on snapdragons. A true-breeding blue plant was 2. Your research team performs cr

Question

on snapdragons. A true-breeding blue plant was 2. Your research team performs crosses all re fertilized with a true-breeding white plant and the F1 progeny we Crossing two F1 I progeny yielded 71 periwinkle snapdragon plants; 18 blue snapdragon s and 3I white snapdragon plants. Several people in the research team think the due to a single gene with two alleles that exhibit incomplete dominance and the other members believe the results are due to an interaction between two genes. . You want to test the hypothesis that the results are due to one gene with two alleles that have incomplete dominance. sing letters of your own choosing and strategies discussed in class, define th i. U e genotype that corresponds with each phenotype. Then write out the genotypes of the blue and white parent plants, the F1 progeny and the F2 progeny. 1. blue parent plant: 2. white parent plant: 3. Fl progeny: 4. F2 blue progeny: 5. F2 white progeny 6. F2 periwinkle progeny: Predict the phenotypes and phenotypic proportions of the progeny from the F2 results (cross of the F1 progeny). Also, state the expected number of progeny for each phenotype from the cross. ii. Perform a chi-square test to test the hypothesis that the F2 progeny correlate with one gene with two alleles that exhibit incomplete dominance. ili. 1. State your calculated chi-square value 2. How many degrees of freedom are there? 3. What is the critical chi-square value? 4. What is your conclusion?

Explanation / Answer

2) b) In this part of the question thy want to test if this numbers adjust to an interaction between 2 genes. So I will use two letters that indicate two different genes. A and B. But B depends on A.

So, letter A will be for the gene that decide if the flower will be white or colored.

If we have an aa individual, it will be white. No matter what letter B brings.

If we have an individual AA or Aa, it will be colored. If we have bb, it will be periwinkle. If it is BB or Bb it will be blue.

i) I decided to use A and B, keep in mind.

1. blue parent plant : A_B_

2.white parent plant : aa B_/ or bb

3.F1 progeny: Aa bb

4.F2 blue progeny: A_ B_

5.F2 white progeny: aa Bb

6.F2 periwinkle progeny: A_bb

ii)

Observed frequencies:

Blue: 18

White:31

Periwinkle:71

To get the estimated frequencies you need to convert this probabilities given (9:3:4) ratio in numbers, so you can compare them to the observed ones.

We do this first, getting the probabilities and multiplying the total.

Total: 18+31+71 = 120

9/16 periwinkle: 0.5625 x total (120)=67.5

3/16 blue: 0.1875 x total (120)=22.5

4/16 white: 0.25 x total (120)=30

This are our expected numbers of individual of each type of phenotype.

iii)

Now we use the Chi-square test: ?2 in order to compare if the observed ones adjust to the ratio you encountered.

Null hypothesis: Observed phenotypes are the expected ones by the predicted ratios found in the section above.

Chi-square test: ?2

?2 = ? (observed individuals – expected individuals) 2

                         expected individuals

?2 = ? (71-67.5) 2 + (18-22.5)2 + (31- 30)2 =

                 67,5          22.5              30

?2= 1.113

The 5% significance level for 1 degree of freedom is 3.84

1.113 < 3.84

This means we do not reject Ho. The observed phenotypes adjust to the expected ratios and so there is an interaction between this 2 genes.

Good luck!

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