1 Mathematical Modelling Worksheet 5 Some problems in Heredity 1. Genetic Selection. Hereditary traits are determined by genes, which occur on every cell of an organism, grouped together on the chromosomes. Except in the reproductive cells genes occur in pairs and appear on paired chromosomes. A particular gene with two alleles R and r. The genes of an offspring result from the pairing of two genes, one from each parent. There are three possible genotypes of the organism relative to this gene: RR rr known as homozygotes. Rr known as the heterozygote. One of the two alleles R of a particular gene is said to be dominant if genotypes RR and Rr are indistinguishable from one another. Provided that the genotype rr is observably different, r is said to be recessive. The gene frequency p of gene R in a population is: PR (a) number of R genes in the population number of R genes number of r genes in the population If the genotypes RR, Rr, rr occur in the population in the proportions l: m: n, where l + m + n =1, we can determine the gene frequencies PR and Pr . 2lP mP 2P 1 l M 2 PR Pr n 1 M 2 2 (b) Suppose that in a large parent population genes R and r are present in proportion p : q, where p + q =1. Assuming these proportions are the same for male and female and that mating is random, we can determine the proportions of the three genotypes RR, Rr and rr. p p R RR r Rr p R Rr q r R q S t qa r t r l RR p 2 , rr m Rr 2 pq, n rr q 2 l 1 m p 2 pq p p q p R 2 n 1 m q 2 pq q p q q r 2 Therefore proportions of the three genotypes RR, Rr and rr in future generations are: RR : p 2 (c) Rr : 2 pq rr : q 2 If a fraction s of the RR and Rr individuals survive to adulthood, while only a fraction (1 – k)s (0<k<1) of rr individuals survive, then zygote rr has selective disadvantage k. Let pn , qn 1 pn be the gene frequencies of genes R and r in the n’th generation adults. We can find the proportions of the genotypes born and surviving to adulthood in the next generation, assuming selective disadvantage p k of the rr individuals. Also by denoting n by y n , we can obtain a recurrence qn relation for yn . 3 Born p 2 : 2 pq : q 2 Adults sp 2 n : 2 sp n q n : 1 k sq 2 n PR 2 sp 2 n 2 sp n q n 2 Sp 2 n 2 Sp n q n 1 k Sq 2 n pn pn qn 2 PR p 2 p 2 2 p n q n 1 k q 2 n n n pn pn qn 2 p n q n q 2 n kq 2 n pn 1 kqn 2(1 k ) Sq n2 2Sp n q n Pr 2 Sp n2 2Sp n q n 1 k Sq n2 q n p n 1 k q n 1 kqn2 pn 1 kqn2 pn PR Pr 1 kqn2 q n kqn2 q n kqn2 pn q n 1 kqn 4 Now pn y n so we get qn yn 1 - kq n pn qn 1 pn yn qn 1 qn yn qn 1 q n y n 1 qn PR Pr 1 and substitute back in. Now we get yn 1 yn k 1 y 1 n y n 1 (d) y n y n 1 yn 1 k y n y n 1 yn 1 k If a recessive gene has initial frequency 0.99 and selective disadvantage k =0.05, how many generations will it take for the gene frequency to drop to; (i) (ii) 0.1 0.01 y n 1 y n y n 1 y n 1 k dy n y n 1 y n dn n 1 n dy n y n 1 y n dn 5 dy n y yn n yn dn y n 1 k 2 yn yn y y 1 k n n y n 1 k y n 1 k 2 dy n kyn dn y n 1 k yn 1 k dy n d n kyn yn ky n dy n 1 k dn n kyn 1 0.95 0.05 dy 0.05 y putting k 0.05 n n n n 20 y n 19 ln y n c n0 y0 p 0 0.01 1 q 0 0.99 99 20 1 19 ln c 99 99 20 1 c 19 ln 87.105 99 99 0 the number of generation ' s it will take for the gene frequency to drop to 0.1 is p n 0.99 9 yn qn 0.1 p n 0.1 , q n 0.9 n 209 19 ln 9 87.105 308.953 n 309 and the number of generation ' s it will take for the gene frequency to drop to 0.01 is p n 0.99 yn q n 0.01 n 2099 19 ln 99 87.105 2154.413 n 2155 6 (e) We can consider the effect on the situation in (c) and (d) if the selective disadvantage is associated with the dominant gene. Sp 2 n : 2 Sp n q n : 1 k Sq 2 n S 1 k p 2 n : 21 k Sp n q n : Sq 2 n 1 2 Sp 2 n : 2 Sp n q n : S q n 1 k dy n y 2 n yn yn dn y n 1 k dy n y 2 n y n 1 k y n y n 1 k 1 dn y n 1 k 1 y n 1 k 1 kyn y n 1 k 1 y n 1 k 1 dy n d n kyn kyn yn 1 dy n kyn kyn dy n kyn n yn 1 y n k ln y n n k 1 y n 1 k ln y n 0 k 0 k 1 2. Sex Linked Characteristics. Sex linked characteristics are not transmitted according to the previously stated laws. One pair of chromosomes determines sex - XX for females and XY for males. If a particular gene occurs only on the X chromosome a female can be classified as RR, Rr or rr, but a male having only one X chromosome can only be classified R or r. (a) Denoting the gene proportions in the female parent population by p and q, and those in the male population by p’ and q’, (p + q = 1 = p’ + q’), we can show that the corresponding proportions in the next female generation are the arithmetic means of the proportions of male and female genes in the parent population. 7 Proportion R in Female = p Proportion r in Female = q Proportion R in Male = p’ Proportion r in Male = r’ l:m:n p + q = 1 = p’+ q’ RR : Rr : rr pp’: p’q + q’p : qq’ PR (b) 2l m 2 pp ' p ' q pq 2l m n 2 pp ' p ' q pq' qq ' p2p'q' p ' q 2 p p' q' q p ' q ' pp'1 p' q 2 p q p' p q p 2 p p' 2 Using the results of 2(a), we can derive and sole a difference equation relating the gene proportion of allele R in the female population in successive generations. From 2(a) we have: p1 p0 p0' 2 pn1 pn pn' 2 pn 1 pn pn 1 2 where pn' pn1 This equation can be rearranged into a recurrence relation, which can be solved. pn 2 pn1 pn1 0 If p n s n s 0 8 s n 2s n 1 s n 1 0 2s 2 s 1 0 Then s 0.5s 1 0 s 1 or s 1 2 n 1 p n A B 2 Using the boundary conditions p0 p1' , p1 p2' we can find the values of A and B. p1' A B 1 p 2' A B 2 Solving these two simultaneous equations we get: 2 ' p1 p 2' 3 21 B p1' p 2' 32 A n 2 ' 1 1 ' ' p n p1 p 2 p1 p 2' 3 2 2