1. 0,1 0,4 0,5 p = (2 x 100) + 400 = 0,3 2 x 1000 q = 1 – 0,3 = 0,7 0,3 0,7 1 - Determine the gene and genotypic frequencies of the following populations:

2. p = (2 x 300) + 200 = 0,4 2 x 1000 q = 1 – 0,4 = 0,6 0,3 0,2 0,5 0,4 0,6 1 - Determine the gene and genotypic frequencies of the following populations:

3. p = (2 x 700) + 200 = 0,8 2 x 1000 q = 1 – 0,8 = 0,2 0,7 0,2 0,1 0,8 0,2 1 - Determine the gene and genotypic frequencies of the following populations: (A(AA)+A(AA))+A(aa)/(total alleles x2)

4. 2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it. Population 1 p = 0,3q = 0,7 f(AA) = p 2 = 0,09 f(Aa) = 2pq = 0,42 f(aa) = q 2 = 0,49 GenotipiXoXa(Xo – Xa) 2 : Xa AA100901,11 Aa4004200,95 aa5004900,20    2,26 Degree of Freedom = 2(p,q)-1= 1 P = 0,10-0,25 Population is in equilibrium

5. Population II p = 0,4q = 0,6 f(AA) = p 2 = 0,16 f(Aa) = 2pq = 0,48 f(aa) = q 2 = 0,36 GenotipiXoXa(Xo – Xa) 2 : Xa AA300160 Aa200480 aa500360 Expected and observed values are too different. The population is not in equilibrium 2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.

6. Population III p = 0,8q = 0,2 f(AA) = p2 = 0,64 f(Aa) = 2pq = 0,32 f(aa) = q2 = 0,04 GenotipiXoXa(Xo – Xa) 2 : Xa AA7006405,6 Aa20032045 aa1004090    140,6 GL = 2– 1 = 1 P <0,01 The population is not in equilibrium 2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.

7. 3 - The frequencies of L M and L N 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 in equilibrium: Group M = L M L M = p 2 = 0,8 x 0,8 = 0,64 x 200 = 128 Group MN = L M L N = 2pq = 2 x 0,8 x 0,2 = 0,32 x 200 = 64 Group N = L N L N = q 2 = 0,2 x 0,2 = 0,04 x 200 = 8

8. Population I f(aa) = 810/1000 = 0,81 Since f(aa) = q 2 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 = 0,6f(A) 1 – 0,6 = 0,4 Population III f(aa) = 490/1000 = 0,49 f(a) = √0,49 = 0,7f(A) 1 – 0,7 = 0,3 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)

9. 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

10. 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

11. 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) = q 2 = 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.

12. 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 Daltonic FEMALE = d d Frequency Daltonic Males = 0,1 Frequency allele d in MALES = 0,1 Daltonic FEMALES = f(d d) = 0,1 x 0,1 = 0,01

13. Fitness relative (W) 2/2 = 1 2/2 = 1 1/2= 0,5 The selection is against the phenotype homozygous recessive 200/100 = 2 400/200 = 2 100/100 = 1

14. 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 1 0 250/1000 = 0,25 500/1000 = 0,5 250/1000 = 0,25 1 x 0,25 = 0,251 x 0,5 = 0,5 0,25 x 0 = 0 [2(250) + 500]/2(1000) = 0,5 [0,5 + 2(0)] /2(0,25 + 0,5 + 0) = 0,33 q1 – q0 = 0,33 – 0,5 = -0,17 1000x2 total of alleles 2x250 aa=250a+250a 2(0.25+0.5) total of freq 2x0 aa=0a+0a

15. 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 1 1 0 0,64 0,32 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  q = -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

16. 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 = 0,4). Try to figure out the variation of genetic frequency of a after a generation of selection genotypesAAAaaa fitness Genotypic frequency Gametic contribution Frequency of q0 Frequency of q1 qq 1 1 – 0,4 = 0,6 1 p 2 = 0,3 2 = 0,09 2pq = 0,42 q 2 = 0,7 2 = 0,49 1 x 0,09 = 0,09 0,6 x 0,42 = 0,252 1 x 0,49 = 0,49 [0,42 + 2(0.49)]/2(0.09+0,42+0,49)=0,7 [0,252 + 2(0,49)]/2(0,09 + 0,252 + 0,49) = 0,74 q1 – q0 = 0,74 – 0,7 = 0,04 The recessive allele will be fixed and the dominant allele will be delete from the population - What is the final result of this selection?

17. - If the starting population has gene frequencies of : 0.8 for A, 0,2 for a, What will be the variation of the genetic frequency and what will be the final result? The most frequent allele will be fixed (A) and the less frequent allele will be deleted.

18. genotypesAAAaaa fitness Genotypic frequency Gametic contribution Frequency of q0 Frequency of q1 qq 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 0,8 1 0,1 640/1000=0,64 320/1000=0,32 40/1000=0,04 0,64 x 0,8=0,51 0,32 x 1 = 0,32 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

19. Genetic Frequencies at polymorphic equilibrium: Previous population: s = 0,20 t = 0,90 q (at equilibrium) = 0,20 = 0,18 0,20 + 0,90 When will be reach the polymorphic balance?

20. 14 – Two population are composed of 100 and 10.000 individuals respectively. Both populations have: frequency of A 0,3 Frequency of a 0,7 Determine the range for which we expect the variation of genetic frequencies during a generation caused by the fate.

21. Risoluzione scheda 11, n. 14 Range for which p1 could vary by chance with a probability of 95%: p1 = p0 ± 2s Population of 100 individuals: Standard deviation Population of 10 000 individuals

22. 15 - In una popolazione la frequenza dell'allele M passa da 0,6 alla generazione G 0 a 0,7 alla generazione G 1. Sapendo che la popolazione è costituita da 200 individui, è possibile che tale variazione di frequenza genica sia dovuta solamente a deriva genetica?

23. Risoluzione scheda 11, n. 15 G0 = f(M) = 0,6 G1 = f(M) = 0,7 Può essere effetto della sola deriva genetica? Dato che p1 è 0,7 non può trattarsi di variazione dovuta solo al caso.