Name:______________________________
Exam 2 • Biology II • Spring 2013
Notice that on the last page of this exam, you will find a genetic code table and some
potentially useful sequences. You can tear it off to make it more useful. This page may
also be used as scratch paper if you like.
Multiple Choice Questions. Circle the one best answer for each question unless otherwise noted. (2
points each)
1. Many people think cystic fibrosis (CF) results from inheriting a CF gene. This is a misunderstanding
because:
A. CF actually results from inheriting a non-functional protein.
B. CF is not a genetic disease: it results from insufficient secretion.
C. CF results from the inheritance of a specific allele.
D. CF results from a mutation, not from the inheritance of a gene.
2. The best illustration of Mendel’s Law of Segregation is:
A. An individual with genotype Aa produces ½ A gametes and ½ a gametes.
B. Mating of two Aa individuals gives a ¼ chance of an aa offspring.
C. An individual with genotype AaBb produces AB, Ab, aB, and ab gametes.
D. An Aa individual would show the phenotype of the dominant A allele.
E. An individual with two co-dominant alleles (e.g., AL and AM) shows both phenotypes.
3. The best illustration of Mendel’s Law of Independent Assortment is:
A. An individual with genotype Aa produces ½ A gametes and ½ a gametes.
B. Mating of two Aa individuals gives a ¼ chance of an aa offspring.
C. An individual with genotype AaBb produces AB, Ab, aB, and ab gametes.
D. An Aa individual would show the phenotype of the dominant A allele.
E. An individual with two co-dominant alleles (e.g., AL and AM) shows both phenotypes.
4. An enzyme that would not be found in the nucleus of a eukaryotic cell is:
A. DNA ligase
B. RNA polymerase II
C. RNA polymerase I
D. topoisomerase (also known as DNA gyrase)
E. none of the above: all of these would be found in the eukaryotic nucleus.
5. A dominant allele is:
A. …the most common allele of a particular gene.
B. …the normal allele.
C. …the allele that determines the phenotype of a heterozygote.
D. …the allele that encodes the non-functional form of a protein.
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6. In order to start translation of an mRNA, eukaryotic Ribosome looks for:
A. the -10 and -35 sequences
B. the Shine-Dalgarno sequence
C. the TATA box
D. transcription factors
E. the start codon
7. If we sequenced the DNA from cells in many different tissues of the human body, we would find that:
A. The CFTR gene is found only in epithelial cells of the lung, pancreas and other secretory tissues.
B. Every cell has one copy of the CFTR gene.
C. Every cell has two copies of the CFTR gene.
D. The CFTR gene is in the nucleus of most cells but in the cytoplasm of lung and pancreas cells.
8. Which of the following is capable of breaking hydrogen bonds between two DNA strands? Choose all
that apply.
A. DNA polymerase III
B. helicase
C. RNA polymerase
D. primase
E. initiator protein
9. For a particular gene, which of these would be the longest molecule?
A. mRNA in the cytoplasm
B. mRNA in the nucleus
C. coding sequence
D. exon
10. Which of the following does not occur during RNA processing in eukaryotes?
A. addition of a poly(A) tail
B. removal of introns
C. removal of RNA primers
D. addition of a G nucleotide to the 5′ phosphate
11. Which investigators showed that one gene encodes one enzyme?
A. Watson and Crick
B. Beadle and Tatum
C. Hershey and Chase
D. Meselson and Stahl
12. Avery, McCarty and MacLeod’s experiments provided evidence that DNA is the genetic material by:
A. …showing that Streptococcus pneumoniae was not transformed by DNAse treated cell extract.
B. …showing that heat-treated S cells could still transform R cells.
C. …showing that radioactive phosphorus did not enter virus-infected bacteria.
D. …showing that radioactive sulfur did not enter virus-infected bacteria.
E. …showing that DNA replication is semiconservative.
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13. Name the regulatory sequence that performs each of the following functions. One answer per blank.
(2 points each)
Site where replication begins.
Origin of Replication
This sequence is bound by the protein TFIID.
Sequence bound by a transcription factor to increase
transcription from a nearby promoter.
Sequence which causes a eukaryotic ribosome to stop
sliding along an mRNA.
Site where RNA Polymerase releases its product and
the DNA template.
Sequence bound by the small subunit of prokaryotic
ribosomes.
14. Bob and his wife Jane like chocolate. Bob’s mother is a
real chocoholic: she loves, loves, loves chocolate (and has
no real interest in seeking treatment). Bob’s father, though,
never cared for chocolate. Bob’s first child (a son) and
second child (a daughter) like chocolate, and son #3 is a
chocoholic like his grandmother, but to Bob’s surprise,
daughter #4 doesn’t like chocolate.
a. Assuming taste for chocolate is a genetic trait, draw a
pedigree for this family in the space at right (3 points).
TATA box
Enhancer
Start codon or stop codon
Terminator
Shine-Dalgarno Sequence
CLCL
CDCD
= chocoholic
I
1
= likes chocolate
2
CLCD
= dislikes
CLCD
II
1
2
III
1
2
CLCD CLCD
3
4
CLCL CDCD
b. Determine dominance and give specific evidence to
support your claim (2 points).
There are three phenotypes; if there is only one gene, then there must be incomplete dominance or
co-dominance. Incomplete dominance makes more sense because we have a single phenotype
(likes) that is between the extremes (dislikes and chocoholic). We know that Tom got a chocoholic
allele from his mother, but he merely likes chocolate, suggesting that he’s heterozygous. Then,
homozygous for one allele could produce chocoholism, while being homozygous for the other
allele apparently results in disliking chocolate.
c. Define appropriate genetic symbols (2 points).
CL = chocoholism; CD = dislikes chocolate
d. Give everyone in the pedigree a genotype (2 points).
15. In the appropriate boxes in the replication fork diagram at
right:
a. Label the leading and lagging strands (2 points).
b. Label the indicated end (3′ or 5′) (1 point).
c. Name the indicated enzyme (1 point).
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16. You are studying six human genes, which we can symbolize simply with letters A-E.
a. If a female has the genotype AABbccDdEE, what are the gametes that she can make? (2 points)
ABcDE, AbcDE, ABcdE, AbcdE
b. Suppose she marries a male who has hairy feet, a trait resulting from the d allele, and is
heterozygous for the other five genes. What is the probability that they will have a child with hairy
feet? (2 points)
The male’s genotype is AaBbCcddEe, so the part of the cross we care about is Dd ×dd and the
probability of getting a dd child is 50% (1/2 chance of getting d from Mom, always gets d from
Dad).
17. The DNA sequence below is a very tiny eukaryotic gene. This entire region is transcribed using the
promoter located on the left. The bottom strand is the template strand.
Promoter
5ACGCAGGACGCCATGCCCGGAACTGAAATGTAAGACTACTAG3
3TGCGTCCTGCGGTACGGGCCTTGACTTTACATTCTGATGATC5
a. Label each end of the DNA molecule as 5′ or 3′. (1 point)
b. What is the primary sequence of the protein produced by this gene? Be sure to label your ends.
(3 points)
Met-Pro-Gly-Thr-Glu-Met
c. What is the sequence of the anticodon of the tRNA that was used to translate the final amino acid?
(2 points)
3-UAC-5 or CAU
d. A “G” nucleotide in the DNA sequence is underlined. Give an example of a missense mutation that
could occur at this nucleotide and how the protein sequence would change. (2 points)
If you change G to an A, you would go from a Gly codon to an Arg codon.
e. Could a nonsense mutation occur at this nucleotide? Please explain. (2 points)
Yes. If you change the G to a T, then we would go from a Gly to a stop codon.
f. Could a silent mutation occur at this nucleotide? Please explain. (2 points)
No. Any substitution at this position will lead to a different amino acid codon.
g. What is the nucleotide sequence of the 3′ UTR? (2 points)
I accepted GACTACTAG and GACUACUAG
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18. The sequence below is the first part of an mRNA from a bacterial cell, starting with the +1 nucleotide:
UGAGCAUCCCAUGGACCAUGAUAGGAGGUCCACAUGGUCCA
a. There are three potential start codons. Circle the one that will actually be used and tell how you
know that this is the correct one. (2 points)
The AUG in blue above is the correct one; this is a bacterial cell, and this is the only AUG
preceded by a Shine-Dalgarno sequence (AGGAGG, underlined above).
b. Suppose this gene were introduced into the nucleus of a eukaryotic cell and correctly transcribed
to produce this same mRNA. Would the same protein be produced? Why or why not? (2 points)
No, because eukaryotes start translation with the first AUG.
19. The bright red color of Chippy the Cardinal’s feathers is controlled by a gene, R, that has three alleles:
RR produces red pigment, RB produces blue pigment, and r produces no pigment (leading to white
feathers). Chippy’s mother was a blue cardinal, and Chippy’s father was the red son of a white
mother.
a. What is Chippy’s genotype? (1 point)
RRr: the red and blue alleles are incompletely dominant or co-dominant, so Chippy’s mother can’t
have an RR allele or she would’ve shown it. Chippy must’ve gotten his RR from his father. But, he
can’t have gotten RB from his mother or he would’ve shown it phenotypically, so he must have
gotten r from his mother.
b. If Chippy has 19 brothers and sisters, what would you predict their genotypes and phenotypes
would be? (4 points)
Chippy’s mother must then be RBr, and his father (who has a white mother) must be RRr. So:
¼ of offspring should be RRr and red (5 of the 20 offspring: Chippy and 4 siblings)
¼ of offspring should be RBr and blue (5)
¼ of offspring should be rr and white (5)
¼ of offspring should be RRRB and purple (5) (could be spotted or striped if co-dominant)
20. For the three RNA polymerases listed below, describe what type of RNA is produced. (1 point each)
Homo sapiens RNA Polymerase III: tRNA
Escherichia coli RNA Polymerase: mRNA, tRNA and rRNA (prokaryotes have only one RNAP)
Yeast RNA Polymerase I: rRNA
21. Gene expression for any given gene in Prokaryotes must eventually be turned off. Describe in detail
how bacteria can turn off gene expression by terminating (you may use illustrations to support
answers):
Transcription (3 points):
Past the end of the coding sequence, in the 3′ UTR, there is a sequence with two segments that when
transcribed into single-stranded mRNA can base-pair and fold up into a stem-loop structure. The
formation of this structure causes the RNAP to pause. Next to this sequence is a run of A’s on the
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template strand, which will pair with U’s in the mRNA, the weakest base-pair. When the RNAP
pauses, these weak base pairs can come apart, terminating the mRNA.
Translation (3 points):
When the ribosome reaches a stop codon, there is no matching tRNA, so the ribosome pauses. A
release factor recognizes the paused RNA and fits into the empty “A” site, stimulating release of the
mRNA and dissociation of the ribosome.
22. It may surprise you to know that your dislike of Brussel sprouts has a biochemical basis! The gene for
the enzyme glucosinolate isomerase (GluIso) has 10 exons that make up the CDS for this protein. If
all 10 exons (especially exons 4 and 5) are present, the protein is localized to the cell membrane and
highly functional: it makes Brussel sprouts taste less bitter. You have isolated molecules from your
own cells to figure out why your GluIso doesn’t appear to work properly, and you have found that
stable GluIso mRNA is made and translated into protein. However, your experiments revealed that
most of the GluIso protein is not associated with the cell membrane but is secreted into the
extracellular environment.
a. What is one modification of the GluIso mRNA that stabilizes it? (1 point)
The best answer is the addition of the 3′ poly(A) tail (stabilizes mRNA because non-coding A
nucleotides can be removed by RNAses without damaging the coding sequence). The 5′ cap also
has some stabilizing effect.
b. Where in the cell would you expect the GluIso protein to be translated? (1 point)
This protein gets inserted into the membrane; membrane proteins are translated by ribosomes
attached to the rough ER so that the protein moves through the ER and Golgi to transport vesicles
and then to the membrane.
c. Briefly describe a mechanism involving RNA processing that could have led to the phenotype
described for your mutant allele (you may use illustrations to support your answer). (3 points)
Exons 4 and 5 are important for localizing the protein to the membrane. Possibly the mutant allele
has a mutation that occurs at the intron 3-exon 4, exon 4-intron 4 or intron 4-exon 5 boundary and
affects splicing such that exon 4 or 5 is spliced out of the mRNA.
23. An albino male (Frank) marries a woman (Marie) with normal pigmentation. Frank is blood type A,
and Marie is blood type O. Frank is also carrier for the sickle cell allele, whereas Marie has sickle cell
anemia. They have 4 kids, where 2 are albino, blood type O, and carriers for sickle cell anemia and
the other two are normal pigmentation, blood type A, and have sickle cell anemia.
a. What are Frank and Marie’s genotypes? (2 points) Gene symbols: Normal pigment (A), albino (a);
blood type A (IA), blood type O (i); Normal hemoglobin (HA), sickle cell allele (HS).
Frank genotype: aa IAiHAHS
Marie genotype: AaiiHSHS
You can figure these out by writing out the recessive alleles first where they are given (albino, blood
type O, and sickle cell). You can also use the same strategy with the kids genotypes. With ratios of
½, it’s pretty clear that you have a heterozygote-homozygote recessive cross, so the genes that aren’t
recessive must be heterozygous.
b. What is the probability that Frank and Marie’s next child will have normal pigmentation, blood
type O , and sickle cell anemia?
Now that we know Frank and Marie’s genotypes, we can calculate the probability for each trait
and use the multiplication rule here!
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For pigmentation: aa x Aa = ½ Normal pigmentation; ½ albino
Blood type:
IAi x ii = ½ Blood type A; ½ Blood type O
A S
Sickle Cell: H H x HSHS = ½ normal Hb; ½ sickle cell.
So the probability of their next child having normal pigmentation, blood type O, and sickle cell
anemia:
P= ½ x ½ x ½ = 1/8. One out of eight offspring will likely have all three of those phenotypes.
c. Skip this. I gave you credit for a poorly written question 
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