Homework Answers
So if the the Genes are not linked then they will follow independent assortment and expected phenotypic ratio is 1:1:1:1
Total number of observed offspring is 1000.
Expected number of wild type phenotype is = 1000×1/4 =250
Expected number of black vestigial phenotype is =1000×1/4=250
Expected number of black,normal phenotype is =1000×1/4=250
Expected number of normal,vestigial phenotype is =1000×1/4=250
Calculated Chi-square value is 397.4
Degrees of freedom = 4-1 =3 (because there are 4 phenotype)
At 3 degrees of freedom the critical value is 7.81 , which is lower than calculated Chi-square value. So the null hypothesis is rejected.
Genes are linked.
Map distance between the Genes = 18.5 map units.
![6t vgt 4,05 (parent] b v vg .LLO bugt loo (Parent) Recombinant) (Recombinant) btvg 85 Totala 1000 Recombination 100 +85 frequ](http://img.homeworklib.com/questions/c33628c0-0b00-11eb-ad52-ef528a4a6ce1.png?x-oss-process=image/resize,w_560)
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