Question

Mendel formulated the law of independent assortment as his second law of inheritance. It involvestwo genes independently segregated during reproduction, each independently determining one aspectof the phenotype. In one experiment, Mendel crossed pea plants producing yellow, round seeds withpea plants producing green, wrinkled seeds. The first generation resulted in only plants producingyellow, round seeds. Self-crossing of the F1 yielded the following phenotypes in F2:

Phenotype Yellow Round Yellow Wrinkled Green Round Green wrinkled Counts 315 101 108 32 Assuming two independent genes, each with dominant allele and a recessive allele, we would like to find a 9:3:3:1 phenotype ratio.

a. What are the null and alternative hypotheses for chi-square goodness-of-fit test?
b. What are the expected counts under the null hypothesis?
c. Calculate the value of the chi-square test statistic.
d. Do the data agree with Mendel’s second law of genetic inheritance?
e. Compare your results by chisq.test in R to perform the test?

Answer 1

Explanation:

total = 315+101+108+32 = 556

total ratio = 9+3+3+1 = 16

1.

h0 : data follows ratio 9:3:3:1

h1: data does not follow ratio 9:3:3:1

2. expected counts under h0

Ei = expected count, i = 1,2,3,4

count 315

ratio = 9

Ei = 9*556/16

= 312.75

count 101

ratio = 3

Ei = 3*556/16

= 104.25

count 1o8

Ei 3 x 556/16

= 104.25

count 32

= 1*556/16

= 34.75

total = 312.75+104.25+104.25+34.75 = 556

c. value of chi² test stat

a. Oi = 315

Ei = 312.75

X² = (315-312.75)²/312.75

= 0.01618705

b. (101-104.25)²/104.25

x² = 0.101318944

c. (108-104.25)²/104.25

= 0.134892086

d. (32-34.75)²/34.75

= 0.217625859

Chi² test stat = 0.01618705+0.101318944+0.134892086+0.217625859

= 0.470023981

n = 4

df = 4-1 = 3

test stat = 0.4700

critical x² = 7.814736

p value = 0.9254

α < pvalue

0.05<0.9254

we therefore fail to reject the h0. we conclude that it follows 9:3:3:1

d. yes it agrees with this law