Answer:
b
Explanation:
<em>Let the wild type trait, the silver-white, be represented by </em><em>A</em><em> allele and the mutant trait, the golden color, be represented by </em><em>a</em><em> allele.</em>
Heterozygous wild-type male fish would be Aa
Golden female fish would be aa
Aa x aa
Aa Aa aa aa
2/4 Aa = silver-white
2/4 aa = golden color
<em>Hence, the percent likelihood of golden offspring is </em><em>2/4 or 50%.</em>
The correct option is b.
Answer:
A. Coal
Explanation:
Coal has the potential to generate energy
Answer:

Explanation:
Given -
Total Population - 
Individuals with homozygous HbA/hbA genotype - 
Individuals with homozygous hbs/hbs genotype -
Individuals with heterozygous hba/hbs genotype - 
Let us assume the given population is in Hardy Weinberg' equilibrium
The frequency of individuals in the given population with homozygous HbA/hba genotype is equal to number of individuals with homozygous HbA/hba genotype divided by total population.

The frequency of hba allele is equal to

The word that goes on to the gap is "cells"
Hope it helped!