Eight percent of a population shows a recessive trait. what would be the frequency of the heterozygotes for that trait if the population is in hardy=w

Population that show recessive trait should be only homozygous recessive. If the population in Hardy Weinberg equilibrium, then the recessive trait gene frequency would be:

homozygous recessive = aa= 8%=0.08

a= √0.08=0.283

a+A=1

A= 1-a = 1- 0.283= 0.717

The percentage of heterozygote in population would be:

heterozygote= 2 Aa= (2) (0.717) ( 0.283)= 0.4058= 40.58%

sand 60

clay 10

silt 30

sandy loam should be dominated by the sand. as the amount of sand increased, the amount of slit needed to anchor the plants' roots to the sand will also be increased. even though the amount is very little, 10% part of clay is still necessary for proper oxidation.