Download Biol 321 Winter 2011 Quiz 4 NAME

Biol 321 Winter 2011 Quiz 4
NAME ______________________________
1. ( 6 pts.) Here is a portion of the Homo sapiens SRY (sex determining region Y) gene
CAGGAGCGCGGAGCCGCGAGCCCCGAGCCCCGAGCCCGGCGCCTGGCTGAGTAGAT
GTCCATGAGGAGCCCCATCTCTÌGCCCAGCTGGCCCTGGATGGCGTTGGCACCATG
GTGAACTGCACCATCAAGTCAGAGGAGAAGAAAGAGCCTTGCCACGAGGCCCCCCA
GGGCTCAGCCACTGCCGCTGAACCTCAGCCTGGAGACCCAGCCCGGGCCTCCCAGG
ATAGTGCTGACCCCCAAGCTCCAGCCCAGGGGAATTTCAGGGGÌCTCCTGGGACTG
TAGCTCTCCAGAGGGTAATGGGTCCCCAGAACCCAAGAGACCAGGAGTGTCGGAGG
CTG………………………….. etc, etc, etc,
a. 4 pts ALL or NOTHING (since one correct and one incorrect primer in a PCR reaction will
give you zero amplification). You are planning to set up a PCR reaction to amplify the region
between the Ì’s. The PCR product should include all (but not extend beyond) the designated
region. List the first five bases of each primer. Your primer sequences must read in the 5’ to
3’ direction.
Primer A: GCCCA
Primer B: CCCCT
b. (2 pts.) What feature of DNA polymerase ensures that, among the vast sea of sequences in
genomic DNA, only the β globin sequences will be amplified? One-two sentence
answer/explanation. Be sure to indicate explicitly (or to draw a diagram showing) the
specific substrate requirement as it relates to this question. (In other words, what does
DNA pol “cradle” in its active site.
Student answer: DNA polymerase requires a primer which is properly base-paired a the 3’
end to the template --- Primer sequences flanking the sequence selectively amplify only the
primer sequence:
Primer: 5’ GCCCA3’
Template 3’ CGGGT-------------------------------------------- 5’
2. ( 4 pts.) By each statement circle True/False/subject Not addressed. Answer false if any part
of the statement is false.
Recall the article entitled Genome Dark Matter
T F N One recent eye-popping observation suggests that although less than 2% of the
human genome is protein-coding, a far larger fraction (>50%) of the genome is transcribed.
T F N People with high starch diets – such as in Japan – have extra copies of a gene
encoding a starch-digesting enzyme as compared with members of hunter-gatherer societies.
Recall your flight over chromosome 11 in the human genome
T F N Although some transposable (mobile) elements in the human genome are molecular
fossils, the majority are still actively moving around.
T F N Over half of the genes in the human genome that code for olfactory receptors are
defunct (inactive due to accumulation of mutations) suggesting that a sophisticated sense of
smell is not as critical to humans as it is to other mammals.
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3. (6 pts.) Circle True or False. If there are two statements, the first statement is true and you
are to decide whether the second statement is True or False. No credit if no explanation.
FALSE Most non-neutral mutations in the proto-oncogene class of genes will result in driver
alleles. One sentence explanation: Most non-neutral mutations will cause a loss-of-function
and in this class of genes would not produce driver alleles; a gain-of-function mutation is
required to produce a driver allele in a proto-oncogene
Student answers: There are many ways to mutate and mess up a gene or protein, so most
mutations probably result in loss-of-function – only a few way to make a gain-of-function
FALSE The tumor suppressor class of genes (often found mutated in cancer cells) are named
for their loss-of-function phenotype. One sentence explanation:
Loss-of-function mutations in this class of genes promote tumors so the name reflects the wildtype function.
FALSE Gain-of-function mutations in a gene named survivin are sometimes found in colon
cancers. From this info you can conclude that the wildtype function of this gene would be to
promote apoptosis (programmed cell death). One sentence explanation:
The wild-type function of this gene must be to inhibit entry into apoptosis (promote survival).
Student answers:
1. If the gene promoted apoptosis, a gain-of-function mutation would signal apoptosis
without upstream signals (not giving cancer)
2. If the wildtype fn is promoting apoptosis, then gaining more fn would result in cell death
not proliferation
4. (4 pts) As reported online in The New England Journal of Medicine (11/23/10), a recent
study recruited people at extremely high risk of becoming infected with HIV. Everyone received
regular counseling about how to reduce their risks of becoming infected. At the end of the trial,
36 out of 1251 people who also received a pill that contained a combination of two anti-HIV
drugs, tenofovir and emtricatbine, became infected. Of the 1248 people who received a placebo
pill, 64 became infected. The structure of tenofovir is shown below.
a. (2 pts all or nothing) On the drawing, clearly number the carbons.
b. (2pts.) Recall that HIV is a retrovirus; as part of its lifecyle, its RNA genome is reversetranscribed into a double-stranded DNA copy. The viral polymerase (reverse transcriptase)
recognizes tenofovir as a legitimate substrate (after it is modified to the triphosphate form) and it
is added to a growing DNA strand. Briefly explain why this compound serves as an anti-HIV
agent. One- two sentences using proper biochemical terminology.
The absence of a 3’ hydroxyl group at the
end of the nascent DNA strand means that
there is no way for a bond to be formed
with an incoming dNTP . o this compound
acts as a chain terminator
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5. (6 pts.) Civilian aircraft mass disasters represent a challenge to the medical-legal personnel
charged with the task of victim identification. In the TWA Flight 800 aircraft disaster, DNA
profiling was used as the principle method of DVI (disaster victim identification). Using this
method, a comparison is made between the DNA profile obtained from the human remains and
those obtained from close biological relatives such as parents or siblings.
The figure on pg 1 of the DATA SHEET shows an STR (microsatellite) fingerprinting profile
from a TWA Flight 800 family pedigree.
a. ( 2 pts.) What does the number by each peak indicate? One sentence.
Each peak represents a specific allele. The number indicates the number tandem repeats
for that particular allele
b. (2 pts) Based on the DNA profile data shown here, could the unidentified passenger be an
offspring of the mother and father? Circle YES
c. (2 pts.) The D3S1358 site is on chromosome 3; FGA on chromosome 4 and VWA is on
chromosome 12.
• If you chose YES above, calculate the probabilty that the parents would have produced a
child of this genotype.
• If you chose NO above, then explain why the probability is zero
¼ (16,16) X ½ (17,18) X ¼ (22,26) = 1/32
Page (pts)
1 (10)
2 (10)
3 ( 6)
4 (14)
Total (40)
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6. (14 pts.) See pg 2 of extra sheet. Both families shown have mutations in the GATA gene.
The affected individuals have similar symptoms including heart defects, immune deficiencies,
deafness and renal malformations. The data for genotyped individuals is shown directly below
their pedigree symbol. N1 and N2 represent normal control individuals. A codon table is on the
extra sheet.
a. (1 pt.) Examine panel a. At the protein level, what type of mutation is seen in this
family? One word. No explanation. Frameshift (49 bp deletion is not a multiple of 3)
b. (1 pt.) Answer the same question for family 12/99. Nonsense (Argà STOP)
Circle True/False/Not enough info to decide. 1 pt if no explanation is required. 2 pts if an
explanation is required. For the latter, no credit given if there is no explanation
TRUE The mutation in Family 12/99 could have been originally generated by environmental
exposure to a base analog mutagen such as 2 aminopurine.
TRUE In family 12/99, the mutant allele is codominant at the molecular level of assessment
but dominant (completely or incompletely) at the organismal level of assessment.
FALSE The GATA gene is likely to be X-linked.
One sentence defense of your answer: Affected males in both families carry two alleles (wt and
mutant) so the GATA gene cannot be located on the X chromosome
FALSE The mutant phenotype is likely to result from a gain-of-function in the GATA gene.
One sentence defense of your answer: Frameshift and nonsense mutations are much more
likely to cause a loss-of-function rather than a gain-of-function.
FALSE The mutant GATA allele is recessive in Family 26/99 and dominant in Family 12/99
One sentence defense of your answer: Affected individuals are het in all cases
TRUE Based on the data presented here, GATA mutations are completely penetrant.
TRUE OR N Setting niceties aside, if individual II #1 from family 26/99 mates with I #2 from
family 12/99, the probability of a normal offspring is 1/4.
FALSE If the parents in either family have additional children, there is a 50:50 probability of
an affected offspring. One sentence defense of your answer Affected son in panel a is the
result of new mutation and not the product of a het parent. The statement is true though for
family 12/99
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