Home > 6 Name _________________________ Biology 201 (Genetics) Exam #2 Read the question carefully before answering. Think before you write. Be co
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Biology 201 (Genetics)
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All of gametes from tester will be gh and � of gametes from heterozygote will be gh
so � of
progeny will be gghh
GH gh gh gh gh
All of gametes from tester will be gh. Since genes are completely linked there is no crossover and � of gametes from the heterozygotes will be gh
So � of
progeny will be gghh
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.
Crossover
type
(see questions below) |
Phenotype of test cross progeny | Number |
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 flowers | 2 |
NCO | Flattened fruit, Hairy fruit | 306 |
SCO | Flattened fruit, Hairy fruit, Clustered flowers | 63 |
Total progeny = 1000 |
See above (because most frequent classes)
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
See above (because least frequent of the 8 expected classes)
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
single crossover events in the heterozygote parent.
see above
HF = (110+96+2+2)/1000*100 = 21% = 21 map units apart
FL = (73+63+2+2)/1000*100 = 14% = 14 map units apart
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.
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.
Inversions: If and only if there is crossover during meiosis in a heterozygote with an inversion, � 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)
.
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
#G=#C and #A=#T
cytosine
= 20% adenine = 30% thymine = 30%
5’
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’
phosphodiester
bonds between the phosphate and the sugar
hydrogen bonds
between the nitrogenous bases on each of the strands
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.
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