1. In mice, one of the gene determining coat color is the Agene, which has multple alleles. two of these are an Allele forblack hair (a) and allele for yellow hair(Ay) . (Ay) is completely dominantto a; however, for reasons that are unclear ( at least to me),AYAY mice always die before embroyicdevelopment is completed.
a) two mice with yellow hair mate with each other. what are thegenotype of the mice?
b) for this cross, give the expected genotype (s) and phenotype(s) for pups that survive embroyic development. A punnet square mayhelp you to solve this part of the probelm.
2. The lubber grasshopper is very large, and as nymph is blackwith red and yellow stripes, Assume that individuals with thegenotype RR have red stripes; that individuals with phenotype rrhave yellow stripes; and that individuals with genotype Rr haveboth red and yellow stripes.
a) if you cross two grasshoppers that, as nymphs, had both redand yellow stripes, then give the expected genotype (s) andphenotype (s) for the offspring. A punnet square may help you tosolve this part of the problem.
b) If you cross a grasshopper with both red and yellow stripeswith one that has just red stripes, then give the expected genotype(s) and phenotype (s) for the offspring. Again, a punnet square mayhelp you to solve this part of the problem.
3. The common grackle is a blackbird that is common over most ofthge eastern and central united states. Suppose that, in grackles,the alleles for long tail (L) is cpmpletly dominant to the allelesfor short tail (l). A female with a short tail mates with along-tailed male, and that mating produces a brood of four chicks-of which just one chick survives, and that a short tail.
a) what is the genotype of the female parent?
b)what is the genotype of the male parent?
c) what is the genotype of the one survival chick?
d) imagine that, instead of just one chick surviving, all foursurvived. give the expected genotype (s) and phenotype (s) for thechicks. Apunnet square may help you to solve this part of theproblem.
4. in human, tongue rolling is controlled by a single gene (theR gene); for that gene, the rolling alleles (R) is completelydominant and the non-rolling (r) is reccessive. Also in human,curly hair is controlled by a single gene (the H gene); for thegene, the curly-hair allele (H) is completely dominant and thestraight-hair allele (h) is recessive.
a) there is a curly-haired non-tongue rolling man, and of theman's parents has straight hair. what is man's genotype?
b) there is a straight-haired tongue-rolling woman,and one ofthe woma's parents is a non-roller. what is the woman'sgenotype?
c) the man from part a marries the woman from part b. give theexpected genotypoe (s) and phenotype (s) of their children. Be sureto use a punnet square to solve this part of the problem.
5) In the breeding season, male anole lizards court females bybobbing their heads up and down while displaying a colorful throatpatch. suppose that the allele for fast head-bobbing (F) iscompletely dominant to the allele for slow bobbing (f), and thatthe allele for red throat patch (R) is completely dominant to theallele for a yellow throat patch (r). further, suppose that afemale homozygous recessive for both traits mates with a malehomozygous dominant for both traits.
a) for the offspring of this pairs- the F1 generationgive the expected genotype (s) and phenotype (s).
b) imagine that one of the F1 individuals mates withan unrelated individual of identical phenotype. if this happens,then the offspring of this mating would be expected to have whatgenotype (s) and phenotype (s)? be sure to use a punnet square toslove this part of the problem.
