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Population Genetics Worksheet (Type in the blanks and submit this worksheet thro

ID: 51432 • Letter: P

Question

Population Genetics Worksheet

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Student Name:

What objects did you use?

Dominant Color (A)                          Recessive Color (a)

Starting p                                 Starting q

Demonstration #1 – Proper Hardy-Weinburg Conditions

Generation 1

Offspring – AA                                  Aa                               aa

Total “A alleles”                                 Total “a alleles”

New p                                     New q

New number of dominant objects                      New number of recessive objects   

Generation 2

Offspring – AA                                  Aa                               aa

Total “A alleles”                                 Total “a alleles”

New p                                     New q

New number of dominant objects                      New number of recessive objects   

Explanation / Answer

Allele frequencies in populations are inferred using Hardy and Weinberg equilibrium. Hardy and Weinberg equilibrium states that in a population, all dominant and recessive alleles comprise all alleles for that gene.

This was mathematically represented as p+ q = 1.0

Where,

p = frequency of dominant alleles

q = frequency of recessive alleles.

Example:

Consider In a population all are heterozygous for a trait, and their genotype is Bb. "B" is the dominant allele that gives blue coloured eyes, and "b" is recessive allele that gives brown coloured eyes. So, initially the values of p and q are equal to 0.5. (initial p = initial q = 0.5).

Generation 1:

The genotypes of the offspring (First generation) will be, 25% BB, 50% Bb, and 25% bb. Consider the number of individuals are 100.

Calculating the number of A alleles: In the population, 25 members have BB genotype, means they have 50 alleles, 50 of Bb population contains 50 B alleles. Thus, the number of A allels is 100.

The frequency of A allele is equal to the total number of alleles = 100/ 200 = 0.5

Similarly, the total number of "a" alleles is 00 and the "a" allele frequency = 50+50/200 = 0.5

Thus, the new p = new q = 0.5 (The genotypic frequencies change with generations, but the allele frequencies remain constant).

Generation 2:

As the population has reached Hardy- Weinberg's equilibrium, the genotypic frequencies do not change, they remain constant as generation 1.

Thus, the genotypes of the offspring (second generation) will be, 25% BB, 50% Bb, and 25% bb.