Question

GENETICS:

Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced in the following frequency: 30 = tall; 20 = short What frequency of tall and short plants is expected? Compute a Chi-square value associated with the appropriate test of significance. How many degrees of freedom are associated with this test of significance?

Answer 1

Ans. Let the dominant allele for height = T

               The recessive allele for height = t

Genotype of heterozygous tall plant = T t

                        Gametes produced by it = T, t

Genotype of homozygous small plant = t t

                        Gametes produced by it = t

Now, Tt x tt

            [T, t] x [t] = Tt, tt      

Total number of progeny = 2 , (Tt and tt)

Phenotypic ration = Tall : short = 1: 2

So, if the total number of progeny were 50, the theoretical outcomes would have been 25 tall and 25 short (1:1 ratio).

So,

Ans. a. Theoretical (expected) phenotypic frequency = 25 : 25

Ans. b. See figure

Ans. c. degree of freedom = 1

Expected Observed 25 27.50 0.23) 25 22.50 (0.28) 50 30 27.50 (0.23) 20 22.50 (0.28) 50 Tall Short 45 100 X2-1.010. df-1,2Hdf-1.01, P(x2 > 1.010): 0.3149