Presentation on theme: "In cattle the polled (hornless) condition (P) is dominant over the horned (p) phenotype. A particular polled bull is bred to three cows. 1) With cow."— Presentation transcript:
1In cattle the polled (hornless) condition (P) is dominant over the horned (p) phenotype. A particular polled bull is bred to three cows. 1) With cow A, which is horned (), a horned () calf is produced; 2) With a polled cow B a horned calf is produced;3) With a horned cow C () a polled calf is produced. What are the genotypes of the bull and the three cows, and what phenotypic ratios do you expect in the offspring of these three matings?
2Therefore, you know that cows A and C are homozygous recessive (pp)Because the polled bull (Pp or PP) produced horned calves, you know it must be heterozygous (Pp)Because the polled bull (Pp) and polled cow B produced a horned calf (pp), cow B must also be heterozygous (Pp)
3Phenotypic ratios expected from each mating:Cow A or C (pp) x bull (Pp)F1: 1/2 polled (Pp) : 1/2 horned (pp)Cow B (Pp) x bull (Pp)F1: 3/4 polled (1/4 PP + 1/2Pp) : 1/4 horned (pp)
4In humans, differences in the ability to taste phenylthiourea are due to a pair of autosomal alleles. Inability to taste is recessive (t) to ability to taste (T). A child who is a nontaster (tt) is born to a couple who can both taste the substance. What is the probability that their next child will be a taster?Note that unaffected parents (tasters) have an affected child (nontaster).Therefore, the parents must both be heterozygous (Tt)A cross between two heterozygous individuals will produce 3/4 tasters (1/2 Tt + 1/4 TT) and 1/4 nontasters (tt)Therefore, the probability that their next child will be a taster is 3/4.
5Parents: Roan (RR') x Roan (RR') In shorthorn cattle the heterozygous condition of the alleles for red coat color (R) and white coat color (R') is roan coat color. If two roan cattle are mated, what proportion of the progeny will resemble their parents in coat color?Parents: Roan (RR') x Roan (RR')F1: 1/4 Red (RR) : 1/2 Roan (RR') : 1/4 White (R'R')Therefore, 1/2 of the progeny will resemble their parents.
6In humans, widow's peak (W) is dominant over a continuous hairline (w) and short fingers (S) are dominant over long fingers (s). If a man that is homozygous for continuous hairline and short fingers marries a woman that is homozygous for widow's peak and long fingers, what will be the phenotypes expressed for these two characteristics by this couple's offspring?Note that this is a dihybrid cross between two homozygous parents: ww SS x WW ssTherefore, all of the progeny will have a widow's peak and short fingers (Ww Ss)
7If one of these children marries a mate with the same genotype, what will be the phenotypes expressed by their children and in what ratios?Note that this is a "self" cross of double heterozygous parents: Ww Ss x Ww SsTherefore, the phenotypic ratios expected among the F2 are:9/16 widow's peak, short fingers3/16 widow's peak, long fingers3/16 continuous hairline, short fingers1/16 continuous hairline, long fingers