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Title: Fruit Fly Genetics: Wild female X Brown Eye Dumpy wing male Post by: Jedi_Knight on Nov 6, 2015 Using Pearson biologylabsonline website. Performed cross of wild female X brown eye, dumpy wing male and backcrossed F1 wild female offspring with P1 male.
These were the results: Results of Cross #2 BACKCROSS Parents (F1 femals & P1 male) (Female: +) x (Male: BW;DP) Offspring (backcross offspring) Phenotype Number Proportion Ratio Female: + 1419 0.1429 1.391 Male: + 1461 0.1471 1.432 Female: BW 1020 0.1027 1.000 Male: BW 1034 0.1041 1.014 Female: DP 1098 0.1105 1.076 Male: DP 1048 0.1055 1.027 Female: BW;DP 1441 0.1451 1.413 Male: BW;DP 1412 0.1422 1.384 Total 9933 ------------------------------------------------------------------------- Results of Cross #2 BACKCROSS Ignoring Sex Parents (F1 female and P1 male) (Female: +) x (Male: BW;DP) Offspring (Backcross offspring) Phenotype Number Proportion Ratio + 2880 0.2899 1.402 BW 2054 0.2068 1.000 DP 2146 0.2160 1.045 BW;DP 2853 0.2872 1.389 Total 9933 Here is what I don't understand, and the answer I need: How do these genes show independent assortment and what evidence does these data provide to support the answer to this question? Title: Re: Fruit Fly Genetics: Wild female X Brown Eye Dumpy wing male Post by: psyche360 on Nov 7, 2015 Numbers should be equally probably is independent assortment is taking place.
Without really looking at the genes in this particular question, let's say you have: Aa, Ab, Ba, Bb You will need two generations of breeding. (Aa, Ab, Ba, Bb) X (Aa, Ab, Ba, Bb) on the second generation , all 16 will be equally probable at a ratio of 9:3:3:1 phenotype. |