Genetics Please HELP?
1) In fruit flies, long wings (L) is dominant over vestigial wings (l). If a vestigial winged fly is crossed with a homozygous long winged fly, what genotype and phenotype possibilities are expected in the F1 generation? In the F2 generation?
2) In humans, being right-handed (R) is dominant over being left handed (r), and normal vision is dominant over color blindness (which is sex-linked). Two right-handed parents with normal vision have a son who is left-handed and color-blind. Determine the genotypes of the son and both parents. What can you tell me about the phenotypes of the parents?
3) A normal woman whose father was a hemophiliac marries a normal man. Determine the genotype of the woman and the man. Determine the possible genotypes for their children. What percent of male and female children would be hemophiliacs? What percent normal? Would any be carriers?
4) In hogs, a white belt around the middle (M) is dominant over being beltless (m), and syndactyly, or fused hooves (F) are dominant over split, or normal, hoof (f). A uniformly colored hog homozygous for fused hooves is mated with a homozygous belted hog with split hooves. Determine the genotypes of the parents and the possible genotypes and phenotypes of the offspring.
- 9 years agoFavorite Answer
1)F1 generation 'll be Heterozygous Longed winged in F2 generation Crossing F1 F1 generation 'll lead to reappearence of Both homozygous long winged 25% and homozygous vetigal winged flies 25% and heterozygous dominant Long winged 50%.
2) R is for right hand and r is for left handed. Suppose Normal eyesight gene is represented by E and Colour blind by e b/c the genes for eyesight are present over X chromosome and Y does not has any gene for eyesight.Therefore the genotype of son 'll be rrX(e)Y.the genotype of parents should be heterozygous Right handed and Heterozygous normal eyesight i.e RrX(E)X(e) (for mother)
For father it should be RrX(E)Y.
3)Represent haemophilic genes with h and normal with H.The genotype of woman 'll be X(H)(h) b/c father also contributed an X chromosome 'ving h gene on it.The genotype of her children with a normal man 'll be X(H)X(H),X(H)Y,X(H)X(h),X(h)Y.
It means that half of their daughters 'll be normal totallyi.e 25%,half carriers 25%,half of their boys 'll be normal 25% and half 'll be Hemophilics 25%.
4)The genotype of uniformaly coloured hog is mmFF and the other is MMff and the genotype of all childrens 'll be MmFf.Source(s): i am also a student of biology.