Question

11. cerin African population, N of the population is born with side cellaneria Calculate the percentage of individuals who the selective de ce genre increased his this population in Hardy Weinbergulirim? Justify your answer. Your explanation should included square goodness of t test Mire: The expected numbers of individuals whould follow the M-Weveti . ond ] 92= .04 A 9-20 209 26-20 1.60) -32 3a Phenotype Genotype served Expected lel loerle Stre non sono 12. In unibed , approximately one child in 10,000 is born with s phenylketonuriala yndrome that affects individuals mapes for the recente del tal Calculate the ency of this all in the population Wawr Curly Total ad -92 + Toon 0001 9:/92 +1.0001 I Calculate the frequency of the normales 01 pra=1 P = 1-0-01 -0.99 2 pq = 26.01(99).0198X10,000 = 198 X-(-e) 20150 1 250 HO 15.1 Id Calculate the number of carriers of the brak within the 11. In Caucasian human hair wightness or curlines is thought to be powered by a single pair of alles showing partial dominance Individuals with their are mo r the sale while there with that at home for the individuals wich wavy bar are hetery ) des population of 1,000 individua2 were found to have strathair. 662 lad wavy hair, and 390 Badcurly hal a Calculate the l egends of the band calles alles Phenotype Genotype Straight Wawy Male Curly seles 490 302 362 786 1 852 2000 Totat .43 1 .. 2000

Answer 1

Answer-

11) 32% populations enjoy selective advantage of sickle cell anemia.

12) Allele frequency of a = 0.01; A =0.99; Number of carriers = 198

13) a) Allele frequency of is = 0.426; ic = 0.574

b) The population is not in Hardy-Weinberg Equilibrium as the expected allele frequency differs from observed frequency.

Explanation for all questions is given below:

11) sickle cell anemia is autosomal recessive trait. Genotype frequency of aa = q2 = 45 9 = 1 = 0.2 we know, P+4=1 :P=1-4 =1-0.2 =0.8 (Aa) will have selective Heterozygous individuals advantages. (P+9)=1 (P+4) ²=1 - p²+ 204 + 92 =1 CAA Maj (aa) Genotype frequency of Aa= 204 = 2x 0.880.2 . = 0.32 = 327 12) Phenylketo nuda (PKU) is autosomal recessive trait. Here, senotype freg = 42= = 0.0ood loooo 4= 0.0001 -0.01 (a) : Allele frequency of allele, a = 9=10.01 (b) we know, (p+9) -1 → P=1-9 = 1-0.01 = 0-99 Allele frequency of normal allele, A= P=10.99 @ be we know, P+4=1 p2 +299 + 92=1 (AA) (Aa) (ad) Genotype frequency of carier, Aa - 2p4 = 2X0-99X0.01 = 0.0198 out of 10000 Number of carries = 0.0198* 10000 = 198

(13) Number Phenotype senotype straight hair Pis is 245 (NAA) Wally hair isic 362 (NAC) Curly hair icic 393 (Naa) Total = 1000 ANAA + NAG (2x245) + 362 Allele frequency of is =- AN 2000 = 490 +362-852 2000 2000 A = 0.426 Allele frequency of ic = 2 Naa + Naa = (2x393) +362 AN 2000 786+362 1148 2000 2000 a=0.574 we know, Ata=1 » (A+ a)² = 1 A2 + 2 Aa + a²=1 (isis) Cisie) (icie) Expected frequency of is is = A²= (6.426)2 = 0.18€ isie - 24a - 2x 0.426 X0.574 - 0.489 icic- a2 = (0.574)2 = 0.33 out of 1000 progenies, Expected number of isis = 0.181X1000 = 1811 isic = 0.489 x 1000 = 489 Icic = 0.33 X 1000 = 330

Phenotype genotupo) Obseuled Expected CO) (E) o-E) I Co-E2 (6-E² straight (isis) 245 | 181 4096 22.63 waley (isic) 362 489 1-127 16129 32.98 Curly licic) 393 330 3969 12.03 Total 1000 1000 x² = 67.64 x² = 67.64 Toto types of phene Total phenotypes = 3 (straight, waley cuorly) .. Degree of freedom = 3-1=2 probability realue = p<0.01 Conclusion - If p value is less than 0.05, then the null hypothesis is rejected. Here Plealle is less than 0.01; so, the null hypothesis is rejected. that means the deviation between observed and expected number is huge. So, the population is not in Hardy-Weinberg Equilibrium.