Assignment:
Clinical Genetics
How do I solve the genetics problem Homework 2C question 8? Here's step by step instructions.
1. Make a list of the alleles being considered, and identify them with an upper and lower case letter.
2. Refer to the description (phenotype) of the parents, and write down each possibility. Do the genes separately. If the phenotype is recessive, you know the genotype. If the phenotype is dominant, it can be either homozygous or heterozygous.
3. Examine the progeny. If any of them have the recessive trait, both parents have at least 1 copy. The ratio should be 3:1 dominant to recessive.
4. Is there another gene to examine? Look at the parents potential genotypes. What progeny would you get from each one? Did you get it?
5. Now, put the genes together for each parent.
Here's the problem given:
1. In mice, the allele C for colored fur is dominant over the allele c for white fur, and the allele V for normal behavior is dominant over the allele v for waltzing behavior, a form of discoordination. Give the genotypes of the parents in each of the following crosses. (Hint: list the phenotypes and number of the progeny and look at their ratios. Look at the C separately from the V).
Following the step by step instructions, here's the solution
1. Make a list of the alleles being considered, and identify them with an upper and lower case letter.
C = colored c = white
V = normal behavior v = waltzing
2. Refer to the description (phenotype) of the parents, and write down each possibility. If the phenotype is recessive, you know the genotype. If the phenotype is dominant, it can be either homozygous or heterozygous.
Parent 1 is colored and normal. Could be CC or Cc. Could be VV or Vv.
Parent 2 is white and normal. Must be cc. Could be VV or Vv.
3. Examine the progeny. If any of them have the recessive trait, both parents have at least 1 copy. The ratio should be 3:1 dominant to recessive.
Progeny are:
29 colored, normal
10 colored, waltzing
Waltzing is recessive, so each parent must have a copy of v (waltzing). Since both parents have normal behavior, they musteach be Vv.
4. Is there another gene to examine? Look at the parents potential genotypes. What progeny would you get from each one? Did you get it?
The other gene is color. Parent 2 is known at [cc]. Parent 1 is either CC or Cc. Since none of the progeny are cc, Parent 1 must be CC.
5. Now, put the genes together for each parent.
Parent 1 is colored and normal. Must be CC Vv.
Parent 2 is white and normal. Must be cc Vv.
6. A geneticist crossed wild, gray-colored mice with white (albino) mice. All of the progeny were gray. These progeny were intercrossed to produce to produce F2, which consisted of 198 gray and 72 white mice.
a. Diagram the crosses (include the numbers that were given, and find the total number)
b. Evaluate the ratio of the progeny
c. Propose a hypothesis to explain these results. (This means: what are the proportions of the phenotypes you expect in a monohybrid cross)
d. Compare the results with the predictions of the hypothesis and state your conclusion. (This means, use Chi square analysis). This will help, complete each blank:
i. What is the number of gray mice you expected to get? White mice?
ii. What is the number of gray mice you observed? white mice?
iii. What is the "degrees of freedom" for this cross?
iv. What is the critical value for that number of degrees of freedom?
v. What is your conclusion?
7. In mice, the allele C for colored fur is dominant over the allele c for white fur, and the allele V for normal behavior is dominant over the allele v for waltzing behavior, a form of discoordination. Give the genotypes of the parents in each of the following crosses. (Hint: list the phenotypes and number of the progeny and look at their ratios. Look at the C separately from the V).
a. Colored, normal mice mated with white, normal mice produced 29 colored, normal and 10 colored, waltzing progeny. _Genotypes of the parents are
b. Colored, normal mice mated with colored, normal mice produced 38 colored, normal, 15 colored, waltzing, 11 white, normal and 4 white, waltzing progeny. _ Genotypes of the parents arec. Colored, normal mice mated with white, waltzing mice produced 8 colored, normal, 7 colored, waltzing,
9 white, normal, and 6 white, waltzing progeny.