part b For cross II, determine whether genetic linkage or independent assortment

Question

part b

For cross II, determine whether genetic linkage or independent assortment is more strongly supported by the data.

a. The chi-square is 225, P values are below 0.01.This supports linkage of the colorless and waxy genes.

b. The chi-square is 225, P values are above 0.01.This rejects linkage of the colorless and waxy genes.

c. The chi-square is 225, P values are above 0.01.This supports linkage of the colorless and waxy genes.

d.The chi-square is 225, P values are below 0.01.This rejects linkage of the colorless and waxy genes

part c

Calculate the recombination frequency for cross I. Enter your answer to two decimal places (example 0.12).

part d

Calculate the recombination frequency for cross II. Enter your answer to two decimal places (example 0.12).

part e

Are the results of these two experiments mutually compatible with the hypothesis of genetic linkage? Are the results of these two experiments mutually compatible with the hypothesis of genetic linkage?

a. Yes, both sets of data are compatible with the hypothesis of genetic linkage.

b. Only sets of data for cross IIis compatible with the hypothesis of genetic linkage.

c. No, both sets of data are not compatible with the hypothesis of genetic linkage.

d. Only sets of data for cross Iis compatible with the hypothesis of genetic linkage.

part f

Merge the two sets of progeny data and determine the combined recombination frequency. Enter your answer to two decimal places (example 0.12).

In Cross 1, pure-breeding colored, starchy kernel plants (C1WxC1Wx) were crossed to plants pure-breeding for colorless, waxy kernels (c1 wx/c1 wx). The Fi of this cross were test-crossed to colorless, waxy plants. The test-cross progeny are as follows: In experiments published in 1918 that sought to verify and expand the genetic linkage and recombination theory proposed by Morgan, Thomas Bregger studied potential genetic linkage in corn (Zea mays) for genes controlling kernel color (colored is dominant to colorless) and starch content (starchy is dominant to waxy). Bregger performed two crosses. Number Phenotype Colored, waxy 310 Colored, starchy 858 Colorless, waxy 781 Colorless, starchy 311 Part A For cross I determine whether genetic linkage or independent assortment is more strongly supported by the data O The chi-square is 464, P values are below 0.01.This rejects linkage of the colorless and waxy genes. O The chi-square is 464, P values are above 0.01.This rejects linkage of the colorless and waxy genes. O The chi-square is 464, P values are above 0.01.This supports linkage of the colorless and waxy genes O The chi-square is 464, P values are below 0.01.This supports linkage of the colorless and waxy genes. 2260 In Cross 2, plants pure-breeding for colored, waxy kernels (Cl wx/C1 wx) and colorless, starchy kernels (c1 Wx/c1 Wx) were mated, and their Fi were test-crossed to colorless, waxy plants. The test-cross progeny are as follows: Submit Number Phenotype Colored, waxy 340 Colored, starchy 115 Colorless, waxy 92 Colorless, starchy 298 Part B For cross II, determine whether genetic linkage or independent assortment is more strongly supported by the data O The chi-square is 225, P values are below 0.01.This supports linkage of the colorless and waxy genes O The chi-square is 225, P values are above 0.01.This rejects linkage of the colorless and waxy genes. 845

Explanation / Answer

Answer:

Part 1) For the given data set:

The expected phenotypic ratio based on test cross is: 1:1:1:1

Total number of F2 progenies = 2260

Since there are 4 phenotypes with equal proportion, therefore expected number of plants for each phenotype = 2260/4 = 565

Chi Square test:

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

P-value corresponding to a Chi square value of 464 with 3 degree of freedom is: P-Value is < 0.00001. The result is significant at p < 0.05.

Therefore, the correct option is: The chi-square is 464, P values are below 0.01. This supports linkage of the colorless and waxy genes.

Part II:

For the given data set:

The expected phenotypic ratio based on test cross is: 1:1:1:1

Total number of F2 progenies = 845

Since there are 4 phenotypes with equal proportion, therefore expected number of plants for each phenotype = 845/4 = 211.25

Chi Square test:

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

P-value corresponding to a Chi square value of 225 with 3 degree of freedom is: P-Value is < 0.00001. The result is significant at p < 0.05.

Therefore, the correct option is: The chi-square is 464, P values are below 0.01. This supports linkage of the colorless and waxy genes.

Phenotype ObservedNumber Expected Number (O-E) (O-E)^2/E Colored, waxy 310 565 -255 115 Colored, starchy 858 565 293 152 Colorless, waxy 781 565 216 83 Colorless, starchy 311 565 -254 114 Total 2260 2260 Chi Square (sum) 464

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