brid test cross data (AaBb x aabb). Do the data suppor 2. The following are dihy

Question

brid test cross data (AaBb x aabb). Do the data suppor 2. The following are dihy the hypothesis of independent assortment? Complete the table, calculate Xand answer the questions based on your calculations. Phenotype O-E | (O-E) (0-EHE 132 1029 109 109 109 4 36 A-B- 23 S29 4. 8s 4.85 A-bb 86 -23 aaB- 90 19 3 61 3.71 aabb 126 17 2 89 2 6 16 a. In interpreting this X value, you have_3_degrees of freedom. Totals 4 34 1708 2 b. What is the probability that the deviations from the expected are due to chance alone? c. In this case do you accept or reject the hypothesis that these data approximate the dihybrid test cross ratio expected with independert assortment? pe d. Complete the following table for the data given above. Determine whether each gene is behaving individually as you would expect in a monohybrid test cross. Each trait considered individually should be expected to approximate a 1:1 ratio as a consequence of Mendel's law of segregation. Accept or reject hypothesis? Hypothesis 1 A-: 1 aa 1 B-: 1 bb x value P value e. In view of the X values obtained for each trait individually, how might you account for the dihybrid test cross ratio obtained?

Explanation / Answer

d)

Hypothessis

X2

P

Accept or reject hypothesis

1A:1aa

15.74

>0.005

Reject

1B:1bb

15.74

>0.005

Reject

Note : Observed and expected values are taken from previous table.

Since our calculated value of X2 exceeds the critical value of X2 we reject the hypothesis, and we can say that there is deviation from expected value.

e) The 1:1:1:1 phenotypic ratio is the classic Mendelian ratio for a test cross in which the alleles of the two genes assort independently into gametes (AaBb × aabb): shown below

The ratio is 4:4:4:4 , that is 1:1:1:1

Hypothessis

X2

P

Accept or reject hypothesis

1A:1aa

15.74

>0.005

Reject

1B:1bb

15.74

>0.005

Reject

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