Question

A corn breeder has a plant that is dominant for 4 traits that are known to assort independently (i.e. it’s phenotype is A_ B_ C_ D_). This plant has desirable properties, so the breeder wishes to determine its exact genotype. The breeder performs a test cross and obtains the following results in the F1 generation.

Phenotype Observed # Phenotype Observed # A_ B_ C_ D_ 61 A_ bb cc D_ 0 aa B_C_D_ 79 A_ bb C_ dd 0 A_bb C_D_ 0 A_ B_ cc dd 73 A_ B_ cc D_ 59 aa bb cc D_ 0 A_ B_ C_ dd 76 aa bb C_ dd 0 aa bb C_ D_ 0 aa B_ cc dd 81 aa B_ cc D_ 97 A_ bb cc dd 0 aa B_ C_ dd 74 aa bb cc dd 0
Based on these results, the breeder is confident that they know the original plant’s genotype. Make a hypothesis about the genotype, and test it with a χ2 test. Show this test on a separate page. Please be detailed, thanks!

Answer 1

A test cross is done the check the genotype of the plant. As we can see from the cross result that the gene A is heterozygous and is present in Aa i.e. heterozygous state.

Gene B is dominant state and is present in BB i.e. homozygous dominant form

gene C is heterozygous and is present in Cc i.e. heterozygous state.

gene D is heterozygous and is present in Dd i.e. heterozygous state.

So the genotype of the test plant is AaBBCcDd.

Chi-Square test.

Total number of progeny = 600

Since it is a test cross so the ratio of all the genotype should be in 1:1:1:1:1:1:1:1:1

So each progeny should have 75 number.

For tetrahybrid cross the degree of freedom = number of alleles- 1

= 8-1 = 7

For degree of freedom 7 and p-value = 0.05 the chi sqy=uare value is 14.07.

So our test cross is consistent with the Chi-square value and fail to reject the null hypothesis