Question

5. You are studying inheritance of various characters in marigolds including flower color (purple or blue), seed color (white or brown) and seed shape (round or oblong). When you cross a plant that is true breeding for blue flowers, white and oblong seeds with a plant that is true breeding for purple flowers, brown and round seeds you find that all of the offspring have blue flowers, and round, white seeds. You then cross these F1 plants with each other and you get the following results:

Blue, brown, oblong - 106

blue, white, oblong - 327

blue, white, round - 1002

blue, brown, round - 338

purple, brown, oblong - 42

purple, white, oblong - 122

purple, white, round - 318

purple, brown, round - 102

a) Which traits are dominant and which are recessive?

b) What are the genotypes of the true breeding parent plants?

c) Use a chi square goodness of fit test to determine whether or not the inheritance of these traits by the F2 plants (the offspring of the F1 crosses) meet what you expect if each of the traits are inherited in a Mendelian fashion.

Answer 1

A - When we cross two true-breeding plants then in the F1 generation only dominant allele shows their expression.

Given that in the phenotype of F1 generation plant was blue flowers, and round, white seeds. It means that Blue flower is dominant over the purple flower. Round seed shape is dominant over the oblong seed shape and White seed are dominant over the brown seed.

So dominant alleles are Blue flower, Round and white seed

Recessive alleles are Purple flower, Oblong and brown seed.

B - Let A be the allele for blue flower and a for purple flower.

B be the allele for Round seed and b for oblong seed.

C be the allele for White seed and c for brown seed.

Given that parent plants were true-breeding, it means that they were homozygous for all the alleles.

Genotype of true breeding for blue flowers, white and oblong seeds - AAbbCC

The genotype of true-breeding purple flowers, brown and round seeds - aaBBcc.

C - It is a trihybrid cross, in trihybrid cross the phenotype ratio in F2 generation is 27:9:9:9:3:3:3:1.

Calculated Chi-square value is 3.544

Degree of freedom = number of phenotype -1

Number of phenotype = 8

Degree of freedom df = 8-1 =7.

For 3 degree of freedom and p-value =0.05 chi-square value is 14.7

3.544< 14.7

Since the calculated value of chi-square is 3.54 which is less as compared to the standard value of chi-square for 7 degrees of freedom, so w expect if each of the traits is inherited in a Mendelian fashion.

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