Sum7LS4Week3a(3slides)

Sum7LS4Week3a(3slides) - Population Genetics Allele...

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1 Population Genetics • Allele frequencies • Hardy-Weinberg Equilibrium Chapter 26 read: 738 - 747 problems:1, 3, 5, 7, 9,13,17, 21 Chapter 27 read: pp. 765 - 768 problems: 3, 5 • Fitness ( w ) = the ability of an individual to survive and reproduce Natural Selection at the Level of the Gene Fitness Fitness in genetics: The ability to survive and reproduce -In a population with a stable size… -The average fitness ( ϖ ) = 1 -In a declining population… ( ) < 1 -In a growing population… ( ) > 1
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2 Example • We have a single gene required for beak size in Darwin’s finches with two segregating alleles: B big beaks , b small beaks -In an environment where seeds are predominantly large, big-beaked finches survive better than the small-beaked ones, therefore… w ( BB or Bb) > w ( bb ) Relative fitness (relative to the most fit genotype)…where we arbitrarily define the following fitnesses: w ( BB or Bb ) = 1 w ( bb ) < 1 Genotype BB Bb bb Initial freq p 2 2pq q 2 fitness ( w ) 1 1 1-s S (selection coefficient): measure the intensity of natural selection acting on the genotype in the population Selection Coefficient S = 0 in normal fitness S = 1 lethality or infertility Let’s assume that the initial frequency of B is p = 0.5 and b is q = 0.5, and that s = 0.1 Genotype BB Bb bb Initial freq p 2 = 0.25 2pq = 0.5 q 2 = 0.25 (at fertilization) fitness (w) 1 1 1-0.1 = 0.9 p = 0.5, q = 0.5, s =0.1 BB = (0.25) x 1 = 0.25 Bb = (0.5) x 1 = 0.50 bb = (0.25) x (0.9) = 0.225 Relative contribution to next generation: Proportional contribution to next generation: -Divide each of the relative contributions by their sum (0.25 + 0.50 + 0.225 = 0.975). BB =0.25/0.975 = 0.256 Bb =0.50/0.975 = 0.513 bb = 0.225/0.975 = 0.231
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3 -Allele frequency in new generation: q’ = 0.231 + (1/2) (0.513) = 0.487 (less than initial 0.5) p’ = 0.256 + (1/2) (0.513) = 0.513 (more than initial 0.5) Natural Selection: Genetic Explanation • Initial generation: BB = 0.25, Bb = 0.50, bb = 0.25 • In next generation: (relative) BB = 0.25, Bb = 0.50, bb = 0.225 (proportional) BB = 0.256, Bb = 0.513, bb = 0.231 p = 0.5, q = 0.5, s =0.1 Genotype AA Aa aa # of progeny 1280 400 400 fitness (w) 8/8=1 4/8=0.5 2/8=0.25 S= 1 - w 0 0.5 0.75 # of adults 160 100 200 # progeny/#adults 8 4 2 “Relative Fitness” Chapter 8, pages 173 - 182, 185 - 188 Problems: 9, 12, 13, 16, 18, 19 Bacterial Genetics
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4 Why study bacterial genetics? • Understand bacteria that cause diseases (pathogens) • Model system for basic life processes (gene expression, etc) • Used widely for genetic engineering –Cloning and manipulating genes done in bacteria What is E. coli and why is it toxic? 3 lawsuits over E. coli deaths settled By Mary Engel, Times Staff Writer April 23, 2007 • The federal Centers for Disease Control and Prevention reported that the September 2006 outbreak caused 205 illnesses and three deaths in 26 states and Canada. • Escherichia coli ( E. coli ) –discovered in 1885 by Theodor Escherich –main species of bacteria living in the lower intestines of mammals –assists with waste processing, vitamin K production, and food absorption. • E.coli O157:H7 –discovered in 1982 and found to have acquired three new genes that make it toxic to humans but harmless to cattle and other animals •Vero toxin is a small protein that acts to cleave the host cell's rRNA (ribosomal RNA) and hence disrupt protein biosynthesis by ribosomes •Hemolysin: kills red blood cells by forming pores in their membranes •New adhesins: attach better to intestinal cells E.coli
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Sum7LS4Week3a(3slides) - Population Genetics Allele...

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