Genes purple eyes (PR) and dumpy wings (DP) are being analyzed for possible link
ID: 79286 • Letter: G
Question
Genes purple eyes (PR) and dumpy wings (DP) are being analyzed for possible linkage. Below are the data coming from a P0cross, and then a test cross in the F1 generation. Use chi-square analysis to determine whether the genes are linked. Hint: your null hypothesis is that the genes are linked. Determine p value to either accept or reject it. Results of Cross #1 Parents (Female: PR,DP) x (Male: F1 Offspring Phenotype Number Proportion Ratio Female: 510 0.51 1.04 Male: 490 0.49 0.96 Total 1000 Results of Cross #2 Ignoring Sex F1 Test Cross (Female: x (Male: PR;DP) F2 Offspring Number Proportion Ratio Phenotype 383 0.3869 3.016 PR 127 0.1283 1.000 DP 129 0.1303 1.016 PR,DP 351 0.3545 2.764 Total 990Explanation / Answer
Chi-square analysis for F1 cross----
Female drosophila --observed---510, expected--500,
male drosophila--observed---490, expected--500,
Chi-square analysis=(O-E)2/E,
female drosophila=(510-500)2/500 = (10)2/500=100/500=0.2.
male drosophila= (490-500)2/500=100/500=0.2,
Total parent number in F2 cross=
total=0.2+0.2=0.4, total chi-square -value for F1 cross=0.4.
From given table for given value of 0.4, the corresponding p-value is--0.5 for degree of freedom equal to value 1.
The p -value is greater than 0.05 so it is not possible to reject the null hypothesis for F1 cross for independent assortment of genes but rejected for linked genes
progeny from F2 cross--
+ drosophila and PR/DP parents chi-square value--observed=383+351=734, expected=495,
chi-square value=(734-495)2/495=115.39/100=1.15.
The p value from table is 0.5 with degree of freedom equal to 2 for Chi-square value 1.15 and degree of freedom 2 , so the value is greater than 0.05 , so it is not possible to reject null hypothesis for independent assortment of genes but rejected for linked genes.
chi-square values for F2 cross----
purple eyes and dumpy wings..=
progeny in F2 cross , expected-127 +129=256, observed=495,
chi-square value=(256-495)2/ 495= 239 x239/495 =57121/495=115.39/100=1.15, total 1.15+.1.15=2.3
From table for value 2.3 , the p-value is 0.1, so null hypothesis cannot be accepted , if p value is less than 0.05, then it can be rejected for independent asssortment and accepted for linked genes.
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