When pure-breeding red cows are bred with pure breeding white cows, the offspring are roan. The roan cows also have a gene for horn vs. hornless cattle. The allele for horns dominates the allele for hornless. If a bull and cow are heterozygous for both genes, what are the possibilities for each possible phenotype?
In a cross between a female AaBbccDdee and a male AabbCcDdee, what proportion of the progeny will be the same phenotype as the female parent?
For the firsts, I think the phenotypes would be AaBb and AABB, but I'm not sure. And for the second one, I'm guessing it's 3/16 because it's a trihybrid cross.
For the first question, let's break it down:
Pure-breeding red cows (RR) bred with pure-breeding white cows (WW) yield roan cows (RW). The roan cows can have either the gene for horns (H) or the gene for hornless (h). The allele for horns (H) is dominant over the allele for hornless (h).
If a bull (AaBbCcDd) and cow (AaBbCcDd) are heterozygous for both genes, the possibilities for each phenotype are as follows:
1. For coat color (red, white, or roan):
- Red (RR): 1/4 probability (from bull: Aa or aa, from cow: AA)
- Roan (RW): 1/2 probability (from bull: Aa or aa, from cow: Aa or aa)
- White (WW): 1/4 probability (from bull: aa, from cow: aa)
2. For horn type (horned or hornless):
- Horned (Hh or HH): 3/4 probability (from bull: Hh or HH, from cow: Hh or HH)
- Hornless (hh): 1/4 probability (from bull: hh, from cow: hh)
Therefore, the possible combinations for each phenotype are:
- RRHH: red, horned
- RRHh: red, horned
- RrHH: red, horned
- RrHh: red, horned
- RRhh: red, hornless
- Rrhh: red, hornless
- RWHH: roan, horned
- RWHh: roan, horned
- RWhh: roan, hornless
- WWHH: white, horned
- WWHh: white, horned
- WWhh: white, hornless
For the second question, let's calculate the proportion of progeny with the same phenotype as the female parent (AaBbccDdee).
Considering a trihybrid cross, the proportion can be determined by multiplying the probability of inheriting each trait for each gene:
- For gene A: 1/2 probability to inherit "Aa" (from male parent Aa) and 1/2 probability to inherit "Aa" (from female parent Aa); thus the probability is (1/2) * (1/2) = 1/4.
- For gene B: 1/2 probability to inherit "Bb" (from male parent bb) and 1/2 probability to inherit "Bb" (from female parent Bb); thus the probability is (1/2) * (1/2) = 1/4.
- For gene C: 0 probability to inherit "cc" (from male parent Cc) and 1 probability to inherit "cc" (from female parent cc); thus the probability is 0 * 1 = 0.
- For gene D: 1/2 probability to inherit "ee" (from male parent ee) and 1/2 probability to inherit "ee" (from female parent ee); thus the probability is (1/2) * (1/2) = 1/4.
To find the overall proportion, we multiply the probabilities for each trait:
Proportion = (1/4) * (1/4) * 0 * (1/4) = 0
Therefore, none of the progeny will have the same phenotype as the female parent (AaBbccDdee).
To determine the possibilities for each phenotype of the offspring when a bull and cow are heterozygous for both genes, we need to consider the principles of Mendelian inheritance.
For the first gene (for coat color):
- The red coat color is dominant, represented by the allele "A."
- The white coat color is recessive, represented by the allele "a."
When a pure-breeding red cow (AA) is bred with a pure-breeding white cow (aa), all offspring will be heterozygous for coat color (Aa), resulting in the roan coat color.
For the second gene (for horns or hornlessness):
- The allele for horns is dominant, represented by the allele "B."
- The allele for hornlessness is recessive, represented by the allele "b."
When a bull and cow are both heterozygous (AaBb) for the second gene, the possible genotypes and phenotypes of their offspring can be determined using Punnett squares:
b B
--------------
a | ab aB
A | Ab AB
The possible genotypes are: AAbb, AABb, AaBB, AaBb.
The corresponding phenotypes are:
- AAbb: Red coat color, hornless
- AABb: Red coat color, horned
- AaBB: Roan coat color, horned
- AaBb: Roan coat color, horned
Therefore, based on the given information, the possibilities for each possible phenotype are AAbb, AABb, AaBB, and AaBb.
For the second question:
In a cross between a female AaBbccDdee and a male AabbCcDdee, we can determine the proportion of progeny that will have the same phenotype as the female parent by considering each gene independently and using the principles of Mendelian inheritance.
Let's analyze each gene and its corresponding phenotypes:
- For the first gene:
The female parent is Aa, while the male parent is Aa. There is a 1/4 chance of producing progeny with the same phenotype as the female parent (AA or aa) and a 1/2 chance of producing progeny with the heterozygous phenotype (Aa). Therefore, the proportion of progeny with the same phenotype as the female parent is 1/4 + 1/2 = 3/4.
- For the second gene:
The female parent is Bb, while the male parent is Bb. There is a 1/4 chance of producing progeny with the same phenotype as the female parent (BB or bb) and a 1/2 chance of producing progeny with the heterozygous phenotype (Bb). Therefore, the proportion of progeny with the same phenotype as the female parent is 1/4 + 1/2 = 3/4.
- For the third gene (not provided in the question):
The female parent is cc, while the male parent is Cc. In this case, all the progeny will have the heterozygous phenotype Cc, since the allele for hornlessness (c) is recessive, and the allele for horns (C) is dominant.
- For the fourth gene (not provided in the question):
The female parent is ddee, while the male parent is ddee. Again, all the progeny will have the same phenotype as the female parent, as there is no variation in the alleles.
Since each gene is inherited independently, the proportions calculated can be multiplied together to determine the overall proportion of progeny with the same phenotype as the female parent:
(3/4) * (3/4) = 9/16
Therefore, the proportion of progeny with the same phenotype as the female parent is 9/16.