Question

A true breeding red eyed fly with long bristles was crossed to a true breeding white eyed fly with stubble bristles. All F1 progeny are red eyed with stubble bristles. An F1 female is crossed to a tester male and the following numbers of progeny are observed: 56 red eyed, stubble bristles; 72 red eyed, long bristles; 61 white eyed stubble bristles; and 68 white eyed, long bristles. You are asked to perform a chi-squared test for this result.

1. What is the best null hypothesis for this test?

2. What is the expected phenotypic ratio?

3. What is the expected number of individuals in each class?

4. What is the calculated Chi-squared value for independent assortment?

5. If the Chi-squared value had been 0.584 (it's not, by the way.) what is the most accurate conclusion?

Answer 1

Answer:

Cross: True breeding red eyed fly with long bristles x true breeding white eyed fly with subble bristles

Let Eye Color be represented by R (red) and r (white)

Let Bristle type be represented as: l (Long) and L (stubble)

All F1 progeny are red eyed with stubble bristles (RrLl)

F1 female (RrLl) is crossed to a tester male (rrll)

Different gametes formed are: RL, Rl, rL, rl and rl, rl, rl, rl

Punette square for this cross is:

	rl	rl	rl	rl
RL	RrLl	RrLl	RrLl	RrLl
Rl	Rrll	Rrll	Rrll	Rrll
rL	rrLl	rrLl	rrLl	rrLl
rl	rrll	rrll	rrll	rrll

	rl	rl	rl	rl
RL	Red, Stubble	Red, Stubble	Red, Stubble	Red, Stubble
Rl	Red, Long	Red, Long	Red, Long	Red, Long
rL	White, Stubble	White, Stubble	White, Stubble	White, Stubble
rl	White, Long	White, Long	White, Long	White, Long

Therefore, expected Phenotype ratio is: 1:1:1:1

Total number of fly = 257

Expected number for each phenotype = 257/4 = 64.25

• Null Hypothesis: H0: There is no significant difference between observed and expected frequencies (based on independent assortment).
• Alternative Hypothesis: H1: there is a significant difference between observed and expected frequencies .

Phenotype	Observed (O)	Expected ('E)	(O-E)	(O-E)^2/E
Red, Stubble	56	64.25	-8.25	1.059
Red, Long	72	64.25	7.75	0.935
White, Stubble	61	64.25	-3.25	0.164
White, Long	68	64.25	3.75	0.219
Total	257	257	Chi Square (sum)	2.377

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

P-value corresponding to chi square value of 2.377 with 3 degree of freedom is P-Value = 0.49793. The result is not significant at p < 0.05. Therefore, Null hypothesis is not rejected.

Q:If the Chi-squared value had been 0.584 (it's not, by the way.) what is the most accurate conclusion?

P-value corresponding to chi square value of 0.584 with 3 degree of freedom is  P-Value = 0.900085. The result is not significant at p < 0.05. Therefore, Null Hypothesis is not rejected.