Are the following progeny numbers consistent with the results expected from self
ID: 54814 • Letter: A
Question
Are the following progeny numbers consistent with the results expected from selfing a plant presumed to be a dihybrid of two independently assorting genes, H/h ; R/r? (H hairy leaves; h smooth leaves; R round ovary; r elongated ovary.) please show your work and explain your answer. hairy, round 180 hairy, elongated 60 smooth, round 52 smooth, elongated 30 Are the following progeny numbers consistent with the results expected from selfing a plant presumed to be a dihybrid of two independently assorting genes, H/h ; R/r? (H hairy leaves; h smooth leaves; R round ovary; r elongated ovary.) please show your work and explain your answer. hairy, round 180 hairy, elongated 60 smooth, round 52 smooth, elongated 30 Are the following progeny numbers consistent with the results expected from selfing a plant presumed to be a dihybrid of two independently assorting genes, H/h ; R/r? (H hairy leaves; h smooth leaves; R round ovary; r elongated ovary.) please show your work and explain your answer. hairy, round 180 hairy, elongated 60 smooth, round 52 smooth, elongated 30Explanation / Answer
A dihybrid cross for these two traits would be as follows:
Hh Rr x Hh Rr
9 H-R- (hairy and round)
3 H-rr (hairy and elongated)
3 hhR- (smooth and round)
1 hhrr (smooth and elongated)
Expected numbers can be known using Chi-square test:
Phenotype Class H-R- = 9/16 (322) = 181. ......... 2 = (180-181) 2 /181 = 0.0055
Phenotype Class H-rr = 3/16 (322) = 60. ......... 2 = (60-60) 2/ 60 = 0
Phenotype Class hhR- = 3/16 (322) = 60. ........ 2 = (52-60) 2/ 60 = 1.06
Phenotype class hhrr = 1/16 (322) = 20. ........ 2 = (30-20) 2 / 20 = 5
The total of the chi-square = 5 + 1.06 + 0 + 0.0055 = 6.0655
Degree of freedom = 4-1 = 3
The P-value is 0.1084.
P > 0.05. The result is significant.
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