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2. The following are dihybrid test cross data (AaBb x aabb). Do the data support

ID: 193087 • Letter: 2

Question

2. The following are dihybrid test cross data (AaBb x aabb). Do the data support the hypothesis of independent assortment? Complete the table, calculate X, then answer the questions based on your calculations Phenoty pe O-F A-B- 132 A-bb 86 aaB- 90 aabb 126 Totals 434 a. In interpreting this X2 value, you have degrees of freedom b. In this case do you accept or reject the hypothesis that these data approximate the dihybrid test cross ratio expected with independent assortment? What is the probability that the deviations from the expected are due to chance alone? c. 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. Hypothesis 1 A-:1 aa 1 B-: 1 bb x value P value Accept or reject hypothesis? e. In view of the X1 values obtained for each trait individually, how might you account for the dihybrid test cross ratio obtained?

Explanation / Answer

Answer:

Based on the given data:

The given cross is: AaBb x aabb

The different combination of parental gametes is: AB, Ab, aB, ab and ab, ab, ab, ab

Punnett square for the given cross is:

Different phenotypes are their frequency are:

Therefore, the phenotypic ratio is: 1:1:1:1

Expected number of each phenotype = 434/4 = 108.5

Chi Square test for the given cross is:

Degree of freedom = Number of categories - 1 = 4 - 1 = 3

P-Value corresponding to Chi Square value of 15.73 for 3 degree of freedom is: P = 0.001288. The result is significant at p < 0.05.

a) In interpreting the Chi square value, you have 3 degree of freedom.

b) We reject the Null Hypothesis. These data doesnt approximate the dihybrid test cross ratio expected with independent assortment.

c) Probability that the deviations from the expected are due to chance alone:

ab ab ab ab AB AaBb AaBb AaBb AaBb Ab Aabb Aabb Aabb Aabb aB aaBb aaBb aaBb AaBb ab aabb aabb aabb aabb
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