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Chapter 22. Population Genetics Hi everybody!! TWO more classes !!!! . Where we’re going. WHAT is population genetics Allelic frequencies, and calculating them p& q, P,H,&Q (I’ll explain later) Hardy-Weinberg principle- application of a little algebra “Fun” things you can do w/ H-W.

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chapter 22

Chapter 22

Population Genetics

Hi everybody!!

TWO more classes !!!! 

where we re going
Where we’re going
  • WHAT is population genetics
  • Allelic frequencies, and calculating them
  • p& q, P,H,&Q (I’ll explain later)
  • Hardy-Weinberg principle- application of a little algebra
  • “Fun” things you can do w/ H-W.
  • THINK!!!!
what s population genetics it s about populations
What’s population genetics?It’s about… populations!!!
  • Questions asked: 
  • What sort of alleles are in a population, and what are the frequencies? 
  • What differences exist between populations, and why (this is the more interesting Q)?
  • How genetically diverse is a population?
  • How does a population in, say, Europe and Asia differ? Why?
definitions
Definitions:
  • Population: Group of interbreeding individuals of the same species, exist together in time and space. 
  • Gene Pool: Genetic constitution of a population- essentially all the genes available, all the alleles available, and the proportions of those alleles that are present. So, in total, we have a larger gene pool than does any one individual.
populations have genetic variation
Populations have genetic variation.
  • This variation shows up in the gene pool, as the variety and frequency of various alleles at a locus. 
  • Studying genetic variation:
    • morphological differences: 
    • chromosomal variations:
    • VNTR differences, other genetic differences at the DNA level.- Even whole genome sequences!
calculating allelic frequencies not so easy tongue roll
Calculating allelic frequencies- not so easy (tongue roll…)
  • Let’s consider a population- CCR5-1, and CCR5Δ32, two alleles. Gets complicated with > 2!
slide7

Frequency: fraction of the total- between 0&1

  • genotypic frequency: P,Q, & H:
  • P= frequency of 1/1 homozygotes
  • Q= frequency of Δ32/ Δ32 homozygotes
  • H= frequency of 1/ Δ32 heterozygotes.
slide10

Two ways to determine allele and genotypic frequencies

P= 0.79, Q= 0.01, H= 0.2

p= P+1/2H

q= Q + 1/2H

hardy weinberg law
Hardy-Weinberg law:
  • Deals with an ideal, non-evolving population, not under selection. It’s also full of no’s: no migration, no mutation, no selection, either for survival or for mates. 
  • Allows you to determine the frequency of genotypes, given the allelic frequency. This includes the predicted genotypes, if a population is allowed to breed at random.
slide13
H-W
  • 1) The frequency of alleles does not change from 1 generation to another
  • 2) After one generation of random mating, offspring genotype frequencies can be predicted from the parent allele frequencies and would be expected to remain constant from that point on.
and if not in h w equilibrium
And if not in H-W equilibrium:
  • Some things, such as selection, or other things, are going on.
this produces the basic law
This produces the basic law:
  • of p & q represent allelic frequencies, then the genotypic frequencies are (p+q)2, or p2 + 2pq + q2 , where p2= P, q2 =Q, and 2pq= H.
  •  Problem 2 in book, p. 512
slide17

Predicted frequencies: Problem 2:

  • Initial population: 0.2AA, 0.6 Aa, 0.2 aa;
  • p= 0.5, q= 0.5, thus, after random mating, you would get:
  • p2 + 2pq + q2 ; AA (=P)= .52= 0.25; aa= .52 = 0.25,
  • Aa= 2 (.5X.5)= 0.5.
  •  What if you can’t tell all the genotypes?
if you if you assume the population is in h w equilibrium then you can use q to determine q
Ifyou :If you assume the population is in H-W equilibrium, then you can use Q to determine q:
  • Q= q2; so q= √Q
  • Problem 1: 88 tasters 37 non-tasters ; P+H= 88/125= 0.70; Q= .296; q = √Q =0.54. Notice that q is quite large.
  • p= 0.46.
  • verify: .462 + 2 (.46*.54) + .542:
  • .462= .21; 0.21 X 125= 26
  • .542 X125= 36
  • 2 (.46*.54)X 125= 62
  •  So, let’s do a tongue curling exercise
slide19

If a gene is sex-linked, then Q is q2 in females, q in males; the male:female ratio is then q/q2, or 1/q.

  • So if 1 in 10 males is color-blind, then 1 in 100 females will be color-blind.
  • Of course, you could ask about selection….
slide20

How can you tell if a population is in H-W equilibrium? by using chi2 analysis:

  • Problem 6: assume 1000 individuals:
  • 1/1 1/ Δ32Δ32/ Δ32
  • .6 .351 .049
  • 600 351 49
  • p= .6+ .351/2= .7755
  • q= .049 + .351/2 = .2245
  • expected= P= p2= .77552= 601
  • expected H= 2pq= 348
  • Expected Q= 51

You’d think there’d be 2 degrees of freedom, but according to your book there’s only one. Has to do with the fact that there are only 2 alleles.

slide21

Sickle cell anemia; population of 1000: Note we have data, NOT assuming H-W equilibrium !!!

  • AA: 756; AS= 242; SS= 2
  • p= 756+ 242/2 * 1000= .877
  • q= 242/2 + 2 * 1000= .123
  • Expected .8772 + 2 (.877*.123) + .1232
  • 769 215 15
  • O E d d2 d2/E
  • 756 769 13 169 0.22
  • 242 215 27 729 3.39
  • 2 15 13 169 11.26
  • This indicates NOT in HW-equilibrium.

Note that SS is underrepresented, AS is overrepresented

so what if a population is not in h w equilibrium
So what if a population is NOT in H-W equilibrium???
  • Five reasons
    • Selection
    • Mutation (not very good w/o selection)
    • Migration
    • Inbreeding
    • Drift
selection
Selection
  • Not a hard concept- Been treated extensively with math.
  • Fitness: w- value between 0 and 1, 1 = all off- spring of that genotype survive, 0.8= 80% of offspring of that genotype survive, etc.
  • Results in loss of less fit homozygotes- SS anemia example.
  • Raises questions- why is ∆32 so prevalent in European populations??? Why is cystic fibrosis still around??
slide24

Notice that 0.8 is a pretty strong selective DISadvantage.

The next slide shows how to calculate what q will be after a certain number of generations, if being homozygous means that your fitness is 0

slide25

(Note that this equation deals with the situation when the homozygous individual has 0 fitness.)

  • qg = qo
  • 1+gqo
  • qg = q after g generations, qo= q at the start. qo= 0.05= 5% of us are carriers. So, after 80 generations, the frequency of q should be:
  • 0.05
  • q80 = 1+80(.05), or 0.01. To my knowledge, this hasn’t happened; so the allele must be advantageous in some way to heterozygotes, at least in the past. Possibly resistance to typhoid fever?? Or mutation rate is high!
selection can be natural or artificial
Selection can be natural or artificial
  • Think of current environmental changes that affect fitness.
  • Artificial- breeding; (??)antibiotic and insecticide resistance;
  • Natural- often environmental changes change the fitness of certain genotypes; global warming (my prediction: don’t expect much if any); moose on Isle Royale, black squirrels, road kill deer, etc.
slide27

This is most common

Think of warming, antibiotic resistance, prey size, etc.

Think of a population moving into a cold and warm place, living low on a mtn and high up…

2 mutation
2. Mutation:
  • Mutation has little effect upon allele frequencies, unless the mutant allele has a selective advantage. Fig. 22-16 illustrates this. This illustration raised the question as to why genes aren’t more polymorphic than they are. When combined with selection, it has the power to change a population’s allelic frequencies, which results in evolution (actually, is the definition of evolution)
slide29

Back-mutation becomes an issue; a-> A happens also; I don’t think this graph considers this.

calculating mutation frequencies
Calculating mutation frequencies:
  • easier to do for dominant mutations:
  • # of mutants per live birth/2- since the mutant typically is heterozygous. 1 mutant in 100,000 births, 1 mutant allele of 200,000 alleles.
3 migration
3. Migration-
  • If genetically different individuals migrate into a region- different in terms of the allelic frequencies- AND if they interbreed, then the frequency of alleles will change. The book gives the example of the B blood type as being a supposed remnant of invading Mongols.
4 genetic drift
4. Genetic drift-
  • This is the result of a small population to start with. As illustrated, an allele can become fixed in a population, or predominant, even if the gene has no selective advantage. See story of albinism in Navajo Indians, p. 506.
slide35

Drift is an IMPORTANT factor, OFTEN overlooked! We WANT differences to be due to selection!

From Weaver & Hedrick, Basic Genetics, 2nd Ed., p. 420

The allelic frequencies over 30 generations for four replicate population of size 20 (flies). The mean is represented by the broken line

5 inbreeding
5. Inbreeding:
  • Reduces the # of heterozygotes; usually reduces fitness.  
  • Fig. 22-19: This shows the effects of inbreeding in a population where each genotype breeds with itself. The number of heterozygotes decreases by ½ with each generation.
slide38

It turns out that being heterozygotic tends to be good for a population- severely inbred populations tend to have more problems- less fertility, more disease, etc., that outbred populations. You’ve all probably heard of diseases in dogs that are more prevalent in certain breeds, such as hip dysplasia in German shepherds. Mongrels, on the other hand, are often healthier.

  • A misunderstanding of this has had historical influences- German leadership in WWII
inbreeding coefficient
Inbreeding coefficient:
  • probability that the two alleles of a given gene in an individual are identical b/c they are descended from the same single copy of the allele in an ancestor. Thus, the inbreeding coefficient would assume that an individual is homozygous at that location.
slide41

One formula for determining the inbreeding coefficient:  

  • F= 1-(H/2pq)(simplification of the formula in your text, p. 507); 2pq is the expected level of heterozygotes, H is the actual. Thus, as the actual # of heterozygotes declines, H/2pq heads toward 0, and the inbreeding coefficient heads towards 1.
some problems

Some problems:

Full & Half-brother-sister:

Half first cousins:

slide44

Aa

Aa

  • F=1/2X 1/2X 1/2X1/2 = 1/16
  • 4X 1/16= 1/4
slide45

Aa

  • F=1/2X 1/2X 1/2X1/2 = 1/16
  • 2X 1/16= 1/8- only 1 common ancestor
slide46

Aa

  • F=(1/2)6= 1/64, for A
  • Total = 2X 1/64= 1/32- only 1 common ancestor
slide47

Problems- see notes.

  • In a population of goats, two alleles determine eye color. A1 A1 goats have brown eyes, A1 A2 goats have grey eyes, and A2 A2 goats have blue eyes. In a population of 100 goats, you have: 
  • Brown eyes- A1A1 55 Grey eyes- A1 A210 Blue eyes A2 A235
  • P,H,Q?
  • p, q?
  • Expected P,H,Q?
  • What is the inbreeding coefficient for this population? (f=1-H/2pq)
  • Χ2 Analysis: Χ2 = Σ (d^2/e)