Download GENETICS 603 EXAM III Dec. 5, 2002 NAME 5 6 7 8 1 2 4 3 I Gene

GENETICS 603
EXAM III
Dec. 5, 2002
NAME
I Gene Fine Structure: Remembering that rII mutants of bacteriophage T4 make large plaques
on lawns of E. coli B but no plaques at all on E. coli K, how would you measure:
A) reversion frequency for a new rII mutation?
use it to infect E. coli B; plate progeny on K; only + revertants will make a colony.
B) recombination frequency between two different rII mutants?
Co-infect the two on B; plate progeny on K; if there is a higher frequency of plaques than from
either olone, recombinaiton must occur.
C) complementation between two rII mutants?
Mixed infection on K (or B) will lead to complete lysis
D) Use the data below that shows the results of crossing mutants 1-8 in all possible combinations
to map the following A cistron mutants (+ indicates recombination, - indicates no
recombination; mutants 1-4 are non-reverting)
1
2
3
4
5
6
7
8
1
-
2
+
-
3
+
-
4
-
5
+
+
+
-
6
+
+
+
+
-
7
+
+
+
-
8
+
+
+
+
+
-
5
6
1
7
8
2
4
3
II Tetrad analysis: Neurospora genes A, B, and D are linked at the distances shown, with the
centromere shown as "•". : A 6 B 4 • 8 D. A cross is made between an A, B, D
parent and an a, b, d parent.
A) What fraction of the asci produced would show second division segregation for each gene?
A20% B 8% D16% (gene -centromere map distance = 1/2 f(2nd division segregation)
B) A very rare ascus had pairs of spores in the following order:
a b D
a B d
A
B
A b D
a
b
A B d
1) which of the loci show second division segregation? B and D
D
d
2)Make a line drawing showing the meiotic events that gave rise to this ascus. (see above)
3) An even rarer ascus had spore genotypes as follows:
a b D; a B d; A b d; A B d
Explain what happened. Gene conversion of D to d
III. Mapping A testcross between a female fruitfly and a homozygous male gave the following
progeny: (F and S are fast and slow migrating alleles of an isozyme locus; the male has a null
allele)
F +
+
50
F +
vg
7
F ry +
356
F ry vg
80
ry 12 F/S 18 vg
S +
+
84
S +
vg
360
S ry +
9
S ry vg
54
Map the chromosome.
What is the value of interference? 16 DXOs of 21.6 expected= 74%, so I = 24%
What progeny and in what frequency would be obtained from the reciprocal cross?
No XO in male therefore 1/2 ry, F, + and 1/2 + S, vg
IV. Cytogenetics: A) When homozygous for mutant alleles of the Ph1 gene on chromosome
5B, homoelogous chromosomes from the various genomes in wheat can pair. Would this
affect fertility? Explain your answer.
Probably would reduce fertility somewhat since chromosomes from different genomes could
"compete for pairing" forming multivalent complexes that would reduce the normal
amphidiploid nature of wheat. ( In fact, it does lower fertility)
B) Colchicine has often been successfully used to double chromosome number in plants. Upon
hearing that the resulting plants often have larger flowers and fruits, an entrepreneur decides
to treat banana plants in the hopes of increasing production. Evaluate his plan.
If he starts with a triploid, he will create a hexaploid, which may have larger fruit (or may not)
but would likely produce some seeds. If he starts with 2N, his best bet os to redo the 3N trick.
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C) A normal maize pollen grain has 10 chromosomes. a) How any chromosomes would be
found in:
i) leaf cells of a triploid
30
ii) root cells of a trisomic? 21
How many chromosomes would be expected in gametes of;
i) the triploid)?
10 to 20
of the trisomic)? 10 0r 11
Which would be least fertile? Why
The tripliod; odds for a balanced gamete = 1/2 to the 10th power
D) Populations of Drosophila are homozygous for the following gene orders:
Pop. 1
1 2 3 4 • 5 6 7 8 9 10
Pop. 2
1 2 5 • 4 3 6 7 8 9 10
Pop. 3
1 7 6 3 4 • 5 2 8 9 10
Pop. 4
1 7 6 3 4 • 5 9 8 2 10
Assuming population 1 has the primitive gene order, list the sequence in which the
populations arose.
1 to 2 to 3 to 4
Define the precise aberration (by name only) for each of the steps.
1 to 2 pericentric inversion; 2 to 3 pericentric inversion; 3 to 4, paracentric inversion
Which of the changes will be most effective so far as contributing to hybrid sterility? Why?
The pericentric inversions, especially the largest lead to loss of viable gametes in both males and
females whenever a crossover occurs since duplications and defeicinies are created. The
paracentric has the same problem, but the bridges formed usually lead to inclusion in the polar
bodies and not the actual egg.
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E) Chromosomes 1 and 2 in the female had the gene order A • B C D and 1 • 2 3 4 5,
respectively, while the male chromosomes were a • b 4' 5' and 1' • 2' 3' c d.
1) Show chromosome pairing in their progeny.
D
d
A
1
a
1'
5'
5
2) Assuming no crossovers:
a) Show the products of alternate segregation; circle those that are viable.
2
ABCD
12345
&2
ab4'5'
1'2'3'cd
b) Show the products of adjacent 1 segregation: circle those that are viable.
2
ABCD &
1'2'3' cd
ab 4'5'
12345
c) Show the products of adjacent 2 segregation: circle those that are viable.
2 ABCD
& 2 12'3'cd
AB4'5'
12345
2) Assuming one crossover between genes 2 and 3:
a) Show the products of alternate segregation; circle those that are viable.
123c d
ab4'5'
12345
ab4'5'
ABCD
12345
ABCD
123cd
b) Show the products of adjacent 1 segregation: circle those that are viable.
123cd
ABCD
12345
ABCD
1'2'3'cd
AB4'5'
12345
AB45
c) Show the products of adjacent 2 segregation: circle those that are viable.
2 ABCD
AB45
123cd
12345
1'2'3'cd
12345
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V. Quantitative Inheritance A) H2 for egg weight in white leghorns is fairly high at 0.6.
Does this means a smart producer could increase production rapidly just by hatching the
largest eggs for several generations? Explain your answer.
Egg weight would increas by selection, but number of eggs would likely decrease; is this a gain?
B) A true breeding pearl millet variety with an average of 10 leaves per plant was crossed to another with
an average of 18. The average leaf number in the F1 was 14, as it was in the F2. The variance in the
F2 was 3, but it was only 1 in the F1.
1) Calculate H2 for leaf number in pearl millet from this cross.
Vp=3 Ve =1 so Vg =2 and H2 = 2/3
2) Progeny in the F2 were found with every possible number of leaves between 8 and 20, though the
extremes were rare. Based on the best available information, what is the number of genes that are
heterozygous in the F1?
8-9-10 ----20 is 13 phenotypc classes, which will occur if there are 6 segregating genes
(the D2 formula only works of parents are at the extemes)
3) Based on your answer to the last question (C2), just how rare would the 8 leaf plants be in the F2?
(1/4)N where N is the number of genes (6 is correct)
4) Again based on your answer to question C2, give a potential genotype for each of the true breeding
varieties:
10 leaf variety
A'A'BBCCDDEEFF
18 leaf variety
AAB"B'C'C'D'D'E'E'F'F' (differejt contributing alleles)
5) Does transgressive segregation occur in this example? Explain
Yes, the F2 has progeny beyond the parents
2
2
C) Which would you expect to be larger, H or h ? Why
In general H2, since it includes Vg components due to dominance and epistasis, which are not
included in Va; However in this example, all the variation is accounted for by additive effects
(1 leaf per contributing allele), so H2 and h2 would be equal.
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VI. Population genetics: A) The ability to taste versus not-taste the chemical
phenylthiocarbamide (PTC) in humans is determined by a single gene with two alleles, where the
allele T (for taster) is dominant to t (for non-taster). In the North American Caucasian
population, non-tasters comprise 36% of all individuals sampled. Assuming the population is in
the Hardy-Weinberg equilibrium, what fraction of tasters will be heterozygous?
The square root of 0.36 is q and is0.6, so p=0.4. The population will have the structure
16 TT : 48Tt :36tt at equilibrium. Thus, there are 3 Tt tasters for every TT taster and the answer is
3/4.
B) A selectively neutral, recessive trait appears in 40% of the males and in 16% of the females in a
large, randomly mating human population. What is the frequency of the recessive allele in the
population? How many females are heterozygous at this locus? How many males are
heterozygous at this locus? You may assume the population is in the hardy-Weinberg
equilibrium.
The difference of 40% of males and 16% of females can only occur if the gene is on the X
chromosome, which means that q=0.4. No males are heterozygous and for females,
2pq=0.4 X 0.6 X 2 = 48% are heterozygous.
C) What allele frequency will generate twice as many homozygous recessives as heterozygotes?
Assume an autosomal locus with two alleles, where one allele is dominant to the other.
when q2 =2(2pq) which means on factoring out q that q=4p or since p=1-q, 5q = 4
Thus q=0.8 which is verifed by noting that p2 + 2pq +q2 = 4:32:64 in this case.
D) The frequencies of the aM and aN alleles (responsible for the co-dominantly inherited MN blood
groups in humans) are the same in the following three populations. The genotype frequencies in
the three populations (show below) however, are not the same. Which of the three populations is
most likely to have been subjected to random mating, genetic drift, or inbreeding? Explain your
answers!
Frequency of
Blood group M
Frequency of
Blood group MN
Frequency of
Blood group N
Population A
0.55
0.30
0.15
Population B
0.49
0.42
0.09
Population C
0.43
0.54
0.03
All have p=0.7 and q=0.3
Population A shows signs of inbreeding since there are extra homozygotes and less
heterozygotes than expected at equilibrium
Population P is a perfect fit for H-W equilibrium
C shows signs of drift (or selection) since it is low in one homozygouls class
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E) A shepherd has a herd where all the males and females have horns. Deciding that he would
prefer all hornless, he sells all the horned males and buys a hornless (HH) ram. In the next
(S1) generation he is surprised to find that the males still have horns. (He didn't know as you
do that horns in sheep may be inherited as a sex-influenced trait; that is heterzygous H'H
males have horns, but H'H females do not.) In any case, he saves only the hornless females
and mates them to the same hornless ram to produce an S2 generation.
Calculate the frequencies of the H' and H alleles in his original, S1 and S2 populations.
Original herd: f(H') =1; f(H) = 0
S1: all are heterozygous, so f(H'H = 0.5, f(H) = 0.5
S2 HH X H'H gives a 1 HH to 1 H'H ratio, so f(H') =1/4 and f(H) =3/4
If he allows the S2 progeny to begin random mating, what will the equilibrium population
look like?
Random mating will give 1/16 H'H' : 6/16 H'H : 9/16 HH; since H'H' and H'H males have horns,
7/16 will have horns and 9/16 will not. since H'H females are hornless, 15/16 of the females
will be hornless.
F) One baby in 10,000 in the US is born with PKU, which until very recently was a recessive
lethal condition. Does it seem likely that the population is in equilibrium with mutation
balanced by selection? (Explain)
This would mean a mutation rate of 1 in 10,000 which seems a bit high; there may have been
selection for heterozygotes in the past.
What will happen to the frequency of PKU births in China over the next 100 years now that the
affected individuals can survive and reproduce?
It will change by the mutation rate, which even if it is 1 in 10,000 means over five generations q
might increase from 0.01 to 0.0105, ignoring reversions; remember gene frequencies will not
change unless forced to do so; relaxing selection does not force a change!
G) Calculate the coefficient of inbreeding the for individual X in the following lineage:
VII H
F
D
G
T
J
E
C
A
B
X
7
3
5
(1/2) (for C) + (1/2) (for F)
Cytoplasmic Inheritance. A) Petite yeast mutants grow anaerobically by fermentation. Predict
the ratio of normal and petite progeny (yeast crosses produce a tetrad of 4 spores) that will be
produced when a wild type "grande" strain is crossed to a petite if:
a) the inability to use O2 results from a mutation in cytochrome C, a nuclear gene
1 peteite: 1 grande per tetrad ( ie the products of meiosis)
b) the inability to use O2 results from a defect in a mitochondrial gene.
since the cells fuse all progeny will have mitochondria, unless the petite mutaion is suppressive,
in which case it out-replicates the normal mt-DNA and all will be petite. ie all like one parent
B) What features of mitochondrial DNA have been useful in tracking human populations?
They have a region that can vary in sequence, but it takes many generations for a new sequence
to become established. The changes act as an "evolutionary-scale clockThe many copies/cell
make detection and sequencing possible, even in fossil remains in some cases. They are
passed maternally only, so provide a constant lineage without all the reassortment and
recombination that would confound origins of nuclear genes.
VIII. Immmunogenetics. Give the basic structure of an immunoglobin "antibody" gene.
V-V V V J J J J -Intron - C region is the general model where on maturation any V can be
attached to any J and the C (V = variable regions, J = joint regions)
How do kappa and lambda light chain genes differ?
Primarily in having different Cs
How do heavy chain genes differ from light chain genes?
Heavy chain genes also have a series of D (diversity) sequences added between the J and C
regions. They also have a series of Cs that are involved in class switching during maturation.
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