You are studying different traits in dogs. You decide to look at six independent

Question

You are studying different traits in dogs. You decide to look at six independently segregating traits. Each trait is determined by a single gene with two alleles.
Gene Symbol        Alleles
D/d             Dark fur (D) is dominant to light fur(d)
T/t           Tall (T) is dominant to short (t)
H/h                              Hairy (H) is dominant to hairless (h)
B/b                              Barking (B) is dominant to yapping (b)
S/s                              Suspicious (S) is dominant to trusting (t)
L/l            Long muzzle (L) is dominant to short muzzle (l)

You found a pair of dog of the following genotypes that have just mated:
Female dog - DdTthhBbSSLl
Male dog -      ddTtHhBBSsLl

a.   How many different types of gametes can the female produce?
b.   How many different types of gametes can the male produce?
c.   How many genotypic classes can result from this cross?
d.   How many phenotypic classes can result from this cross?
e.   What fraction of the progeny will be phenotypically identical the female parent?
f.   What fraction of the progeny will be phenotypically identical the male parent?
g.   What fraction of the progeny will be genotypically identical the female parent?
h.   What fraction of the progeny will be homozygous at all seven loci?

Explanation / Answer

a) No of possible gametes produced will be based on the genotype of parent by counting the number of heterozygous allele pairs (denoted by n), found in that particular genotype.

No of possible genotypes=2^n

Here, the female dog is DdTthhBbSSLl

So, n=4

And the number of possible gametes are 2^4=16

Ans: 16 different types of gametes the female can produce.

b) Here, the male dog is ddTtHhBBSsLl

So, n=4

And the number of possible gametes are 2^4=16

Ans: 16 different types of gametes the male can produce.

c) So, the cross is between DdTthhBbSSLl X ddTtHhBBSsLl

When we need to analyze two parents, it is easier to determine how many possible genotypic combinations are present in each gene pair, and then multiply them all together.

Here, 2 possible D genotypes X 3 possible T genotypes X 2 possible H genotypes X 2 possible B genotypes X 2 possible S genotypes X 3 possible L genotypes

So, 2 X 3 X 2 X 2 X 2 X 3 = 144

Ans: 144 genotypic classes will result from this cross.

d) When we need to analyze two parents, it is easier to determine how many possible phenotypic combinations are present in each gene pair, and then multiply them all together.

Here, 2 possible D phenotypes X 2 possible T phenotypes X 2 possible H phenotypes X 1 possible B phenotype X 1 possible S phenotypes X 2 possible L phenotypes

So, 2X2X2X1X1X2 =16

Ans: 16 different phenotypic classes will result from this cross.

If you have more than 4 queries, please ask a separate question.

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