Question

A plant breeding class scored the seed color and leaf morphology phenotypes of a dry pea population. Seed color was scored based on its genotype (yellow vs. green vs. mixed) and leaf morphology by its phenotype (normal vs. tendrilled). Both traits are controlled by a single gene with complete dominance. The data they collected for leaf morphology were: 76 normal: 18 tendrilled and for seed color: 18 yellow: 24 green: 52 mixed. Combined phenotypes for these traits were 3 yellow, normal; 49 mixed, normal; 24 green, normal; 15 yellow, tendrilled; 3 mixed tendrilled; 0 green, tendrilled. Use a Chi square analysis to, first, determine if each gene separately fits the expected ratios and, secondly, us e the Chi square analysis to determine if the combined phenotypes fit the expected ratio assuming independent assortment. If you reject the hypothesis that the genes assort independently, explain why.

Answer 1

As per the statement, the two pairs are completely dominant over its allele. In a monohybrid cross the ratio of dominant and recessive traits is 3:1. It is 3 for the dominant trait and 1 for the recessive trait.

The total offspring are 94. Out of which 76 is normal and 18 is tendrilled. As per the 3:1 ratio of the monohybrid cross, 71 should be normal, 23 should be tendrilled.

Similarly , if we consider the seed color, 18 yellow, 24 green and 53 are mixed. If it is complete dominance , we do not get mixed phenotype. It has to be either green or yellow which ever is completely dominant. So, here also we should get 71 and 23 dominant and recessive traits respectively. But as monohybrid cross, we are getting 52 mixed colored seeds. This shows that the seed color is not completely dominant. It shows co-dominance because both colors are visible as it is mixed colored.

So , if we consider each gene independently, gene for tendril / normal fits the complete dominance ratios. But the seed color does not fit the complete dominant ratios.

Combined phenotypes of the two traits---

When two pairs of genes assort independently, the phenotypic ratio should be 9:3:3:1.

In the cross given, there are 5 different phenotypes. Which means , we should get yellow - normal,yellow -tendril,green -normal, green-tendril. But the cross gave mixed tendril and mixed green. It does not have the phenotype green normal. When we considered monihybrid cross, we observed green -24 and yellow - 18. Based on this if we consider green as dominant and do chi - square analysis, the result we get is ---

Dihybrid cross ratio--9:3:3:1.

Total offspring---94. Green -normal --52; green- tendril--18; yellow-normal--18; yellow-tendril-6

The Chi-square results show that the P value is much lower than the Chi-square value. We totally reject the Null hypothesis that the cross is independent assortment because the phenotypes of the cross do not accept the hypothesis. We reject null hypothesis. There are no green- tendril plants which should have been there in the offspring if it is independent assortment. Instead we have mixed-normal and mixed-tendril offspring. Mixed type of plants only tells about co-dominance of the trait.