Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A cross was made between two strains of whea. One strain was drought-sensitive b

ID: 34284 • Letter: A

Question

A cross was made between two strains of whea. One strain was drought-sensitive but disease-resistant; the other strain was drought-tolerant but disease-sensitive. In the F1 generation, all wheat pleants were drought-sesitive and disease-sensitive. Plants of the F1 generation were then allowed to self-fertilize. The numbers of plants in the F2 generation were as follows:

87 drought-sensitive and disease-sensitive

31 drought-tolerant and disease-sensitive

30 drought-tolerant and disease-resistant

12 drought-tolerant and disease-resistant

a) Make a hypothesis about the inheritance of these traits.

b) Under this hypothesis, how many plants of each phenotype would you expect to get in the F2 generation in a sample of the same size?

c) Test the goodness of fit between your hypothesis and the data using a chi-square test. What do you conclude from the chi-square test ?

Please show work so I can understand where you're coming from. Points will be rewarded to best answer/explaination. Thank you !

Explanation / Answer

a. HYPOTHESIS : According to Mendel's laws of inheritance, the phenotypic ratio of F2 generation = 9:3:3:1

b. Total number of plants given = 160.

   Expected ratio of drought-sensitive & disease-sensitive plants = 9/16. Therefore, expected number = (9/16)*160 = 90

   Expected ratio of drought-tolerant & disease-sensitive plants = 3/16. Therefore, expected number = (3/16)*160 = 30

   Expected ratio of drought-sensitive & disease-resistant plants = 3/16. Therefore, expected number = (3/16)*160 = 30

   Expected ratio of drought-tolerant & disease-resistant plants = 1/16. Therefore, expected number = (1/16)*160 = 10

c. Chi-square test :

Expected ratio

Chi square value = 0.53

No. of degrees of freedom = (No. of phenotypes) - 1 = 4-1 = 3

Probability of a match = 0.05

Critical value = 7.82

According to the above calculation, our chi square value doesn't exceed the critical value.

Conclusion : The given data correctly fits into the hypothesis.

Drought-sensitive&Disease-sensitive Drought-tolerant & Disease-sensitive Drought-sensitive&Disease-resistant Drought-tolerant & Disease-resistant Observed-number(O) 87 31 30 12

Expected ratio

9/16 3/16 3/16 1/16 Expected number(E) 90 30 30 10 (O-E) -3 1 0 2 (O-E)2 9 1 0 4 (O-E)2/E 0.1 0.03 0 0.4
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote