Question

11. You are a researcher studying mice. You hypothesize allele, brown coat; B-dominant allele, black coat) and tail length (I recessive allele, shor allele, long tail) in mice are independently assorted. You know that both genes are autosomai gen. Youhve that genes that control coat color (b - recessive tail, L-dominant colonies of true- Obreeding access to two with short tails and brown coats. You cross the females and males described above and then you cross the F1s. Your aggregated F2 data is below mice: female mice with long tails and black coats and male mice mice Phenot Black coat, Long tail Black coat, Short tail Brown coat Long tail Brown coat, Short Tail Number 53 12 13 A. What would your expected F2 data be if the coat color and tail length genes are independenthy assorted? (Complete the chart below) Number Phenotype Black coat, Long tail Black coat, Short ta Brown coat Long tail Brown coat, Short Tail B. Conduct a Chi Squared test to determine i and clearly indicate your calculated Xz value and P value. f the genes are independently assorted. Show your work C. Based on your Chi Squared test, are the genes independenty assorted? Explain your answer. short recessive 12. You perform a cross between two allele). Ofthe animals born, 666%are tall and 333% are short. What is the best explanation for these results? heterozygote TI) sheep T tall dominant alele

Answer 1

11) In a dihybrid cross with two unlinked genes, the phenotypic ratio in the F2 generation is 9:3:3:1.

A) Total number of offsprings in F2 is,

= 53+12+13+2

= 80

Expected number of black coat, long tail = 80×(9/16) = 45

Expected number of black coat, short tail = 80×(3/16) = 15

Expected number of brown coat, long tail = 80×(3/16) = 15

Expected number of brown coat, short tail = 80×(1/16) = 5

B)

Phenotype	O	E	O-E	(O-E)^2	(O-E)^2/E
Black coat, long tail	53	45	8	64	1.42
Black coat, short tail	12	15	-3	9	0.6
Brown coat, long tail	13	15	-2	4	0.26
Brown coat, short tail	2	5	-3	9	1.8

Chi square value is,

= 1.42 + 0.6 + 0.26 + 1.8

= 4.08

Degree of freedom = number of phenotypes - 1

= 4 - 1

= 3

For a degree of freedom 3 and 95% significance level, the chi square table value is 7.815 and the p value is close to 0.25

C) The calculated chi square value (4.08) is less than the table value. So, the null hypothesis is true and there is no significant difference between the observed and expected frequencies. So, the genes are assorting independently.