In horses, the coat color black is dominant (B) over chestnut (b). The trotting gait is dominant (T) over the pacing gait (t). If a homozygous black pacer is mated to a homozygous chestnut, heterozygous trotter, what will be the probabilities for genotype and phenotype of F1 generation

Explanation: From the information provided, the genotype of a homozygous black pacer is BBtt while the genotype of a homozygous chestnut, heterozygous trotter is bbTt.

A dihybrid cross between BBtt and bbTt will produce sixteen offsprings in F1 generation. The genotypes of the F1 offsprings is 8 BbTt and 8 Bbtt.

The probability of having a BbTt genotype is 8/16 =1/2

The probability of having a Bbtt genotype is 8/16 = 1/2

The genotype BbTt is heterozygous black, heterozygous trotter.

The genotype Bbtt is heterozygous black, homozygous pacer.

The phenotypes of the F1 offsprings are black trotter and black pacer.

The probability of having a black trotter is 8/16 = 1/2

The probability of having a black pacer is 8/16 = 1/2.

mahima6824soni mahima6824soni · Answer 2 · 2022-02-26T07:42:59+01:00

The genotype and phenotype for F1 generation are BbTt and Bbtt and the probability for F1 generation is 50%.

What is a genotype?

The genotype of an organism is the collection of genes, that contribute to the traits and characters of the organism.

In the F1 generation, a dihybrid cross between BBtt and bbTt will generate sixteen offspring.

1. The F1 offspring had genotypes of 8 BbTt and 8 Bbtt.

The probability of BbTt genotype is [tex]\dfrac{8}{16} =\dfrac{1}{2}[/tex]

The probability of Bbtt genotype is [tex]\dfrac{8}{16} =\dfrac{1}{2}[/tex]

2. Black trotter and black pacer are the phenotypes of the F1 offspring.

The probability of black trotter is [tex]\dfrac{8}{16} =\dfrac{1}{2}[/tex]

The probability of black pacer is [tex]\dfrac{8}{16} =\dfrac{1}{2}[/tex]

Thus, the probability for genotype and phenotype of F1 generation is 50%.

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