Question

activity: monohybrid cross
8 parts | 2.5 of 5 points
the result of the following cross indicates that the genotype of the male parent is _.
p generation
phenotypes: wild type head top flat head top
f, generation
phenotype: all wild type head top
rr
rr
r
rr
r

Explanation:

Step1: Determine dominant and recessive traits

Since all \(F_1\) generation have wild - type head top, wild - type is dominant. Let \(R\) represent the dominant allele (wild - type) and \(r\) represent the recessive allele (flat head top). The female parent has wild - type head top. If the male parent had a dominant allele (\(R\)), for example, if it was \(Rr\) or \(RR\), but since all offspring show the dominant trait. If the male was \(Rr\), there would be a chance of getting recessive offspring (if the female was \(Rr\) as well, but we assume simple monohybrid cross with one gene). The only way all offspring show the dominant trait when one parent is recessive (flat head top, which must be \(rr\) because recessive traits are expressed in homozygous form) and the other is dominant (\(RR\)) is when the cross is \(RR\times rr\).

Step2: Analyze the cross

The cross \(RR\times rr\):
The female (assuming wild - type is dominant) could be \(RR\) (homozygous dominant) and the male (flat head top) is \(rr\) (homozygous recessive). Using Punnett square:

Gametes of female (\(R\))	Gametes of female (\(R\))
Gametes of male (\(r\))	\(Rr\)	\(Rr\)

All \(F_1\) generation will have genotype \(Rr\) (wild - type phenotype).

Answer:

\(rr\)