repressor stays inactive and does not bind the operator.
(B) It is switched off, because tryptophan binds the operator directly.
(C) It is switched off for good, because the repressor stays bound to the operator.
(D) It is switched off, because tryptophan blocks the promoter.

Answer and explanation

Answer: (A). In this repressible system, only repressor with tryptophan bound attaches to the operator. This repressor never becomes active, so the trp genes stay on, wasting energy.

(B) and (D) give tryptophan a job it lacks; it works only through the repressor.
(C) mixes up the trp and lac systems. The trp repressor is inactive on its own.

If you missed this: “The trp operon: a repressible system” in study guide 6.5.

Question 6 (multiple choice · 6.5)

Gene H is silent in liver cells of a fictional mouse but active in its kidney cells. Which finding would best show that the silencing in liver is epigenetic?

(A) In liver, gene H has a base substitution in its promoter.
(B) Gene H has the same sequence in both cell types, its promoter is heavily methylated only in liver, and liver cells express it after a drug removes the methyl groups.
(C) Liver cells have lost gene H from their genome.
(D) The H protein in liver has a different amino acid sequence.

Answer and explanation

Answer: (B). An epigenetic change alters expression without changing the base sequence, and it is reversible. Option B shows both.

(A) and (C) change the DNA itself.
(D) is impossible: a silent gene makes no protein.

If you missed this: “Epigenetic regulation” in study guide 6.5.

Question 7 (multiple choice · 6.6)

In a fictional frog, transcription factor M binds enhancers of 40 genes that muscle cells need. A mutation stops M binding DNA. What is the most likely result?

(A) Only one of the 40 genes is affected.
(B) Every cell stops transcribing all its genes.
(C) The 40 proteins are made with changed amino acid sequences.
(D) Muscle cells transcribe much less of many of the 40 genes and do not specialise fully, though those genes are unchanged.

Answer and explanation

Answer: (D). Activators bound at enhancers raise transcription. One factor can act on many genes, so losing it lowers expression of a whole set, and the cell’s features depend on that set.

(A) and (B) misjudge the reach of one factor: it controls many genes, but not every gene.
(C) confuses regulation with coding. The coding sequences are unchanged; only the amount changes.

If you missed this: “Enhancers and activators” and “From regulation to phenotype” in study guide 6.6.

Question 8 (multiple choice · 6.7)

In one family of a fictional plant, a segment carrying about 40 genes is missing from one copy of chromosome 5. In another family, one of those genes has a single base substitution. Which statement is best supported?

(A) The substitution has the larger effect, because single-base changes are always harmful.
(B) Neither change can affect the phenotype, because a second chromosome 5 is present.
(C) The deletion is more likely to change the phenotype in several ways, because it removes one copy of many genes.
(D) The deletion causes a frameshift in all 40 genes.

Answer and explanation

Answer: (C). A change in chromosome structure alters the amount of product from many genes at once, so it often leads to disorders.

(A) is wrong twice: a substitution can be silent, and no mutation is always harmful.
(B) assumes one copy is always enough; halving many products can matter.
(D) misuses “frameshift”: the genes are missing, not misread.

If you missed this: “Changes in chromosome structure” in study guide 6.7.

Question 9 (multiple choice · 6.7)

A fictional plant has 2n = 14. In one flower, no chromosomes separate in meiosis, so a gamete gets a full diploid set. It fuses with a normal gamete. How many chromosomes are in each offspring cell?

(A) 14
(B) 15
(C) 21
(D) 28

Answer and explanation

Answer: (C). The diploid gamete brings 14 and the normal gamete 7: 21 chromosomes, three full sets, so the offspring is triploid.

(A) is the normal number.
(B) is a trisomy: one extra chromosome, not an extra set.
(D) treats both gametes as diploid.

If you missed this: “Changes in chromosome number” in study guide 6.7.

Question 10 (short answer · 6.4)

Part of a fictional mature mRNA is 5′-CUAUGGCUUGGAAACAUUAGCC-3′.

Codons you need: AUG Met (start) · CUA Leu · GCU Ala · UGG Trp · AAA Lys · AAG Lys · CAU His · UAG stop

(a) Write the polypeptide it encodes, and state where translation starts. (b) Give the anticodon of the tRNA that reads the third codon translated, written 3′ → 5′. (c) An error changes the fourth codon from AAA to AAG. Predict and explain the effect.

Answer and explanation

(a) Translation starts at the first AUG (nucleotides 3–5), not at CUA. Reading in triplets: AUG GCU UGG AAA CAU UAG gives Met–Ala–Trp–Lys–His, then stop.

(b) The third codon is UGG; the antiparallel anticodon is 3′-ACC-5′.

(c) No change. AAA and AAG both code for lysine: the code is redundant.

If you missed this: Worked example 1 in study guide 6.4.

Question 11 (short answer · 6.8)

A fictional gene region is amplified by PCR. The normal allele gives a 620 bp product. A deletion allele lacks 150 bp between the primer sites.

(a) Calculate the deletion allele’s product size. (b) Predict the gel bands for people with two normal alleles, two deletion alleles, and one of each. Which band stays nearer the wells, and why? (c) A control lane with no DNA added shows a faint 620 bp band. Explain what this means.

Answer and explanation

(a) 620 − 150 = 470 bp.

(b) Two normal alleles: one band, 620 bp. Two deletion alleles: one band, 470 bp. One of each: both bands. The 620 bp band stays nearer the wells: all DNA moves toward the positive electrode, but longer fragments are held back more by the gel.

(c) A reagent was contaminated with this DNA, so sample lanes may hold extra DNA too. Repeat the test with fresh materials.

If you missed this: “Gel electrophoresis” and “PCR” in study guide 6.8.

Your next step

Mark each question, then use this table. A short answer missed in one part counts as missed.

Topic	Question(s)	If you missed it, read
6.1 DNA and RNA structure	1	Study guide 6.1
6.2 DNA replication	2	Study guide 6.2
6.3 Transcription and RNA processing	3	Study guide 6.3
6.4 Translation	4, 10	Study guide 6.4
6.5 Regulation of gene expression	5, 6	Study guide 6.5
6.6 Gene expression and cell specialization	7	Study guide 6.6
6.7 Mutations	8, 9	Study guide 6.7
6.8 Biotechnology	11	Study guide 6.8

How to use your result

Wrong answers first. Read the named section, then do that topic’s practice questions. A lucky guess counts as a miss.
Missed Questions 2 and 10(b)? Both depend on strand direction. Start with Topic 6.2, then Topic 6.4.
Missed Questions 5, 6 and 7? For each control step in Topic 6.5 and Topic 6.6, ask “which molecule binds which, and where?”
No misses? Go to the mixed unit review.