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Practice questions · Topic 6.3

Unit 6 · Practice for the Topic 6.3 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Transcription and RNA processing, summed up

Video coming soon

Three RNAs and their jobs; RNA polymerase at the promoter; one template strand, read 3′ to 5′, the RNA built 5′ to 3′; the transcript released and the gene read again; a cap and a tail; introns out, exons joined; alternative splicing; a prokaryote's message against a eukaryote's.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.
Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one gene one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it.
Question 1

A student says: “The gene for a protein is in the nucleus, so the ribosomes that build the protein must be inside the nucleus too.”

Which statement about the student's claim is correct?

Question 2
One monospace row: 3′-AAC-5′3′-AAC-5′
The anticodon, written from its 3′ end.

A strand is written from its 5′ end to its 3′ end, and both ends are marked. The anticodon is written from its 3′ end to its 5′ end, so each of its bases sits opposite its partner on the codon. A tRNA's anticodon reads 3′-AAC-5′, as drawn.

Which codon on an mRNA does this tRNA pair with?

Question 3

Suppose a drug jams the rRNA of a ribosome's large subunit and leaves the small subunit untouched.

Which of the following can the ribosome still do?

Question 4

RNA polymerase is about to copy a gene into RNA.

Which of the following does it need to begin?

Question 5
Four monospace rows with a note beside each: 5′-GACTTACGGA-3′, the upper strand; 5′-TCCGTAAGTC-3′, the lower strand; 5′-GACUUACGGA-3′, RNA from gene P; 5′-UCCGUAAGUC-3′, RNA from gene Q5′-GACTTACGGA-3′the upper DNA strand5′-TCCGTAAGTC-3′the lower DNA strand5′-GACUUACGGA-3′RNA from gene P5′-UCCGUAAGUC-3′RNA from gene Q
The two DNA strands of the shared stretch and the RNA of each gene, each written from its own 5′ end.

A strand is written from its 5′ end to its 3′ end, and both ends are marked. Two genes, P and Q, overlap on one stretch of a chromosome. The two DNA strands of the stretch and the RNA each gene gives from it are shown.

Which strand is the template for each gene?

Question 6

Two RNA polymerases are copying one gene at the same time. The first started 20 seconds before the second.

Which statement is correct?

Question 7
One monospace row: 3′-CGT ACA GGT-5′3′-CGT ACA GGT-5′
The template strand, written from its 3′ end.

A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A template strand reads 3′-CGT ACA GGT-5′, as drawn.

Written with both ends marked, what does the RNA read?

Question 8

A ribosome begins to read a eukaryotic mRNA.

Which statement names the end the ribosome recognizes first and what that end carries?

Question 9

A eukaryotic gene is transcribed into a pre-mRNA 3,600 nucleotides long. Its introns total 2,750 nucleotides.

Calculate the coding length of the mature mRNA.

Question 10

A liver cell's mature mRNA and a bacterium's mRNA are compared.

Which of the following is carried by the liver cell's mature mRNA and by no bacterium's mRNA?

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.
Free response 1 · Analyze Data · 4 points
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A eukaryotic gene is drawn as a gene map with every exon and intron length written on it, in nucleotides. Beneath it are the first nine bases of the template strand. A skin cell's mature mRNA from this gene keeps every exon. A gland cell's mature mRNA from the same gene has a coding length of 735 nucleotides.
A gene map with the exon lengths written above the filled boxes and the intron lengths written below the line between them, the promoter labeled; beneath it one monospace row of nine DNA bases with its ends marked 3′ at the left and 5′ at the right133059021507603405promoter3′-TGA CCG TTC-5′the first nine bases of the template strand
The gene map with its lengths, and the first nine bases of the template strand.

(a) Identify the RNA that RNA polymerase makes from the nine template bases drawn, written with both ends marked. (1 point)

Hint: Pair each template base with its RNA partner, U opposite A. The RNA's 5′ end sits under the template's 3′ end, at the left.

A full-credit answer: The RNA reads 5′-ACU GGC AAG-3′.

Check the box for each point your answer earns

Accept the same nine letters without the spaces.

(b) Calculate the coding length of the skin cell's mature mRNA. (1 point)

Hint: Add the exon lengths only. The introns are cut out.
nucleotides

Write down the values on the map:

exon lengths=330,150,405

Coding length, in nucleotides, = sum of the exon lengths:

330+150+405=885

A full-credit answer: The coding length is 885 nucleotides.

(c) Determine which exon the gland cell's enzymes skipped. Justify your answer with the lengths. (1 point)

Hint: Subtract the gland cell's coding length, 735, from your answer to (b). Find the exon whose length equals the difference.

A full-credit answer: The gland cell skipped exon 2, because 885 − 735 = 150 nucleotides are missing, and exon 2 is 150 nucleotides long.

Check the box for each point your answer earns

Accept a decision that follows correctly from the student's own answer to (b).

(d) Predict how the gland cell's protein differs from the skin cell's protein. (1 point)

Hint: Each exon carries the instructions for one part of the protein.

A full-credit answer: The gland cell's protein lacks the part made from exon 2 and shares the parts made from exons 1 and 3.

Check the box for each point your answer earns

Free-response score: 0 of 4
Free response 2 · Analyze Model · 4 points
The drawing shows a eukaryotic gene with RNA polymerase on it, the RNA as released, and the mature mRNA. Five parts are lettered N to R.
Three rows: a gene drawn as a line with an open box under a large oval, a bent arrow and three numbered filled boxes with long plain stretches between them; beneath it a light row of the same three boxes joined by long thin lines; beneath that a shorter light row of the three boxes touching, with a dot at one end and a row of small tabs at the other, ending well short of the row above. Five small circles carry the letters N, O, P, Q and R; N, O and Q each have a leader to a part, and P and R sit at the left margin of the second and third rows1231235′3′123AAAAAAA…A5′3′ONPQR
A gene, the RNA as released and the mature mRNA, with five parts lettered.

(a) Identify the part lettered N, and describe what it does at the part lettered O. (1 point)

A full-credit answer: N is RNA polymerase.
It binds O, the promoter, separates the two DNA strands at the start site and begins building RNA.

Check the box for each point your answer earns

(b) Explain what determines the length of the row lettered P. (1 point)

A full-credit answer: RNA polymerase copies the whole gene from the start site to the stop sequence, exons and introns alike.
So the length of P, the RNA it releases, is the length of the gene's exons and introns together.

Check the box for each point your answer earns

(c) Identify the part lettered Q, and state what it is made of. (1 point)

A full-credit answer: Q is the 5′ cap.
The cap is one modified guanine nucleotide.

Check the box for each point your answer earns

(d) Explain why the row lettered R is shorter than the row lettered P. (1 point)

A full-credit answer: P carries the introns between its exons.
Enzymes cut each intron out and join the exons end to end.
So R, the mature mRNA, carries the exons only and is shorter by the introns' length.

Check the box for each point your answer earns

The point is for the mechanism (the introns cut out and the exons joined), not for the bare word 'introns'.

Free-response score: 0 of 4
Multiple choice checked: 0 of 10 correct.
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