Unit 6 · Topic 6.3 end-of-topic test
Suggested time: about 46 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.
In a eukaryotic cell, one gene's newly made RNA is marked so that it can be followed. The table shows where the marked RNA is over the next thirty minutes.
Which statement explains the pattern in the table?
An mRNA is drawn with its 3′ end at the left, as shown. A ribosome is about to read it.
At which end of the drawn mRNA does the ribosome start?
The codon 5′-GAU-3′ is drawn.
Which anticodon pairs with this codon?
The table describes four RNAs found in one eukaryotic cell.
Which row describes ribosomal RNA (rRNA)?
An RNA binds a small molecule only when it is folded. In the folded RNA, nucleotides 12–16 lie against nucleotides 31–35, the two stretches lying the opposite way round. A scientist made two changed versions of the RNA and measured the share of molecules that bound the small molecule. The table shows the results.
Which conclusion do the results support?
Genes P and Q sit side by side on one chromosome. A short stretch of DNA is deleted just before gene P. Afterward, RNA polymerase makes no RNA from gene P and still makes RNA from gene Q.
Which stretch was deleted?
Four test tubes each hold a gene with its promoter and the enzyme and nucleotides listed in the table. The last column shows whether new RNA appeared.
Which statement do the four tubes support?
The two DNA strands of a short stretch of a gene, X and Y, and the RNA made from that stretch are shown.
Which strand did RNA polymerase read as the template, and how do the sequences show it?
RNA polymerase is part-way along a gene, with a growing RNA behind it.
Which statement describes the ends at that moment?
A template strand reads 3′-ACG TTA GCC-5′, as drawn.
Written with both ends marked, what does the RNA read?
One gene is drawn with five RNA polymerases along it at once. Each trails the RNA it has made so far.
At which end of the gene is the promoter?
A gene's non-template strand reads 5′-GCTACA-3′, as drawn.
Written with both ends marked, what does the RNA read?
In a eukaryotic cell, enzymes in the nucleus add two things to a pre-mRNA before it leaves.
Which of the following do the enzymes join to the pre-mRNA's 3′ end?
Four copies of one gene's mRNA are made in a test tube. Each copy has or lacks a 5′ cap, and each copy has or lacks a poly-A tail. Each copy is put into a eukaryotic cell. The table shows how much protein each cell made in the first ten minutes and how long each mRNA lasted.
Which row is the mRNA with a 5′ cap and no poly-A tail?
In a eukaryotic cell, a gene's pre-mRNA is 3,200 nucleotides long. Its mature mRNA has a coding length of 3,200 nucleotides.
Which statement about this gene is correct?
The drawing shows a eukaryotic gene with RNA polymerase on it, the RNA as released, and the mature mRNA. Four parts are lettered.
Which lettered part is a stretch that enzymes cut out before the mRNA leaves the nucleus?
The table describes four mRNAs.
Which mRNA was made by a bacterium?
(a) Identify the RNA that RNA polymerase makes from the nine template bases drawn, written with both ends marked. (1 point)
A full-credit answer: Under each template base its RNA partner: G gives C, T gives A, C gives G, C gives G, A gives U, A gives U, T gives A, G gives C, C gives G.
The RNA's 5′ end sits under the template's 3′ end.
So the RNA reads 5′-CAG GUU ACG-3′.
Check the box for each point your answer earns
Accept the same nine letters without the spaces.
Common slip: Writing 5′-GUC CAA UGC-3′: the template's own letters with U for T, not their partners.
(b) Calculate the coding length of the liver cell's mature mRNA. (1 point)
Write down the values on the map:
Coding length, in nucleotides, = sum of the exon lengths:
A full-credit answer: The liver cell kept every exon, so its coding length is the exon total.
270 + 195 + 495 = 960 nucleotides.
(c) Determine which exon the kidney cell's enzymes skipped. Justify your answer with the lengths. (1 point)
A full-credit answer: The kidney cell's mature mRNA is 765 nucleotides, 195 short of the full coding length of 960.
Exon 2 is 195 nucleotides long, and no other exon is.
So the kidney cell skipped exon 2.
Check the box for each point your answer earns
Accept consistent working from an incorrect (b) total: the exon whose length equals the student's own (b) total minus 765, with that subtraction shown.
Common slip: Subtracting an intron length. Introns are cut out in both cells; only exon lengths decide the coding length.
(d) Explain why the two cells' pre-mRNAs are the same length although the two mature mRNAs differ. (1 point)
A full-credit answer: RNA polymerase copies the whole gene, exons and introns, into the pre-mRNA in both cells.
So both pre-mRNAs are the full length of the gene, 2,250 nucleotides.
The two mRNAs differ only afterward, when the kidney cell's enzymes cut out exon 2 along with the introns.
Check the box for each point your answer earns
Common slip: Saying the kidney cell's gene is shorter. The DNA is the same in both cells; splicing, not the gene, differs.
(a) Describe how the leaf cell's enzymes treat the pre-mRNA of this gene differently in the light and in the dark. (1 point)
A full-credit answer: In the light the enzymes cut out only the introns and join all four exons.
In the dark the enzymes cut out exon 3 along with the introns beside it and join exons 1, 2 and 4.
Check the box for each point your answer earns
(b) Explain why the protein made in the dark is related to the protein made in the light but differs from it. (1 point)
A full-credit answer: Ribosomes read each mature mRNA into a protein.
Both mRNAs carry exons 1, 2 and 4, so both proteins share the parts made from those exons.
The dark mRNA lacks exon 3, so the dark protein lacks the part made from exon 3.
Check the box for each point your answer earns
Common slip: Saying the dark protein is unrelated. Three of the four exons are shared, so most of the protein is the same.
(c) Predict what a comparison of this gene's DNA in a leaf cell in the light and a leaf cell in the dark would show. (1 point)
A full-credit answer: The gene's DNA is the same in both cells: the same exons and the same introns, in the same order.
Check the box for each point your answer earns
Common slip: Predicting that exon 3 is missing from the gene in the dark. The exon is missing from the mRNA, not from the DNA.
(d) Justify your prediction. (1 point)
A full-credit answer: Splicing changes which stretches of the RNA copy are kept.
Enzymes cut exon 3 out of the pre-mRNA, not out of the gene.
So the gene's DNA keeps exon 3 in the dark as in the light, and the same gene gives two different mRNAs.
Check the box for each point your answer earns