1 / 7

Therefore, you know that cows A and C are homozygous recessive (pp)

0. 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 A, which is horned (), a horned () calf is produced;  2) With a polled cow B a horned calf is produced;

rusty
Télécharger la présentation

Therefore, you know that cows A and C are homozygous recessive (pp)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 0 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 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?

  2. 0 Therefore, 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)

  3. 0 Phenotypic 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)

  4. 0 In 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.

  5. 0 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.

  6. 0 In 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:  wwSS   x  WWss Therefore, all of the progeny will have a widow's peak and short fingers (Ww Ss)

  7. 0 If 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:  WwSs   x  WwSs Therefore, the phenotypic ratios expected among the F2 are: 9/16 widow's peak, short fingers 3/16 widow's peak, long fingers 3/16 continuous hairline, short fingers 1/16 continuous hairline, long fingers

More Related