Question

You need to list equations used and provide steps of problem solving. Providing an answer itself is not enough for full grade.

The ability to taste PTC is due to a single dominant allele “T”. You sampled 215 individuals, and determined that 150 could detect the bitter taste of PTC. Calculate the frequency of dominant and recessive alleles. Calculate the frequency of homozygous dominant, homozygous recessive, and heterozygous individuals.

Answer 1

At very first we know that, Hardy Weinberg principal

It states that the variation in population will remain constant from generation to generation in the absence of disturbance like mutation, migration, genetic drift, selection etc.  

The Hardy Weinberg equation is based on allele at a particular locus. Let's consider there are 2 alleles at single locus A & a There frequency is p & q respectively

They put an equation from the principal

i.e p+q=1 this equation is useful for allele frequency calculation.

please refer the principal again to understand this equation.

The expansion of this equation is i.e. (p+q)²=1 is

p²+2pq+q²=1 this equation is helpful to calculate the genotype frequency.

I hope indeed you understand it now come to the question

PTC is the chemical which is bitter in taste normally it's a dominant trait to taste it and recessive those who do not taste it so there are taster and non-taster for PTC.

In the given problem there is a population of 215 individuals out of the 150 are PTC Taster. That means those remain left are PTC non-taster i.e 215-150=65 are non-taster. As the information from question single Dominant allele is enough to taste PTC

Now let's calculate allele frequencies first

We know the recessive population for PTC i.e non-taster =65 out of 215 let's understand

As the individual is diploid so

TT is the homozygous Tt is the heterozygous and tt is the recessive set of allele

Here individuals with tt are non-taster and are recessive there number is 65

The important thing to remind here is the frequency of t is q and tt is q²

So

q2=65/215

q²=0.30

q=0.547 (0.55)

From the equation

p+q=1 we can calculate p also by putting the value of q

p=1-0.55

p=0.45

These are allele frequencies from this data it is easy to calculate genotype frequency for homozygous dominant, homozygous recessive and heterozygous.

Let's use the equation in expanded form

p² + 2pq + q² = 1

(0.45)²+2(0.45*0.55)+(0.55)²=1

0.20+0.50+0.30=1

(The values are taking round figures)

So the frequency for TT i.e homozygous dominant is 20%

For Tt heterozygous is 50% and for homozygous recessive is 30%.

Hopes this calculation and information will helpful to you

Thank you