Phenogram and hardy weinberg equillibrum

Phenogram and hardy weinberg equillibrum

1. Mariam Zaheer Assistant Professor Zoology Department Govt. Graduate College of Science Wahdat Road, Lahore

2. A diagram depicting taxonomic relationship among organisms based on overall similarity of many characteristics without any regard to evolutionary history. Phenogram

3. TAXA A B C D E A X B 35 X C 70 45 X D 55 20 40 X E 95 30 50 90 X Smallest value = BD=20 Average dissimilarity of BD= 20/2=10

4. Average similarity between : BD and A = AB+AD = 35+55/2= 90/2=45 BD and C= BC+CD = 45+40/2= 85/2=42.5 BD and E = BE+ED = 30+90/2= 120/2=60 Smallest value = 42.5 Average dissimilarity of BD and C = 42.5/2=21.25 TAXA BD A C E BD X A 45 X C 42.5 70 X E 60 95 50 X

5. TAXA BDC A E BDC X A 57.5 X E 55 95 X Average similarity between: BDCand A= BDA+CA/2=45+70/2=115/2=57.5 BDCand E= BDE+CE/2=60+50/2=110/2=55 Smallest value= BDCE=55 Average dissimilarity of BDCE=55/2=27.5

6. TAXA BDCE A BDCE X A 75 X Similarity of BCDE and A =BCDA+EA/2=57.5+95/2 =152.5/2 =76.25 Average dissimilarity of ABCDE=76.25/2 =38.125

7. Allele and genotype frequencies in a population tend to remain constant in the absence of disturbing influences. Disturbing influences: Non-random mating Mutations Selection Limited population size Random genetic drift Migration Hardy Weinberg Equilibrium

8. To determine of the frequencies of alleles: p + q = 1 To determine of the frequencies of genotypes: p2+2pq+q2 = 1 The Equations A gene has two alleles, A and a The frequency of dominant allel A = p The frequency of recessive allel a = q The frequency of homozygous dominant genotype AA = p2 The frequency of homozygous recessive genotype aa = q2 The frequency of heterozygous genotype Aa = 2 pq

9. In males the chances of color blindness are more than 100/300,000. Calculate the genotypes frequencies. Solution: q2 = 100/300,000=0.00033 √ q2= √ 0.00033=0.0182 For calculating “p” p+q =1 p=1-q p= 1- 0.0182=0.98 P+q =1 0.98+0.0182=1 0.99≈1 Problem No. 1

10. For calculating genotypic frequencies: p2+2pq+q2=1 By putting values (0.98) ² + 2(0.98)(0.0182) + ( 0.0182) ² =1 0.9604 + 0.035672 + 0.00033=1 0.99 ≈ 1 Result The frequency of homozygous dominant genotype = p2= 0.9604 The frequency of homozygous recessive genotype = q2= 0.00033 The frequency of heterozygous genotype = 2pq= 0.035672

11. In a population the genotype frequency of BB is 900, bb is 700 and heterozygote is Bb 1600. Calculate gene frequency and genotype frequencies of this population. Genotypes BB Bb bb Total Individuals 900 1600 700 3200 Genes 1800 3200 1400 6400 Frequency of allel B (p)=No. of dominant + ½ No. of heterozygotes (Bb) homozygotes(BB) =1800+1/2(3200) = 3400 Frequency of allel b (q)=No. of recessive + ½ No. of heterozygotes(Bb) homozygotes(bb) =1400+1/2(3200) = 3000 Problem No. 2

12. Relative frequency of allel B(p) = 3400/6400= 0.531 Relative frequency of allel b (q)= 3000/6400=0.468 Genotype Frequencies: p ² +2pq+q ² =1 BB=B² B ² +2Bb+b ² =1 bb=b² (0.531)² +2(0.531)(0.468)+(0.468)² =1 0.281+0.497+0.219=1 0.99 ≈1 Result: BB(p)²=0.281 Bb( pq )=0.497 bb(q)²=0.219

13. There is a population of 1300 individuals on island, 13 are affected with disease cystic fibrosis. Find the no. of homozygous dominant, homozygous recessive , heterozygous and normal individuals. Solution: cc =q ²=13 Total individuals =1300 C=dominant Find CC=? Cc=? c=recessive Relative frequency of cc(q) ² =13/1300=0.01 Frequency of c= √ q2= √ 0.01=0.1 As p+q =1 P=1-q = 1-0.1 = 0.9 Frequency of CC(p) ² =(0.9) ²=0.81 Problem No. 3

14. Frequency of heterozygotes=2pq =2(0.9)(0.1)=0.18 No. of dominant homozygotes(p)² =frequency of p²xTotal individuals =0.81x1300=1053 No. of heterozygotes= frequency of 2pq xTotal individuals =0.18x1300=234 No. of Normal individuals = dominant homozygotes+hetrozygotes = CC+Cc =1053+234=1287