Question

Observing the phenotypic ratio of the offspring generation of a mated pair can increase our understanding of inheritance of traits. In wolves, gray fur and yellow eyes are dominant, black fur and blue eyes are recessive, and both traits are governed by simple (complete) dominance In the Ely, MN area there exists an alpha pair: a male dihybrid (double heterozygote) wolf displaying both dominant phenotypes mated with a female wolf recessive for both traits. Researchers hypothesize the two genes (fur color and eye color) assort independently from one another because they are found on different chromosomes. Complete the following SHOW ALL WORK

If the hypothesis is true, then what is the phenotypic ratio you expect to see in their offspring generation?

Complete the following Punnett square to show how you arrived at your prediction.

First, record the genotype of each of the parents: ♂ ________________   x    ♀ ________________

♂ ♀

What are the four possible offspring phenotypes and expected proportions of each according to the hypothesis?

Offspring Phenotypes	Expected Proportion
_____________________________	__________
_____________________________	__________
_____________________________	__________
_____________________________	__________

This pair produce 24 offspring over their lifetime as parents: 7 have the dominant phenotype for both traits, 4 have the recessive phenotype for both traits, 5 wolves have gray fur and blue eyes, and 8 have black fur and yellow eyes.

Perform Chi-square analysis to determine whether the collected data supports or refutes the hypothesis. Compare your Chi-square calculated value to the appropriate Chi-square critical value.

Phenotypic Category	Expected Proportion	Observed Number	Expected Number	(O – E)	(O-E)2	(O – E)2/E
Total				X2 calculated =

Do you accept or reject your hypothesis? ____________________

What could be a reason for why the observed numbers are not exactly the same as the expected numbers

Answer 1

70 Cnal laue

Phenotypic Category	Expected Proportion	Observed Number	Expected Number	(O -E)	(O-E)2	(O – E)2/E
Gray, yellow	1	7	6	1	1	0.16
Gray, blue	1	5	6	-1	1	0.16
black, yellow	1	8	6	2	4	0.66
black, blue	1	4	6	-2	4	0.66
Total		24		X2 calculated =		1.666666667
Degrees of freedom		3

Table value @ 0.05 probability and 3 DF is 7.81. The Calcualted values is less than table values. Hence hypothesis accepted.