1. POPULATION GENETICS
The study of the allele frequencies in a population
The Hardy Weinberg equilibrium
“Allele and genotype frequencies in a population tend to remain constant in the absence of disturbing influences”
-non-random mating
-mutations
-selection
-limited population size
-random genetic drift
Conditions for HW equilibrium are never met in nature.
The equations
p + q = p2 + 2pq + q2 =1
p=frequency of allele A p2=frequency of AA
q=frequency of allele a q2=frequency of aa
2pq=frequency of Aa

2. An Example
Assume a population in which 36% of the population are homozygous for a certain recessive allele a. Assume the population is at equilibrium.
What is the frequency of the recessive allele in this population?
36%aa
q2=0.36
q=f(a)= 0.6
What is the frequency of the dominant allele in this population?
q =0.6
p = f(A) =1-0.6=0.4
What percentage of the population are homozygous for the dominant allele A?
p2=f(AA)=0.4 x 0.4=0.16 16%
What percentage of the population are heterozygous for the trait?
2pq=f(Aa)=2 x 0.6 x 0.4 =0.48  48%
Why do we have to start the problem with the percentage or numeber of the homozygous recessive in the population?
Because only the recessive phenotype indicates only the homozygous recessive genotype…the dominant phenotype indicates heterozygous and homozygous

3. In this case we know the number of heterozygous
1 - Determine the gene and genotypic frequencies of the following populations:
0, , ,5
0, ,7
In this case we know the number of heterozygous
We think about the number of alleles
p =f (A) = (2 x 100) = 0,3
2 x 1000
q = f(a)=1 – 0,3 = 0,7

4. 1 - Determine the gene and genotypic frequencies of the following populations:
0, , ,5
0, ,6
p = (2 x 300) = 0,4
2 x 1000
q = 1 – 0,4 = 0,6

5. 1 - Determine the gene and genotypic frequencies of the following populations:
0, , ,1
0, ,2
p = (2 x 700) = 0,8
2 x 1000
q = 1 – 0,8 = 0,2
(A(AA)+A(AA))+A(Aa)/(total alleles)

6. Population 1 p = 0,3 q = 0,7 f(AA) = p2 = 0,09 f(Aa) = 2pq = 0,42
2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.
Population 1
p = 0,3 q = 0,7
f(AA) = p2 = 0,09
f(Aa) = 2pq = 0,42
f(aa) = q2 = 0,49
Genotipi Xo Xa
(Xo – Xa)2 : Xa AA
100 90
1,11 Aa
400
420
0,95 aa
500
490
0,20
c2 = 2,26
Degree of Freedom = 2(p,q)-1= 1
P = 0,10-0,25
Population is in equilibrium

7. Expected and observed values are too different.
2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.
Population II
p = 0,4 q = 0,6
f(AA) = p2 = 0,16
f(Aa) = 2pq = 0,48
f(aa) = q2 = 0,36
Genotipi Xo Xa
(Xo – Xa)2 : Xa AA
300
160 Aa
200
480 aa
500
360
Expected and observed values are too different.
The population is not in equilibrium

8. The population is not at equilibrium
2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.
Population III
p = 0,8 q = 0,2
f(AA) = p2 = 0,64
f(Aa) = 2pq = 0,32
f(aa) = q2 = 0,04
Genotipi Xo Xa
(Xo – Xa)2 : Xa AA
700
640
5,6 Aa
200
320 45 aa
100 40 90
c2 = 140,6
GL = 2– 1 = 1
P <0,01
The population is not at equilibrium

9. 3 - The frequencies of LM and LN alleles in a group of 200 black Americans were 0,8 e 0,2, respectively. Calculate the expected frequencies for individuals with M, N and MN blood group.
If the population is at equilibrium:
Group M = LM LM = p2 = 0,8 x 0,8 = 0,64 x 200 = 128
Group MN = LM LN = 2pq = 2 x 0,8 x 0,2 = 0,32 x 200 = 64
Group N = LN LN = q2 = 0,2 x 0,2 = 0,04 x 200 = 8

10. The individuals with phenotype A can be AA or Aa.
We are sure that individuals with phenotype a are: aa
We start from this information (aa) to figure out f(A)
Population I
f(aa) = 810/1000 = 0,81
Since f(aa) = q2 f(a) = √0,81 = 0,9
f(A) 1 – 0,9 = 0,1
Population II
f(aa) = 360/1000 = 0,36
f(a) = √0,36 = 0,6 f(A) 1 – 0,6 = 0,4
Population III
f(aa) = 490/1000 = 0,49
f(a) = √0,49 = 0,7 f(A) 1 – 0,7 = 0,3

11. f(aa) = 0,09 f(a) = √0,09 = 0,3 f(A) = 1 – 0,3 = 0,7
5 - What is the expected frequency for dominant homozygous and heterozygous genotypes in a population at equilibrium in which the homozygous recessive genotype frequency is 0,09?
f(aa) = 0,09
f(a) = √0,09 = 0,3
f(A) = 1 – 0,3 = 0,7
f(AA) = 0,7 x 0,7 = 0,49
f(Aa) = 2 x 0,7 x 0,3 = 0,42

12. f(Rh-) = 0,0025 = f(dd) f(d) = √0,0025 = 0,05 f(D) = 1 – 0,05 = 0,95
6 - if in a population at equilibrium the frequency of Rh- phenotype is 0,0025, which is the expected frequency of heterozygous individuals?
f(Rh-) = 0,0025 = f(dd)
f(d) = √0,0025 = 0,05
f(D) = 1 – 0,05 = 0,95
f(Dd) = 2 x 0,95 x 0,05 = 0,095

13. Males white eyes = f(w) = 0,3
7 In Drosophila melanogaster the w recessive sex-linked allele, is responsible of the white colour of the eyes. In a population the frequency of this allele is 0,3. Which frequencies of male and female with white-eyes is expected if the population is at equilibrium?
Male white eyes = w Y
Female white eyes = w w
Frequency:
Males white eyes = f(w) = 0,3
Females white eyes = f(ww) = q2 = 0,3 x 0,3 = 0,09
If alleles are X-linked, females may be homozygous or heterozygous, but males carry only a single allele for each X-linked locus. For X-linked alleles in females, the H-W frequencies are the same as those for autosomal loci.
In males, however, the frequencies of the genotypes will be p and q, the same as the frequencies of the alleles in the population.

14. Daltonic MALE = d Y (XY) hemyzygote freq dY = freq allele d
8 - Daltonism is due to a sex-linked recessive allele. In a population the daltonic male frequency is 0,1; which is the one of daltonic females?
Daltonic MALE = d Y (XY) hemyzygote freq dY = freq allele d
Daltonic FEMALE = d d (XX)
Frequency Daltonic Males = 0,1
Frequency allele d in MALES = 0,1
Daltonic FEMALES = f(d d) = 0,1 x 0,1 = 0,01
Healthy carriers= 2*0.1*0.9=0.18

15. The selection is against the phenotype homozygous recessive
Fitness involves the ability of organisms to survive and reproduce in the environment in which they find themselves . The consequence of this survival and reproduction is that organisms contribute genes to the next generation.
Fitness= Its lifetime reproductive success
Fitness absolute /100 = /200 = /100 = 1
Fitness relative (W) /2 = /2 = /2= 0,5
The selection is against the phenotype homozygous recessive

16. As expected the frequency of a allele decreases
10 - A population is composed of: AA 250, Aa 500, aa 250 individuals, when a selection factor acts against the homozygote recessive and its fitness decreases to 0.
The variation of genic frequency (selection effect) could be figured out taking into consideration the contribution of the gametes of different genotypes in the next generation.
Consider the previous population, try to figure out the variation of genetic frequency of a after a generation of selection. 1  0 
 250/1000 = 0,25
 500/1000 = 0,5
 1 x 0,25 = 0,25
1 x 0,5 = 0,5 
 0,25 x 0 = 0
1000x2 total of alleles
2x250 aa=250a+250a
 [2(250) + 500]/2(1000) = 0,5
2( ) total of freq
2x0 aa=0a+0a
[2(0) + 0,5] /2(0,25 + 0,5 + 0) = 0,33
 q1 – q0 = 0,33 – 0,5 = -0,17
As expected the frequency of a allele decreases

17. 11 – In a population there is a selection against the homozygous recessive with fitness = 0 but the genotypic frequencies are:
AA 0,64; Aa 0,32; aa 0,04,
Try to figure out the variation of genetic frequency of a after a generation of selection
0, , ,04
0,64 x 1 = 0,64 0,32 x 1 = 0,32 0,04 x 0 = 0
[0,32 + 2(0,04)]/2(0,64 + 0,32 + 0,04) = 0,20
[0,32 + 2(0)]/2(0,64 + 0,32 + 0) = 0,16
q1 – q0 = 0,16 – 0,20 = -0,04
Previous population Dq = -0,17 (> of q0 (0,5))
This population q0 = 0,2  With the same selection coefficient, the effect of selection against homozygous recessive change on varying of genic frequency

18. - What is the final result of this selection?
12 – In a population the gene frequency of A is 0,3 and the frequency of a is 0,7, and there is a selection against the heterozygous
(s =coefficient of selection=1-w= 0,4).
Try to figure out the variation of genetic frequency of a after a generation of selection
genotypes AA Aa aa
fitness
Genotypic frequency
Gamete contribution
Frequency of q0
Frequency of q1 Dq
– 0,4 = 0,
p2 = 0,32 = 0, pq = 0, q2 = 0,72 = 0,49
1 x 0,09 = 0, ,6 x 0,42 = 0, x 0,49 = 0,49
[0,42 + 2(0.49)]/2(0.09+0,42+0,49)=0,7
[0, (0,49)]/2(0,09 + 0, ,49) = 0,74
q1 – q0 = 0,74 – 0,7 = 0,04
- What is the final result of this selection?
The recessive allele will be fixed and the dominant allele will be delete from the population

19. 13 – The population is composed of: 640 AA, 320 Aa and 40 aa,
The fitness of homozygous dominant is 0,8 and the fitness of homozygous recessive is 0,1.
Try to figure out the variation of genetic frequency of after a generation of selection
genotypes AA Aa aa
fitness
Genotypic frequency
Gametic contribution
Frequency of q0
Frequency of q1 Dq
0, ,1
640/1000=0, /1000=0, /1000=0,04
0,64 x 0,8=0, ,32 x 1 = 0, ,04 x 0,1=0,004
[320+2(40)]/2(1000) = 0,2
[0,32+2(0,004)]/2(0,51+0,32+0,004) = 0,19
0,19 – 0,20 = -0,01

20. Considering people that become part of the family through marriage don’t have the mutant allele, calculate the probability that an affected (sick) child is born from the indicated cross ?
Daltonism is a recessive x-linked disease.
A girl has a daltonic father and a normal mother but the mum’s father is daltonic.
Which is the probability that the girl is daltonic?
Draw the tree

21. Genes A and B are concatenated at 20 mu
Genes A and B are concatenated at 20 mu. Which gametes are produced by an individual: AaBb?
If this individual is cross with a homozygous recessive, which phenotypic classes and which frequencies will we obtain?

22. For each of the following crosses:
determine the genotype of the individuals used for the cross;
determine if there is concatenation of the analyzed genes ;
- if genes are concatenated, determine their map distance;
- draw a schematic picture of the map of crossed individuals.
Parental Phenotypes
AB x ab
Progeny Phenotypes:
AB 40
Ab 63
aB 58
Ab 37

23. Gene A, B and C are located in this order on the same chromosome
Gene A, B and C are located in this order on the same chromosome. The distances are A-B 20 mapunits and B-C 10 mapunits. If an individual triple heterozygous is crossed with a recessive homozygous, which phenotypic classes are expected in the progeny and with which frequencies?
If the interference is 1?

24. Analysing a sample of individuals we have these phenotypic results
MM 60
MN 92
NN 20
Determine allelic frequencies of M and N
Determine if the population is at equilibrium (use statistical analysis)

25. 7 Gene A, B and C are located in this order on the same chromosome
7 Gene A, B and C are located in this order on the same chromosome. The distances are M-N 2 map units and N-O 8 map units (genes are in cis). If an individual triple heterozygous is crossed with a recessive homozygous, which phenotypical classes are expected in the progeny and with which frequencies?
Draw the map
M N O m n o
M n o x m n o 2 8
Parental MNO mno
Recombinants Mno mNO
Recombinants Mno mnO
DR MnO mNo
Freq DR= 0.02 x 0.08 =0.0016/2=0.0008
Freq Ric 1= =0.0184/2=
Freq Ric 2 =0.08 – = /2 =0.0392
Parental = =0.9016/2= 0.450