Allele Frequency Worksheet Answer Key
Prior to starting these questions….
The principle of allele frequency is based upon a few assumptions, both mathematical and biological. For these questions we will only consider the math and how it can relate to gene frequency.
Remember there is a distinct difference between single gene or allele frequency and genotype frequency!
 Lets assume then the symbol for an allele is “A” or “a”, in a population these are the only two options.
 Since each allele must be in a pair, the sum of “A” and “a” equals one or ( A + a ) = 1, we are now talking about the individual allele frequency. Likewise if we want to predict phenotypes we need to consider pairs of alleles. The total allele frequency will always add up to one…this is an important point.
 From Punnet squares we know that we can predict probability for a heterozygous cross. In a population we assume some display of both alleles. If we follow this assumption then the square of both parents alleles is equal to ( A+a) 2
 O.K. now here comes your quadratic equation…. Bet you never thought you would use it in math… ( A + a ) 2 = A 2 + 2 Aa + a 2 .
 This is a lot like a punnet square because A 2 is equivalent to the homozygous dominant, 2 Aa is the heterozygous, and a 2 is the recessive.
O.K. now strategies for questions…lets go through each one
1.
Facts:
 We are told that the dominant allele frequency is equal to .3, lets give Dominant traits the symbol “P” from now on, just so we don’s get our “A’s” and “B’s” confused…
2.
Formulas
 P=.3 , how do we calculate Q or recessive allele frequency?
 Q= 1- (P) or in this case…Q=1-.3= .7
3.
Work
 Now we have both allele frequencies substituting into the quadratic equation.
( P + Q ) 2 = P 2 + 2PQ + Q 2

= ( .3 ) 2 + 2 (.3)(.7) + (.7) 2
4. Answers:
P 2 = homozygous tall = (.3) 2 x (700) = frequency times population
2PQ= heterozygous = (2)(.3)(.7) x 700= frequency times population
An easier way!
Consider the following…The tall phenotypes are equal to homozygous plus heterozygous, ….if we take the total population (700) and subtract just the recessive
(.7) 2 x (700) then we have the total tall phenotypes. Remember…it is easier to subtract the recessive phenotype than to add both the hetero and homozygous frequencies.
Question Two:
Given: Q=5% or .05
So P= (1-.05) =.05
Time to number crunch!
Total frequency equals ( P + Q ) 2 = ( .05) 2 + 2 (.05)(.05) + (.05) 2
Looking for all dominant phenotype or P 2 + 2 PQ
Quick method!
Total population minus (recessive frequency)(population)
2300- (Q) 2 (2300) = answer
Question Three:
Given… it is a bit more sneaky…
50/1650 = Q 2 or this is the percent of the population that is recessive!
= .03

.03 = Q which is .173
P = 1-.173 = .827
(P + Q) 2 = (.827) 2 + 2 (.827)(.173) + (.173) 2

Purebred Dominant frequency = P 2 = (.827) 2
Hetero
Recessive
= 2PQ = 2 (.827)(.173)
= Q 2 = .03
Question 4:
Given: 50% = P 2 = .5
P=

.5 = .707
So
Q= 1- .707 =.293
P 2 = (.707) 2
2PQ = 2 (.707)(.293)
B) Given Q 2 = 90% or .9
Q =

.9 =
P = 1 -

.9 =
Phenotypes
P 2 = ( 1 -

.9) 2
2PQ= 2 ( 1-

.9)(

.9)
Q 2 = (.9) 2
C) Given
Q 2 = 10% or .1
Q =

.1 =
P = 1 -

.1 =
P 2 = ( 1 -

.1) 2
2PQ= 2 ( 1-

.1)(

.1)
Q 2 = (.1) 2

Side Two:
Question 1:
Given Q=.01 ( these are lethal)
P= 1 - .01= .99
Population = 4000
Recessive Genotype = Q 2 = (.01)(.01) = frequency of lethal recessive
So … (.01) 2 (4000) = afflicted
2 (.99)(.01)(4000) = carriers
Question 2:
Given Total Population=16,000 a) 1000/1600 = Q 2 so Q=

1/16 …we reduce the fraction folks!
P = 1-

1/16 b) 10240/1600= P 2 so P =

10240/16000
Question 3:
40 Albinos = Q 2 a) = Q
= 40/16,00
=

40/16,00 b) 100% c) P 2 + 2PQ + Q 2 ( these leave)
So total 16000 – 40 = P 2 + 2PQ
Know that P = 1-Q so 1-

40/16,00 = P
(1600 – 40) – P 2 (1600 –40) = population of 2PQ