Question

A brown Golder Retriever with short hair (dog X) is of an unknown genotype. To identify the genotype, a test cross was carried out between this dog and another Golder Retriever with light-color and long hair, both of which are known to be recessive traits. 50% of the F1 offspring from this cross are brown with short hair and the other 50% are light-coated with short hair.

A) What is the genotype of dog X? Explain your answer (you can use a Punnett square to check your answer)

B) A cross between two individuals from the F1 generation, a brown dog with short hair (dog Y) and a light-coated dog with short hair (dog Z), is carried out and yield the F2 generation. Using a Punnett square, determine the yield of the F2 generation (both genotypes and phenotypes) and include the probability or percentage for each of the genotype and phenotype.

Answer 1

Answer A:

We observe that both the offspring of the test cross have short hair, which shows that receive gene is missing from our unknown genotype. At the same time we have 50% with brown hair (Dominant) and 50% with light hair (recessive). Considering these the Genotype of the unknown brown Golden Retriever (Dog X) will be BbSS. See Punnet square below:

here we use the convention -
Brown colour - B (Dominant) / Light colour - b (recessive)
Short hair - S (Dominant) / Long hair - s (recessive)

Answer B:

The progeny F1 generation from the above cross having genotypes BbSs and bbSs are crossed to get the F2 generation as shown in the Punnet square below:

The phenotypic probabilities and percentages of the F2 generation are:

• Brown with short hair = 3 /8 or 37.5 %
• Brown with long hair = 1/8 or 12.5 %
• Light with short hair = 3/8 or 37.5 %
• Light with long hair = 1/8 or 12.5 %

The genotypic probabilities and percentages of F2 generation are:

• BbSS - is 1/8 or 12.5%
• BbSs - is 2/8 or 25 %
• bbSS - is 1/8 or 12.5 %
• bbSs - is 2/8 or 25 %
• Bbss - is 1/8 or 12.5 %
• bbss - is 1/8 or 12.5%