Suppose two genes determine shape and color of peas. Gene G codes for color and G...

Suppose two genes determine shape and color of peas. Gene G codes for color and G is dominant. A pea that is GG or Gg is green, a pea that is gg is yellow. The gene R codes for shape. A pea that is RR or Rr is round, a pea that is rr is oval. If we cross two peas that are Gg and Rr we expect the following distribution of peas: (9/16) are green and round, (3/16) are green and oval, (3/16) are yellow and round and (1/16) are yellow and oval. The data we observe are as follows: 40 green and round peas, 25 green and oval peas, 20 yellow and round peas and 15 yellow and oval peas.

(a) State a test statistic procedure that can be used for this problem.

(b) State the null and alternative hypothesis.

(c) Test if the the peas crossed could reasonably have been heterozygous for both genes (Gg,Rr) at α = .05.

Expert Solution

Suppose two genes determine shape and color of peas. Gene G codes for color and G is dominant. A pea that is GG or Gg is green, a pea that is gg is yellow. The gene R codes for shape. A pea that is RR or Rr is round, a pea that is rr is oval. If we cross two peas that are Gg and Rr we expect the following distribution of peas: (9/16) are green and round, (3/16) are green and oval, (3/16) are yellow and round and (1/16) are yellow and oval. The data we observe are as follows: 40 green and round peas, 25 green and oval peas, 20 yellow and round peas and 15 yellow and oval peas.

(a) State a test statistic procedure that can be used for this problem.

Chi square goodness of fit is used

(b) State the null and alternative hypothesis.

Ho: shape and color of peas are in the expected proportions.

H1: shape and color of peas are not in the expected proportions.

(c) Test if the peas crossed could reasonably have been heterozygous for both genes (Gg,Rr) at α = .05.

observed	expected	expected
40	56.25	`=(9/16)*100
25	18.75	`=(3/16)*100
20	18.75	`=(3/16)*100
15	6.25	`=(1/16)*100

Goodness of Fit Test

	observed	expected	O - E	(O - E)² / E
	40	56.250	-16.250	4.694
	25	18.750	6.250	2.083
	20	18.750	1.250	0.083
	15	6.250	8.750	12.250
Total	100	100.000	0.000	19.111

	19.111	chi-square
	3	df
	.0003	p-value

Calculated chi square =19.111 which is > critical chi square with 3 df at 0.05 level 7.81.

Ho is rejected.

We conclude that peas crossed could reasonably have been heterozygous for both genes

orchestra answered 1 year ago

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