Download Exam 2

Name _________________________
Biology 201 (Genetics)
Exam #2

Read the question carefully before answering. Think before you write. Be concise.

You will have up to one hour to take this exam. After that, you MUST stop no matter
where you are in the exam.

If I can not read your handwriting, I will count the question wrong.

Sign the honor pledge if applicable.

Good luck!
I pledge that I have neither given nor received unauthorized assistance during the completion
of this work.
Signature: __________________________________________________________
1) A cat with the hypothetical genotype GGHH is crossed to a cat that is gghh, and then the F1
progeny are testcrossed. What percent of the progeny of the test cross will be gghh if the two
genes are:
a) Unlinked GGHH x gghh  GgHh so test cross is GgHh x gghh
All of gametes from tester will be gh and ¼ of gametes from heterozygote will be gh
so ¼ of progeny will be gghh
b) completely linked GH x gh  GH so testcross is GH x gh
GH gh
gh
gh
gh
All of gametes from tester will be gh. Since genes are completely linked there is no crossover
and ½ of gametes from the heterozygotes will be gh
So ½ of progeny will be gghh
c) 10 map units apart GH x gh  GH so testcross is GH x gh
GH gh
gh
gh
gh
Since g and h are 10 map units apart, this means that crossover between G and H occurs 10%
of the time which means that 90% of the time there is no crossover. In this heterozygote, the
noncrossover gametes would be GH and gh, each occurring 45% of the time. So since this
was a test cross, the progeny would be gghh 45% of the time.
2) In tomatoes, the mutant gene f causes flattened fruit, the gene h causes hairy fruit, and the
gene l causes clustered flowers. A plant heterozygous for these three genes was test crossed
and the progeny are shown below.
Crossover type
Phenotype of test cross
Number
(see questions below)
progeny
SCO
Wild type
73
NCO
Clustered flowers
348
DCO
Hairy fruit
2
SCO
Hairy fruit, Clustered flowers
96
SCO
Flattened fruit
110
DCO
Flattened fruit, Clustered
2
flowers
NCO
Flattened fruit, Hairy fruit
306
SCO
Flattened fruit, Hairy fruit,
63
Clustered flowers
Total progeny = 1000
a) In the left column in the table above, indicate (as NCO) which progeny result from
noncrossover events in the heterozygote parent.
See above (because most frequent classes)
b) What is the arrangement of the alleles in the heterozygote parent?
Since clustered flowers (FHl/fhl) and flattened hairy fruits (fhL/fhl) were the two progeny
types that resulted from noncrossover events in the heterozygote parent, that means that
the heterozygote parent must have been FHl
fhL
c) In the left column in the table above, indicate (as DCO) which progeny result from
double crossover events in the heterozygote parent.
See above (because least frequent of the 8 expected classes)
d) What is the order of the genes?
One of the DCO was hairy (FhL/fhl). The only order of genes in the heterozygote that
would result in this progeny type as a DCO was:
HFl
hfL
e) In the left column in the table above, indicate (as SCO) which progeny result from
single crossover events in the heterozygote parent.
see above
f) What is the distance between the genes?
HF = (110+96+2+2)/1000*100 = 21% = 21 map units apart
FL = (73+63+2+2)/1000*100 = 14% = 14 map units apart
3) There are four main types of chromosomal mutations: deletions, duplications, inversions, and
translocations.
a) Which one type of mutation is the most detrimental to an individual and why is it the
most detrimental?
Deletions are the most detrimental mutation because there is a high likelihood that you
will unmask a lethal allele in the heterozygote or that you will have the loss of essential
genes in the homozygote. Although translocations and inversions may affect gamete
formation, they are usually not detrimental to the person with the mutation.
b) Which two mutations result in fertility reductions?
The best answers were translocations and inversions, as the question was addressing
mutations in part but not all of the chromosome. However, I did accept deletions or
duplications for the reasons listed below.
c) Considering your answer to b): There are different reasons for the reduction in fertility
for the two types of mutations. Briefly, explain the differences in 3 sentences or less.
Inversions: If and only if there is crossover during meiosis in a heterozygote with an
inversion, ½ of the chromosomes will have deletions or duplications. The progeny
resulting from gametes carrying these defective chromosomes most likely will not
develop.
Translocations: Homologous chromosomes in heterozygotes line up in cross formation
during meiosis. When adjacent segregation occurs (50% of the time), all chromosomes
will have deletions or duplications The progeny resulting from gametes carrying these
defective chromosomes most likely will not develop. Thus, the individual appears
semisterile because 50% of gametes carry defective chromosomes.
Deletions: XO individuals are sterile. Alternatively, deleted genes were required for
fertility.
Duplications: XXX and XXY individuals are sterile. Alternatively, genetic imbalance
affects expression of fertility genes (just saying imbalance was not correct unless you
specifically told me how this affected fertility) .
4) You have isolated a virus with both DNA and RNA in it. Briefly describe one experiment
that you would do to determine whether DNA or the RNA was the genetic material?
Answer #1: Selectively labeled the virus DNA with radioactive thymine (or deoxyribose) in
tube#1 and label the virus RNA with radioactive uracil (or ribose) in tube #2  Infected E.
coli with labeled virus  After infection, removed empty virus heads  Looked to see whether
the labeled DNA or the labeled RNA was in E. coli  Assuming only the genetic material was
injected, if you find radioactive thymine (or deoxyribose) in bacteria, DNA carried the genetic
info. If you find radioactive uracil (or ribose) in bacteria, RNA carried the genetic info.
Answer #2: I accepted variations of the Avery, MacLeod, and McCarty experiment, even though
virus can not be transformed like Streptococcus (but you would not have known this yet). Isolate
2 tubes of the virus  Treat sample 1 with RNase to deactivate the RNA and sample 2 with
DNase to deactive the DNA. Add samples to two tubes of avirulent virus Look for
transformation of avirulent virus to virulent virus by infecting host cell with the treated samples
and seeing which infection results in virus replication indicating genetic material was present. If
the RNase treated sample from the virulent can transform the avirulent virus to virulent virus,
that means RNA is not the genetic material,. If the DNase treated sample from the virulent can
transform the avirulent virus to virulent virus, that means DNA is not the genetic material.
Answer #3: Isolate 2 tubes of the virus  Treat sample 1 with RNase to deactivate the RNA and
sample 2 with DNase to deactivate the DNA. Artificially infect host cell with samples to see
which infection results in virus replication indicating genetic material was present. If the RNase
treated sample can still allow virus production, that means RNA is not the genetic material. If
the DNase treated sample can still allow virus production, that means DNA is not the genetic
material.
Answer #4: Separate RNA and DNA from virus. Add DNA to one cell and RNA to another.
See which one produces viruses. I accepted this answer although technically it will not work for
RNA viruses since they carry a special enzymes that are needed for their life cycles.
Answer #5: There were a couple of other creative answers that I accepted.
Not acceptable: Treat with UV light and see if mutations occur. This is not acceptable because
both DNA and RNA absorb UV 260nm
5) A sample of DNA was found to have 20% guanine. What are the percentages of and names
of each of the remaining nitrogenous bases.
#G=#C and #A=#T
cytosine = 20% adenine = 30% thymine = 30%
6) Label the 5’ and 3’ ends of this molecule and indicate which way DNA replication would
occur if this were a template for DNA replication.
5’
3’
3’
5’
7) What holds each of the nucleotides together on one strand of the helix?
phosphodiester bonds between the phosphate and the sugar
8) How are the two strands of the double helix held together?
hydrogen bonds between the nitrogenous bases on each of the strands
9) Eukaryotes have several unique problems for DNA replication, relative to prokaryotes. List
two. Briefly indicate how eukaryotes solve the problems. No more than two sentences for
each problem.
a) Euks have more DNA than proks  multiple origins of replication and multiple DNA
polymerases
b) Euks have slower DNA polymerases than proks  multiple origins of replications
multiple DNA polymerases
c) Euks have linear chromosomes (as compared to circular in proks). This results in a gap
being left at the end of the lagging strand after the RNA primers were removed and no 3’
OH for DNA pol I to use to start filling in gap  telomerase polymerizes a sequence at
the end of the leading strand using the leading strand 3’OH and an RNA template internal
to the telomerse molecule. The polymerization of the dNTPs forms a hairpin loop. The
3’OH of the loop allows DNA polI to use to fill in gap. The loop in later cleaved off.
10) Many different mutants have been isolated in the various genes that encode the proteins that
are involved in DNA replication in Escherichia coli. These mutant proteins work like
normal proteins at 30˚C so DNA replication occurs, but when you shift the temperature to
37˚C, the mutant proteins stop working and DNA replication ceases. Mutants in any of the
subunits of DNA polymerase III or in the DnaB helicase exhibit an immediate cessation of
DNA replication after the shift to 37˚C (quick stop phenotype). In contrast, mutants in DnaA
exhibit a delay in the cessation of DNA replication after the shift to 37˚C (slow stop
phenotype). Based on your knowledge the function of the normal proteins in DNA
replication, explain why each mutation gives the quick or slow stop phenotype when shifted
to 37˚C. Specifically tell me what the protein does at 30˚C (i.e. what the normal protein
does) and then why when you shift the temperature to 37˚C do you see the quick stop or slow
stop of DNA replication.
a) DNA polymerase III – quick stop
At 30C, the mutant protein still works like normal protein, catalyzing the addition of the
dNTPs to the growing nucleotide chain. At 37C, the mutant protein stops working and since
it is the main polymerase in DNA replication, DNA replication stops immediately.
b) DnaB helicase – quick stop
At 30C, the mutant protein still works like normal protein, opening up the double stranded
template DNA so that DNA polIII can make new DNA strand. At 37C, the mutant protein
stops working and since opening of the helices has to occur in order for DNA polymerization
to occur, you see DNA replication stop immediately.
c) DnaA – slow stop
At 30C, the mutant protein still works like normal protein, binding to the origin of replication
(oriC) to initiate DNA replication by initially opening up the DNA helix. At 37C, the mutant
protein stops working. All of those chromosomes where replication has already been
initiated can continue to be replicated until replication is finished, since DnaA is only
required to initiate DNA replication and not for the actual polymerization. Thus, you see
DNA replication stop slowly, reflecting all the already initiated replicons finishing
replication. Once they have replicated, there will be no more replication because DnaA is
required to initiate.
11) The primary structure of a protein is determined by
a) the sequence of amino acids.
b) covalent bonds formed between amino acids R groups.
c) hydrogen bonds formed between the components of the peptide linkage.
d) a series of helical domains.
e) pleated sheets.
12) If a human has the genotype XXXY, they have
a) 3 Barr bodies and are male
b) 2 Barr bodies and are male
c) 3 Barr bodies and are female
d) 2 Barr bodies and are female
e) 1 Barr body and are male
13) A couple has a baby that is monosomic for the X chromosome. This most likely occurred
during meiosis because:
a) there was nondisjunction of the sister chromatids
b) there was an pericentric inversion in the X chromosome and crossover occurred
c) there was an paracentric inversion in the X chromosome and crossover occurred
d) there was translocation of another chromosome onto the X chromosome
e) either b or c
14) Which of the following abnormal ploidys would have the most significant problem with
chromosomal segregation during meiosis?
a) allotetraploid
b) autotetraploid
c) autohexaploid
d) diploid
e) autotriploid
15) Normally in corn, the waxy and virescent genes are linked. In a certain stock, however, it
was found that these two genes assort independently. Which chromosomal aberration would
explain this?
a) translocation
b) pericentric inversion
c) duplication
d) deletion
e) paracentric inversion
16) In mammals females have two copies of the X chromosome while males have only one copy.
Dosage compensation is accomplished in mammals by
a) females inactivating the X chromosome from the mother in all the cells
b) males inactivating the X chromosome from the father in all the cells
c) females expressing the X encoded genes at half the level of the male
d) males expressing the X encoded genes at 2 times the level of the female
e) females possessing a randomly inactivated X chromosome in all cells