Solved by a verified expert:Biology 2120 Assignment 2

Assignment Part I – Mendelian Genetics (7.5 pts).
Information on the analysis of crosses and Chi-square
statistics are available under the supplemental reading section of the course
website. Below is the Chi-square probability table at 0.05 (5%)

Degrees of Freedom Reject
the null hypothesis if the calculated chi-square is larger than
1 3.841
2 5.991
3 7.815
4 9.488
5 11.070

1. An
organism with genotype AaeeDd can make how many different gametes (list each
one)? (1 pt)

2. You are
studying one trait that is autosomal recessive. You crossed the following true
breeding parents: autosomal dominant female with an autosomal recessive male.
Draw the Punnett square and provide the expected genotypic and phenotypic
ratios in the F2 generation? (1.25 pt)

3. You
hypothesize that you have one autosomal dominant trait. You crossed the
following true breeding parents: wild type female and mutant male. In the F2
generation you observe the following phenotypes: 35 dominant females, 35
dominant males, 20 recessive males, and ten recessive females. Show your
chi-square calculations, and does this fail to reject the stated hypothesis? (1

4. What is
the genotypic ratio of the progeny produced from this mating?
Parental cross: female aaBb crossed with a male AABb

5. You are
studying hair length in wild yaks and perform the standard genetic crosses.
From the following observations provide a hypothesis of how this trait is
inherited (which allele is dominant and whether it is autosomal or sex-linked).
Assume both parents are true breeding. Test your hypothesis with a Chi-square
test. Show all your calculations for the Chi-Square. (1.5 pt)
P: short haired male X long haired female
F1: 145 long haired males
and 155 short haired females
(F2 is a cross of F1 male and female)
F2: 73 long haired females; 77 short haired females; 72 long
haired males, 78 short haired males.

6. You are
analyzing a dihybrid cross where both mutant traits are autosomal recessive in
a Scooby Doo species. You set up the cross with each true breeding parent
exhibiting one wild type trait. You observe A = brown, a = gray and M =
munchies, m = no munchies. In the F2 generation you observe 152 brown munchie
having scoobys, 48 brown no munchie having scoobys, 42 grey munchie having scoobys
and 12 grey no munchie having scoobys.
a. Draw the
Punnett squares for F1and F2. (.5 pt)
b. What is
the expected number of each F2 phenotype (show work) (1 pt)
c. Does
Chi-square support the null hypothesis? (show your work) (.75 pt)