98 POINTS PLEASE HELP
2 answers:
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Answer:
9 out of 16 offsprings
Explanation:
This question involves two genes in guinea pig; one coding for coat texture and the other for coat color. The alleles for smooth coat (S) and black coat (B) are dominant over rough coat (s) and white coat (b).
In this question, two heterozygous guinea pigs were crossed i.e. SsBb × SsBb. The gametes produced by each parent is as follows: SB, Sb, sB, sb. Using these gametes in a punnet square (see attached image), the proportion of offsprings will be:
Smooth black coat (S_B_) = 9
Smooth white coat (S_bb) = 3
Rough black coat (ssB_) = 3
Rough white coat (ssbb) = 1
Hence, according to this question, 9 out of 16 offsprings will have a smooth black coat on average.
The male fruit fly has an XY profile of the sex chromosomes, and the female has an XX profile of the sex chromosomes.
If a male is white-eyed, then it must have a white recessive allele on his only X chromosome and the red eyed female can be homozygous (having two dominant red eye alleles) or heterozygous (having both a dominant and a recessive allele).
If the female is homozygous then all of the offspring will be red-eyed, and if she's heterozygous then half of the offspring will be white eyed. (see the table provided below)
If a male is white-eyed, then it must have a white recessive allele on his only X chromosome and the red eyed female can be homozygous (having two dominant red eye alleles) or heterozygous (having both a dominant and a recessive allele).
If the female is homozygous then all of the offspring will be red-eyed, and if she's heterozygous then half of the offspring will be white eyed. (see the table provided below)