Answer:
1/2
Explanation:
This question involves a single gene coding for feather color in chickens. The black allele (B) is incompletely dominant over the white allele (W). This means that the allele B will not mask the phenotypic expression of allele W, hence, a third intermediate blue phenotype (BW) will be produced.
According to the question, if two blue chickens are crossed i.e. BW × BW, the gametes B and W will be produced by each parent. Using these gametes in a punnet square (see attached image), the following offsprings will be produced:
BB, BW and WW in the ratio 1:2:1
BB is black, BW is blue, WW is white. Hence, the probability of producing a blue chicken from this cross is 1/2.
Answer:
Explanation
Given that 36% are recessive in traits
100-36 = 64% for dominant traits considering a whole population to be 100%
P=dominant allele
q= recessive allele
P2= dominant genotype
q2= recessive genotype
according to hardyweinberg principle, p+q=1
64/100= 0.64 frequency for dominant traits or genotype, therefore
p2=0.64
then
P=√0.64
p= 0.8
Therefore, dominant allele frequency (p) for the population is 0.8
Answer:
The genotypes of the rooster and the chicken are homozygous and that of their offspring is heterozygous.
This case is called codominance, where the offspring receives an allele from each parent, from the rooster and the hen, because there is codominance, so neither allele is recessive and the phenotype of both alleles is expressed so the phenotype of the offspring is checkered.
We can say then that the chicken and the rooster have equal strength between their alleles. in the cases of codominance the laws of mendel do not apply.
Answer:
Cellular respiration, the process by which organisms combine oxygen with foodstuff molecules, diverting the chemical energy in these substances into life-sustaining activities and discarding, as waste products, carbon dioxide and water.
Explanation:
Brainliest please?
