Question

Assuming H-W equilibrium, find both the allele and genotype frequencies. a. In Drosophilia, the allele for normal length wings is dominant over the allele for vestigial wings. In a population of 1,000 individuals, 160 show the recessive phenotype.

Answer 1

Using the Hardy-Weinberg principle, we calculated that in a population of 1,000 Drosophila, with 160 showing the recessive phenotype for vestigial wings, the frequency of the recessive allele 'a' (q) is 0.4, and the dominant allele 'A' (p) is 0.6. The corresponding genotype frequencies are: AA (36%), Aa (48%), and aa (16%).

Step-by-step explanation:

When dealing with a population that is in Hardy-Weinberg equilibrium, it is essential to remember that allele and genotype frequencies are stable and can be predictable. In the case of Drosophila, given that 160 out of 1000 individuals show the recessive phenotype for vestigial wings, we can presume that these are homozygous recessive individuals (aa). Therefore, the frequency of the recessive allele 'a', represented as q, can be found by taking the square root of the recessive phenotype frequency, which is 160/1000 or 0.16. Taking the square root gives us q = 0.4. Following the Hardy-Weinberg equation, p + q = 1, we can then determine the frequency of the dominant allele 'A', represented as p, which would be 1 - q, thus p = 0.6.

With these frequencies, it is then possible to calculate the genotype frequencies for the population. The frequency of homozygous dominant genotype (AA) is p², the frequency of heterozygous genotype (Aa) is 2pq, and the frequency of homozygous recessive genotype (aa) is q². With p = 0.6 and q = 0.4, we have the following frequencies: AA = (0.6)², Aa = 2(0.6)(0.4), and aa = (0.4)².

The Hardy-Weinberg principle is a fundamental concept in biology that provides a mathematical baseline to compare actual populations to an ideal non-evolving population. This allows scientists to understand and predict how evolutionary forces could be acting upon different alleles and genotypes within a population.

Answer 2

Hardy Weinberg Equilibrium

Step-by-step explanation:

• The frequency of the prevailing (typical) allele in the populace (p) is basically 1 - 0.02 = 0.98 (or 98%). The level of heterozygous people (bearers) in the populace.
• 2pq equivalents the recurrence of heterozygotes or transporters, at that point the condition will be as per the following: 2pq = (2) .
• The every individual's genotype, check the quantity of duplicates of every allele, and separation by the total number of gene copies.
• Sickle cell attribute was identified in 20.2% of populace though 1.2% was influenced with sickle cell iron deficiency. The allele recurrence of HbA and HbS alleles were 0.89 and 0.11.
• p2 speaks to the recurrence of the homozygous genotype AA, q2 speaks to the recurrence of the homozygous genotype aa, and 2pq speaks to the recurrence of the heterozygous genotype Aa.
• The entirety of the allele frequencies for all the alleles at the locus must be 1, so p + q = 1.