Determine the Chi Suquare for this table, remember thet yellow mutant is a sex-lincked gene. Show...

Determine the Chi Suquare for this table, remember thet yellow mutant is a sex-lincked gene. Show all the work.

Cross 1: Dumpy Males x Yellow Females
	Males	Females	Total
Wild Type	23	27	50
Mutant A (Yellow )	14	5	19
Mutant B (Dumpy)	6	7	13
Double Mutant AB	4	2	6
Total	47	41

Expert Solution

Cross 1: Dumpy Males x Yellow Females
	Males	Females	Total
Wild Type	23	27	50
Mutant A (Yellow )	14	5	19
Mutant B (Dumpy)	6	7	13
Double Mutant AB	4	2	6
Total	47	41	88 (Grand Total)

Null hypothesis = The genes for color & height are assorted independently.

Phenotype	Number Observed	Number Expected
Wild type male	23	(50 x 47)/88 = 26.70
Wild type female	27	(50 x 41)/88 = 23.30
Mutant A (Yellow) male	14	(19 x 47)/88 = 10.15
Mutant A (Yellow) female	5	(19 x 41)/88 = 8.85
Mutant B (Dumpy) male	6	(13 x 47)/88 = 6.94
Mutant B (Dumpy) female	7	(13 x 41)/88 = 6.06
Double mutant AB male	4	(6 x 47)/88 = 3.20
Double mutant AB female	2	(6 x 41)/88 = 2.80

² = (Observed - Expected)²/Expected

= (23-26.70)²/26.70 + (27-23.30)²/23.30 + (14-10.15)²/10.15 + (5-8.85)²/8.85 + (6-6.94)²/6.94 + (7-6.06)²/6.06 + (4-3.20)²/3.20 + (2-2.80)²/2.80

= 0.51 + 0.59 + 1.46 + 1.67 + 0.13 + 0.15 + 0.2 + 0.23

= 4.94

Degree of freedom (df) = (number of rows – 1) x (number of columns – 1) = (4-1) x (2-1) = 3 x 1 = 3

We find that probability from ² distribution table is 0.1 < P < 0.5.

The probability associated with the calculated chi-square value is between 0.10 and 0.50, indicating a high probability that the difference between observed and expected values is due to chance.

gladiator answered 2 years ago

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