Answer:
we will know that the allelic frequencies are for R 0.95 and r 0.05
Explanation:
We know that the population is in Hardy-Winberg equilibrium, we deduce the following formula:
p + q = 1
p2 + 2pq + q2 = 1
data
R: red flower allele
r: allele blor blanca
p would be equal to the allelic frequency R
q will be equal to the frequency allelic r
2p = RR
2q = rr
2pq = Rr
If there are 25 white flowers in 1000 plants, their frequency will be:
2pq frequency of the Rr genotype
white flower = 25/10000 = 0.0025 = rr = 2q = 0.0025
we deduce that q is equal to 0.05
we replace the data with the previous formula
p + q = 1
p = 1-0.05
we get as a result
p = 0.95
if p = 0.95 and q = 0.05
we will know that the allelic frequencies are for R 0.95 and r 0.05
Answer:
B
Explanation:
I've had this question before got it right.
Answer:
Your question is incomplete but supposing that the mother is a recessive allele and the father is dominant, we deduce that his daughter could have curly hair or else wavy hair that would be a combination of the two since both parents can incompletely dominate
Explanation:
The chances that your child has curly hair is 50%, that is, it would be the most likely, and also the genotype that you would need for this child to have curly hair would be: CC with CS or CC with CC
The wrinkled pea will have two recessive alleles on the locus.
Explanation:
Two forms of same gene are called alleles of that gene.
The allele that expresses itself under heterozygous condition is called the Dominant allele.
The recessive allele can express itself only if they are under homozygous combination i.e. both the alleles on homologous chromosome are recessive.
Here wrinkled pea is a recessive trait and if it is expressed in the phenotype then the genotype of the plant must be homozygous for the recessive allele.
Ascomycota basidiomycota deuteromycota zygomycota
