More Genetics Practice Problems Answers

1. A large ear of corn has a total of 433 grains, including 271 Purple & starchy, 73 Purple & sweet, 63 Yellow & starchy, and 26 Yellow & sweet.

Calculate the chi-squared value for the null hypothesis that the ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two genes for color and sweetness. Explain the chi-square test results.

So, your tentative hypothesis is: This ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1.

Test your hypothesis using chi square and probability values.

Total Kernals = 840

Expected values are as follows:

Purple and Smooth 9/16 of 433 = 243.6

Purple and Wrinkled 3/16 of 433 = 81.2

Yellow and Smooth 3/16 of 433 = 81.2

Yellow and Wrinkled 1/16 of 433 = 27.1

3.08 + .82 + 4.07 + .04 = 8.01

there are 4 phenotypic classes, so you have 3 degrees of freedom (4-1), and we want to be 95% accurate, so we look at the 0.05 column...

This is a poor fit since 8.01 > 7.81.

2. In a certain reptile, eyes can be either black or yellow. Two black eyed lizards are crossed, and the result is 72 black eyed lizards, and 28 yellow-eyed lizards.

Calculate the chi-squared value for the null hypothesis that the black eyed parents were heterozygous for eye color. Explain the chi-square test results.

Your Tentative Hypothesis: The black eyed parents were Bb x Bb.

Expected values:

If you do a punnet square of Bb x Bb, you get an expected percentage of 25% and 75% or 1/4 and 3/4

Since the total number of lizards is 100, the observed is already in the correct ratio

Expected ....................... Observed

Black Eyes = 75 ................ 72

Yellow Eyes = 25 ............. 28

Test your hypothesis using chi square analysis. In this set, because only two values (traits) are examined, the degrees of freedom (df) is 1.

.12 + .36 = .48 which is less than 3.84 (from the above table)

Good FIT

3. Problem: A sample of mice (all from the same parents) shows

· 58 black hair, black eyes

· 16 black hair, red eyes

· 19 white hair, black eyes

· 7 white hair, red eyes

a) What were the genotypes and phenotypes of the parents?

b) Use a chi square-test to support your hypothesis.

Looks like a 9/3/3/1 ratio, meaning the parents are HhEe x HhEe

Total number of offspring = 100

Expected values would be:

9/16 of 100 = 56.25

3/16 of 100 = 18.75

3/16 of 100 = 18.76

1/16 of 100 = 6.25

.05 + .40 + .003 + .09 = .54 = GOOD FIT since 0.54 is less than 7.81

4. A wild-type fruit fly (heterozygous for gray body color and normal wings) was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162.

Calculate the recombination frequency (%) between the linked genes for body color and wing type.

First count the total number of offspring 778+785+158+162 = 1883

In all dihybrid test crosses (a cross between a known heterozygote for two wild type traits and a homozygous recessive individual for both traits) the expected ratio of phenotypes if the genes are on separate chromosomes must be:

wild type, 25%; black-vestigial, 25% black-normal, 25%; gray-vestigial, 25%. These results do not fit the experimental data above (778+785+158+162).

In fact the black-normal (158) and gray-vestigial (162) offspring represent recombinant individuals.

Calculation of recombination frequency: