QUESTION 3 A cross was made between a dihybrid female fhuitty anda recesve male

Question

QUESTION 3 A cross was made between a dihybrid female fhuitty anda recesve male tester. The folowing offspring were observed 366 normal phenotype 341 wingless, white eyes 59 wingless, normal eyes 54 normal wings, white eyes Use the chi-square statistic to test the hypothesis that the genes for wing development and eye coloration are on separate chromosomes. What is the x2 value? OA. 5.1 OB 1.3 OC.7.5 O D. 204 OE. 431 Click Save and Submit to save and submit. Click Save All Answers to save all anstcers 17 MacBook Air

Explanation / Answer

Lets denote wingless trait by w and normal wings by W and normal eyes by E and white eyes by e.

A dihybrid female is cross with a tester male. The tester is suppose to be homozygous recessive , therefore his genotype is wwee and female is dihybrid therefore her genotype is WwEe

If genes for wings development and eye colouration are present on different chromosomes, the allele W or w will assort independently of E or e. Allele for normal wings W and wingless will be present on homologous chromosomes. However genes W and E will be present on different chromosomes lets say chromosome 1 and 2 respectively.

WwEe x wwee

Different gametes produced by WwEe will be WE, We, wE, we

gametes produced by wwee will be we only

Progeny : WwEe ( normal phenotype) , Wwee (normal wings and white eyes) , wwEe (wingless and normal eyes), wwee ( wingless and white eyes)

The expected ratio is 1:1:1:1

Expected number of progeny with a particular phenotype , lets say normal phenotype is 1/4 x total number of progeny

observed

(O)

Chi square value is calculated to be 431

The degree of freedom is number of phenotypic classes observed minus 1. Here we have 4 phenotypic classes ( normal phenotype ,normal wings and white eyes , wingless and normal eyes, wingless and white eyes). Therefore degree of freedom is 4-1 =3

Look into the chi square table for Probability value of our calculated chi-square value under degree of freedom =3 row, The last probability value 0.005 in the table is for chi square value of 12.383. Our value is way beyond this value and not available in the chi-square table. Therefore it must be having a probability value less than 0.005

Phenotype O

observed

(O)

Expected (E ) O-E d2 d2/ E normal phenotype 366 1/4 x 820 = 205 161 25921 126.4 wingless, white eyes 341 1/4 x 820= 205 136 18496 90.2 wingless normal eyes 59 1/4 x 820= 205 -146 21316 103.9 wingless, white eyes 54 1/4 x 820= 205 -151 22801 111.2 total progeny = 820 chi square = sum of all d2/E values = 431.7

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