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Transcript
Answers to Mendelian Genetics Problems (p. 275–76)
1. Alleles segregate in meiosis, and the products of that segregation are
contained within a pod. Each pea is a gamete. In this diagram, the
segregation is incorrectly shown as being between pods, each pod shown as
uniformly wrinkled or round.
2. The probability of getting two genes on the same chromosome is 1/223.
3. Somewhere in your herd you have cows and bulls that are not
homozygous for the dominant gene "polled." Since you have many cows and
probably only one or some small number of bulls, it would make sense to
concentrate on the bulls. If you have only homozygous "polled" bulls, you
could never produce a horned offspring regardless of the genotype of the
mother. The most expedient thing to do would be to keep track of the
matings and the phenotype of the offspring resulting from these matings and
render ineffective any bull found to produce horned offspring.
4. It would not be possible on the basis of the information presented to
substantiate a claim of infidelity. You do not know if the woolly trait is the
result of a single gene product, or even if the trait is dominant or recessive.
Assuming for the moment that it was the effect of a single dominant
allele W, the man would still be a heterozygote for the gene and, when
mated to a recessive homozygous female, would expect to produce woollyheaded offspring only one-half the time.
5. One-half of her offspring would be expected to be affected.
6. Albinism, a, is a recessive gene. If heterozygotes mated you would have
the following:
A
a
A
AA
Aa
a
Aa
aa
Clearly one-fourth would be expected to be albinos.
7. The best thing to do would be to mate the racehorse to several mares
homozygous for the recessive gene that causes the brittle bones. Half of the
offspring would be expected to have brittle bones if the racehorse were a
heterozygous carrier of the disease gene. Although you could never be
100% certain your horse was not a carrier, you could reduce the probability
to a reasonable level.
8. Your mating of d+d+Ww and d+dww individuals would look like the
following:
d+w
d+w
dw
dw
d+W d+d+Ww d+d+Ww d+dWw d+dWw
d+w d+d+ww d+d+ww d+dww d+dww
d+W d+d+Ww d+d+Ww d+dWw d+dWw
d+w d+d+ww d+d+ww d+dww d+dww
Long-wing, red-eyed individuals would result from eight of the possible 16
combinations, and dumpy, white-eyed individuals would never be produced.
9. Breed the fly to one from the white-eyed stock. If half of the offspring are
white-eyed, then your fly is a heterozygote.
10. Both parents carry at least one of the recessive genes, even if only in
the heterozygous condition (in which the trait would not be expressed in the
parent). Since it is recessive, the trait is not manifest until they produce an
offspring who is homozygous.
11. To solve this problem, first look at the second cross, where the
individuals were crossed with the homozygous recessivesepia flies se/se. In
one case, all the flies were red-eyed:
Unknown genotype
Se
Se
Sepia Se
Se/se
Se/se
se
Se/se
Se/se
The only way to have all red-eyed flies when bred to homozygous sepia flies
is to mate the sepia fly with a homozygous red-eyed fly. In the other case,
half of the offspring were black-eyed and the other half red-eyed:
Unknown genotype
Se
se
Sepia se
Se/se
se/se
se
Se/se
se/se
The unknown genotype in this case must have been Se/se, since this is the
only mating that will produce the proper ratio of sepia-eyed flies to red-eyed
flies. Since the ratio of this unknown genotype and one previously
determined was 1:1, we must deduce the genotype of the original flies,
which when mated, will produce a 1:1 ratio of Se/se to Se/Se flies.
Unknown Original 1
Original 2
Se
Se
Se Se/Se
Se/Se
se
Se/se
Se/se
You can see from this diagram that if one of the original flies was
homozygous for red yes and the other was a heterozygous individual, the
proper ratio of heterozygous and homozygous offspring would be obtained.
12. a. It could have originated as a mutation in his germ cell line.
b. Since their son Alexis was a hemophiliac, the disease almost certainly
originated with Alexandra; Nicholas II would have contributed only a silent
Y-chromosome to Alexis’s genome. There is a 50% chance that Anastasia
was a carrier.
13. The ability to smell musk is an example of dominant inheritance.
14. No. IAIo x IBIO→IAIB or IAIO or IBIO or IOIO
15.
c is a rare allele, and Mabel’s child can develop CF only if her husband also
carries the allele:
[probability that husband has c allele]=frequency of c=1/20
[probability that Mabel has c allele]=1/4
16. You could expect 45 chromosomes (44 autosomes plus one extra copy of
an X).
17. The woman’s blood genotype and blood type is AO.
18. It expresses dominant inheritance.
19.
The chances are 25%.
20. Let a=albino. The genotype of the father is Aa.
21.
Parents
Baby
O and O
0
A and B
AB
B and B
BB
AB and O
A
22. It would be very difficult to establish negligence, especially if the child is
a boy. If the child's maternal grandmother was a carrier, there's a 50:50
chance that the mother is a carrier. If the mother is a carrier, there's a
50:50 chance that a boy will express the disease. One would be suspicious
about the causes of Duchenne muscular dystrophy in a girl. In the first
place, symptomatic males may not live to reproductive age, and it would be
rare for a symptomatic male to be virile. A symptomatic girl from a
nonsymptomatic father could result from a genetic anomaly in the father, or
from a mutation unrelated to radiation. The case for mutation on the job is
clearly stronger for the girl, but very weak in either case.