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Suppose that the
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17. Since true breeding fabulous father (AABB) mates with a true breeding plain mother (aabb), the gametes produced in the F1 generation will have the genotype of AaBb. When AaBb is crossed with aaBB, the resultant ratio of genotype will be 1:1:1:1 and phenotypic ratio will be 1:1 in F2 generation.

AB

Ab

aB

ab

aB

AaBB

AaBb

aaBB

aaBb

aB

AaBB

AaBb

aaBB

aaBb

aB

AaBB

AaBb

aaBB

aaBb

aB

AaBB

AaBb

aaBB

aaBb

d. It is not an example of dominant or recessive epistasis.

c. Here no information is given about which allele can mask the activity of which alleles and the phenotypic ratio of the F2 generation do not match with any of the epistatic conditions. Both the enzymes 1 and 2 are necessary for the synthesis of fabulous protein (phenotype), so any one enzyme only can't produce fabulous phenotype. In case of aaBB and aaBb genotypes, the phenotype will only be plain as no production of enzyme 1 can occur rendering production of enzyme 2 useless as it can only act on smashing to syntheisze fabulous protein.

18. The above pedigree shows X-linked dominant inheritance pattern because of the following reasons:

  1. Dominant as there is no generation skipping.
  2. X linked dominant as it follows DDD rule where in a dominant condition if the father is affected, all daughters will be affected too. Thus, it’s an example of X-linked dominant pedigree.
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