A population of laboratory rats in Hardy-Weinberg equilibrium displays a polymorphism for coat color controlled by two alleles at a single locus. The B allele codes for black coat colour and is completely dominant. There are 200 rats in the population: 32 are white and 168 are black. How many of the black rats are heterozygous (Bb) at the coat-color locus?

96

Explanation:

In solving Hardy-Weinberg related question, there are two formulas involved:

p + q = 1   ----------------------------- equation (1)

p² + 2pq + q² = 1 ------------------- equation (2)

where;

p is the frequency of the dominant allele.

q  is the frequency of the recessive allele.

p² is the frequency of individuals with the homozygous dominant genotype.

2pq is the frequency of individuals with the heterozygous genotype.

q² is the frequency of individuals with the homozygous recessive genotype.  

Total population of rats = 200

32 are white (bb)  = q²

168 are black (BB)  = p²

How many of the black rats are heterozygous (Bb) at the coat-color locus?

Frequency of individual (q²) =

Frequency of individual  (q²) =

q² = 0.16

To find q from q²; we take the square root of the both sides

√q² = √0.16

q = 0.4

Using the first equation;

p + q = 1

p + 0.4 = 1

p = 1 - 0.4

p = 0.6

To find how many of the black rats are heterozygous (Bb) at the coat-color locus (i.e 2pq) ; we have

= 2 × p × q

= 2 × 0.6 × 0.4

= 0.48

∴ 0.48 × Total population of rats

= 0.48 × 200

= 96

Hence, there are 96 black rats that are heterozygous (Bb) at the coat-color locus

answer: a) a compound microscope has a greater magnification ability than a simple microscope.

explanation:

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