Answer:
The fraction of heterozygous individuals in the population is 32/100 that equals 0.32 which is the genotipic proportion for these endividuals.
Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the exposed example, the r-6 allelic frequency is 0,2. This means that if r-6=0.2, then the other allele frequency (R) is=0.8, and the sum of both the allelic frequencies equals one. This is:
p + q = 1
r-6 + R = 1
0.2 + 0.8 = 1
Then, the genotypic proportion for the homozygous individuals RR is 0.8 ² = 0.64
The genotypic proportion for the homozygous individuals r-6r-6 is 0.2² = 0.04
And the genotypic proportion for heterozygous individuals Rr-6 is 2xRxr-6 = 2 x 0.8 x 0.2 = 0.32
Answer:
The mRNA transcript that is produced in the nucleus is in immature form, it had to get matured, and then its capping and tailing is done. 5' cap is added and a poly A tail is added at 3' end of this mRNA. This helps it to get matured. moreover for its maturation the non-coding regions (introns) are removed and exons are packed up again to form a mature mRNA.
This mature mRNA then proceeds to its way to the ribosomes for translation of proteins.
