Biology 122, Sample Transmission Genetics Problems
- In mice a black (B/b) female is crossed to a
brown (b/b) male. What proportion of the resulting progeny mice will be
brown? __1/2_________ What
proportion of the black offspring will be true breeding if mated to mice
from a true breeding black line? _____0% or 100%__, depending on your reading of the
question.
0% because all the black mice from the cross
are heterozygous, so if they are crossed to a B/B strain and the progeny
interbred, some brown mice will appear because the b allele will be present in
this stock. If you interpret “true breeding” to mean producing all black
offspring in the 1st generation, then 100% of the black mice (B/b)
will give “black” progeny when crossed to B/B.
2. A gene, C, has three co-dominant alleles (C1, C2, C3), each of which produces a blood serum protein product that can be resolved by polyacrylamide gel electrophoresis. If a C1/C2 individual is mated to a C2/C3 individual, what proportion of the offspring will show two different C gene proteins when their blood serum is analyzed by electrophoresis? ____3/4_______________
C1/C2 x C2/C3
¼ C1/C3
¼ C1/C2
¼ C2/C3
¼ C2/C2
So, ¾ of the
offspring will be heterozygous and therefore show 2 different gene C proteins
3. The following pedigree depicts the inheritance of a rare hereditary disease affecting muscles:
What is the likely mode of inheritance of this disease (e.g. autosomal recessive, X-linked dominant, ect.)? Explain (briefly) your reasoning.
X-linked recessive:
1) recessive because
offspring show disease and parents don’t
2) X-linked because
only males show the disease, apparently inheriting the disease allele on the X
chromosome from their carrier mothers.
4. Mendel crossed Y/Y; R/R (yellow wrinkled) peas with y/y;r/r (green smooth) peas and selfed the F1 to obtain an F2. In the F2, what proportion of the yellow wrinkled individuals were true breeding when selfed?_______1/9___________
The phenotypic ratio will be 9 (“YR”): 3 (“Yr”): 3 (“yR”); 1 (“yr”), a typical result from a dihybrid cross. But only 1/9 of the “YR” plants will be homozygous (YY RR). The rest of the “YR” will be heterozygous for one or both genes and will thus not be true breeding when selfed.
5. A rare, curly-winged mutant of Drosophila was found in nature. A mating of this fly (cross 1) with a true-breeding, normal laboratory stock produced progeny in the ratio 1 curly-winged : 1 normal (both sexes had same ratio). All curly-winged progeny of this cross, when mated to normal progeny of the same cross, again yielded a progeny ratio of 1 curly winged : 1 normal . When mated to one another (cross 2), the curly winged progeny of cross 1 yielded the following progeny: 605 curly winged and 310 normal. This result suggests which of the following (circle the correct letter)?:
- Curly and normal are in the 3:1 ratio expected from intercrossing monohybrid genotypes for a recessive mutant
- Curly and normal are in the 3:1 ratio expected from intercrossing monohybrid genotypes for a dominant mutant
- The curly winged parent in cross 1 is homozygous
- Flies
homozygous for the curly allele do not survive
- The gene for curly is sex-linked
The answer is d
from the following reasoning:
Cy/+ x +/+
½ Cy/+ and ½ +/+ =
1 “Cy” : 1 “+”
cross these two types
and get same result
but Cy/+ x Cy/+
yields:
1/2 Cy/+
¼ +/+
¼ Cy/Cy
we expect a 1”+”
: 3 “Cy” ratio, but instead get a ~ 1:2
ratio.
This would be easily accounted for if the Cy/Cy flies died before hatching into adults
6. Consider the following cross: A/a;B/b;C/c;D/d;E/e x a/a;B/b;C/C;d/d;E/E. If there is complete dominance for each of the genes, What proportion of the offspring will show the dominant phenotype for each gene? ______3/16____________. What proportion will be heterozygous for every gene? ____1/32_____________ Hint: you do not need to do a Punnett square! Just calculate the probability of the stated outcome for each gene separately and perform the correct arithmetic (addition or multiplication) as needed.
½ probability of getting A allele, ¾ probability of getting B
allele, 1.0 probability of getting C allele, ½ probability of getting D allele,
1 probability of getting E allele: ½ * ¾ * 1 * ½ * 1 = 3/16
½ is the probability of being heterozygous for each gene. So, (½)5
= 1/32
7.
a. In a plant
in which 2n
=
24, what is the total number of chromatids present during prophase I of
meiosis?
____48___________
b. In a plant in which 2n = 24, what is
the total number of chromatids present during
prophase of mitosis? ______48_________
Each chromosome has replicated into 2 chromatids in each case
8. In a dihybrid the
genes are located on different chromosomes. Which of the following diagrams
does not represent a stage of meiosis in this organism? Circle the
letter.
The answer is E because although this event
might happen in mitosis, it would not occur in normal meiosis. The consequence
would be a 2N gamete, not a desired outcome in meiosis!
9. Suppose
Red-striped laughing frogs have 32 chromosomes in somatic cells. Answer the
following questions as
they pertain to these vanishingly rare but beautiful
creatures.
(a) How many
homologous pairs of chromosomes does a red-striped laughing frog have? ____16_____ 2N = 32, so 1N =
16 = 16 pairs
(b) How many
chromosomes are there in a haploid set? _______16_______
1N=16
(c) How many
chromosomes are there in a mature sperm cell? _____16________
1N=16
(d) How many
chromosomes are there in a somatic cell just after mitosis? ____32_________ 2N=32
(e) How many
double-stranded DNA molecules are there in a somatic cell just before mitosis?
___64_______ Each of the 32
chromosomes has replicated and consists of 2 chromatids
(each chromatid having a single DNA molecule)
10. In a certain breed of dog, the alleles B and b determine black
and brown coats respectively. However, the allele Q of a gene on a
separate chromosome is epistatic to the B and b color alleles
resulting in a gray coat (q has no effect on color). If animals of
genotype B/b ; Q/q are intercrossed, what phenotypic ratio is
expected in the progeny? Circle the correct letter.
(a) 9 gray, 3
brown, 4 black
(b) 1 black, 2
gray, 1 brown
(c) 9 black, 6
brown, 1 gray
(d) 9 black, 4 gray, 3 brown
(e)
12 gray, 3 black, 1 brown
The answer is E because, given a 9:3:3:1
genotypic ratio resulting from the dihybrid cross, 9
plus 3 (=12) will have the dominant Q allele and therefore will be gray. 3 will
be homozygous for the q allele and have at least 1 B (black) allele. 1 will be
q/q and b/b, i.e. brown.
11. In petunias, the R allele produces red flower pigment, the r allele produces no red pigment. The Sp allele produces uniform color in the flower, the sp allele produces a spotted pattern of pigment in the flower. The following cross is made between a plant with uniform, pink flower color and a plant with uniformly white flowers.
Let’s say that the two genes that control color and pattern in the flower are 15 cM apart on the same chromosome.
1a. Is this a case of epistasis? Explain clearly.
Yes, the phenotypic effect of one gene is overridden by alleles at another locus (gene)
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1b. What will be the proportion of progeny that have spotted, red flowers? ______0%______________
1c. What will be the proportion of progeny that have spotted, pink flowers? ________7.5%____________
1d. What will be the proportion of progeny that have uniformly pink flowers? _______42.5%_____________
There is no way to get red flowers from the above cross since R and
r are clearly codominant
15% recombination (15 cM = 15 map units)
between genes produces 15% recombinant chromosomes
½ of this 15% will be R sp = 7.5%. When a R
sp gamete fuses with an r sp gamete, we have spotted pink
85% of the chromosomes will be non-recombinant so ½ of these (42.5%)
will be R Sp. So 42.5% of progeny will be uniform-pink
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