Explanation

Only II. It must be homozygous recessive genotypes. Hardy-Weinberg Principle It was proposed by GH Hardy an English mathematician and W Weinberg a German physician independently in 1908 (i) It describes a theoretical situation in which a population is undergoing no evolutionary change. This is called genetic or Hardy-Weinberg equilibrium (ii) It can be expressed as $p^{2} + 2pq + q^{2} = 1\$or$\ (p + q)^{2} = 1$ (iii) Evolution occurs when the genetic equilibrium is up set (evolution is a departure from Hardy-Weinberg equilibrium principle) The sum of total of Allelic frequency $(p + q)is = 1$ $p^{2} + 2pq + q^{2}\$or $(p + q)^{2}$ Where, $p^{2} = \%$ homozygous dominant individuals $p =$ frequency of dominant allele $q^{2} = \%$ homozygous recessive individuals $q =$ frequency of recessive allele $2pq = \%$ heterozygous individuals Realize that $(p + q)^{2} = 1$ (three are only 2 alleles) $p^{2} + 2pq + q^{2} = 1$ (these are the only genotypes) Example An investigator has determined by the inspection that 16% of a human population has a recessive trait. Using this information, we can calculate all the genotypes and allele frequencies for the population, provided the conditions for Hardy-Weinberg equilibrium are met Given $q^{2} = 16\% = 0.16$ are homozygous recessive individuals Therefore, $q = \sqrt{0.16} = 0.4 =$ frequency of recessive allele $p = 1.0 - 0.4 = 0.6 =$ frequency of dominant allele $p^{2} = 0.6 \times 0.6 = 0.36$ or 36% are homozygous dominant individuals $2pq = 2 \times 0.6 \times 0.4 = 0.48 = 48\%$ are heterozygous individuals Or $= 1.00 - 0.52$ $= 0.48$ Thus, 84% (36+48) have the dominant phenotype