Fur color in dogs can be black, brown or white. The B gene is responsible for...
Fur color in dogs can be black, brown or white. The B gene is responsible for fur color. Dogs that inherit a dominant B will have black fur. Dogs that inherit 2 recessive alleles, bb, will have brown color. There is a second gene that contributes to fur color in dogs: gene I. Individuals who inherit a dominant allele, I, at this location, will always have white color. Individuals who inherit the homozygous recessive condition, ii, will be able to make coat color pigments (either brown or black).
a.) The term that best fits with how coat color in dogs is inherited is:
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epistasis |
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multiple alleles |
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pleiotropy |
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sex-linkage |
b.) If two dogs with the doubly heterozygous BbIi genotype are mated, what phenotypic ratios are expected for the offspring of this cross?
c.) Suppose dogs with the following genotypes mated: Bbii x bbIi. What proportion of the offspring from such a cross are expected to be white?
Homework Answers
a) here the expression of genes in the B locus is determined by the genotype at the I locus if the I locus has the dominant allele then B locus is not expressed.so this is an example of dominant epistasis, If the dominant allele I is present phenotype is white regardless of the genotype at the B locus so the answer is epistasis.
( multiple alleles means a single gene has more than 2 alleles, pleiotropy means a single gene controls multiple phenotypes, Sex-linkage means the gene responsible for the phenotype is located on the X chromosome).
b) BbIi * BbIi
| BI | Bi | bI | bi | |
| BI | BBII ( white) | BBIi ( white) | BbII ( white) | BbIi ( white) |
| Bi | BBIi ( white) | BBii (Black) | BbIi ( white) | Bbii (Black) |
| bI | BbII ( white) | BbIi ( white) | bbII ( white) | bbIi ( white) |
| bi | BbIi ( white) | Bbii (Black) | bbIi ( white) | bbii (brown) |
expected proportion of black progeny= number of black progeny / total number of progeny=3/16
expected proportion of brown progeny= number of brown progeny / total number of progeny=1/16
expected proportion of white progeny= number of white progeny / total number of progeny=12/16
c) Bbii * bbIi
| Bi | bi | |
| bI | BbIi ( white) | bbIi ( white) |
| bi | Bbii (Black) | bbii (brown) |
proportion of white progenies= number of white progenies/total number= 2/4=1/2
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