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Introduction

Introduction. Science of genetics can be divided into 4 subdisciplines: Transmission genetics : how are genes passed from one generation to the next? Study object: individual Molecular genetics : how relates DNA to phenotype? Study object: cell

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Introduction

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  1. Introduction • Science of genetics can be divided into 4 subdisciplines: • Transmission genetics: how are genes passed from one generation to the next? Study object: individual • Molecular genetics: how relates DNA to phenotype? Study object: cell • Population genetics: how behave traits in populations? Study object: population • Quantitative genetics: what are rules for polygenic inheritance? Study object: population • In this course we will study • Complex transmission genetics and we will use statistical methods for the analysis of genetic experiments • Population and quantitative genetics • Extra-nuclear genetics

  2. Laws of Probability • 1. Rule of multiplication • Rule of multiplication = The probability that independent events will occursimultaneously is the product of their individual probabilities Question: In a Mendelian cross between pea plants that are heterozygous forflower color (Pp), what is the probability that the offspring will be homozygousrecessive? • 2. Rule of addition • Rule of addition = The probability of an event that can occur in two or moreindependent ways is the sum of the separate probabilities of the different ways Question: In a Mendelian cross between pea plants that are heterozygous forflower color (Pp), what is the probability of the offspring being a heterozygote?

  3. Laws of Probability • 3. Conditional probability • The probability of an outcome that depends on a specific condition related to that outcome • Question: In the F2 of a monohybrid cross with plants with purple and white flowers: what is the probability that a plant with purple flowers is heterozygous? • Pa probability of being heterozygous = 1/2 • Pb probability of the condition (purple) = 3/4 • Pc = Pa/Pb = 2/3

  4. Laws of Probability • 4. The binomial theorem • The probability of two alternative outcomes during a number of trials. • Question: In a family of four, what is the probability of having two boys and two girls? • N Binomial Expanded binomial • 1 (x + y)1 = 1x + 1y2 (x + y)2 = 1x2 + 2xy + 1y23 (x + y)3 = 1x3 + 3x2y + 3xy2 + 1y34 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y45 (x + y)5 = 1x5y0 + 5x4y1 + 10x3y2 + 10x2y3 + 5x1y4 + 1x0y5 • N=4, x=0.5, y=0.5 • P(4 boys)= 1x4 = 1/16 • P(2 boys and 2 girls)= 6x2y2 = 6/16

  5. Laws of Probability • The general formula for the binomial expansion is: • Probability =N! ·(px)(qN-x) x!(N-x)! • N=total number of events p is the probability of event P q=probability of the alternative event x is the number of times event P occurs ! means factorial. • The total number of events (N) is 4 children. The number of times (x) that event P (male child) occurs is 2. The probabilities p and q are both 0.5. • Thus: 4! ·(0.52) (0.52) = 24/4×1/16 =6/16=3/8 (2!)(2!) • Of all families with 4 children 3 out of 8 are predicted to have two boys and two girls.

  6. Testing hypotheses using the 2-statistic • Mendel raised 705 purple-flowered and 224 white-flowered plants in the F2generation. Hypothesis: Is this a ratio of 3:1? • Calculate expected values based on the result total (929):3 : 1 = 696.75 : 232.25 • Calculate the differences between observed (O) and expected (E) • Standardise the differences: |O – E |2 2 =  E • Calculate degrees of freedom (DF) = number of differences – 1 • Lookup in 2 table at DF • If p<0.05 then reject hypothesis O E |O–E| 705 696.75 8.25 224 232.25 8.25 929 929 (8.25-0.5)2 (8.25-0.5)2 2 =  +  696.75 232.25 The correction for continuity of 0.5 is only applied when DF=1 2 = 0.086 + 0.259 = 0.345 At DF=1 0.345 lies between 0.016 and 0.455 with 0.900>p>0.500, so accept hypothesis

  7. Testing hypotheses using the 2-statistic • Mendel raised the F2generation of a dihybrid cross and found yellow/round 315, green/round 108, yellow/wrinkled 101 and green/wrinkled 32.Hypothesis: Is this a ratio of 9:3:3:1? • class observed expected deviationround/yellow 315 312.75 +2.25round/green 108 104.25 +3.75wrinkled/yellow 101 104.25 -3.25wrinkled/green 32 34.75 -2.75Total 556 556.00 +0.00 • Standardise the differences: |O – E |2 2 =  = 0.470 E • Calculate degrees of freedom (DF) = number of differences – 1 = 3 • Lookup in 2 table at DF: P>0.9 • If P<0.05 then reject hypothesis.

  8. the 2-statistic

  9. Population genetics • Questions studied in Population genetics: • how much genetic variation is found in natural populations and what processes control the amount of variation? • What processes are responsible for producing genetic divergence between populations? • How do population characteristics like breeding system, fecundity, and age structure influence the gene pool?

  10. Genetic structure of populations • To study the genetic structure of a Mendelian population one must describe the gene pool by calculating the genotypic and allelic frequencies • Genotypic frequencies • tiger moth:f(BB)= 452/497=0.909 (top 2)f(Bb)= 43/497= 0.087 (middle 4)f(bb)= 2/497= 0.004 (bottom 1) • Allelic frequencies:tiger moth:p=f(B)=(2×452+43)/(2×497)=0.953q=f(b)=(2×2+43)/(2×497)=0.047orp=f(B)=f(BB)+½f(Bb)=0.953q=f(b)=f(bb)+½f(Bb)=0.047

  11. Genetic structure of populations • Allelic frequencies with multiple alleles A1, A2 and A3 • (2×count of A1A1)+(A1 A2)+(A1 A3) • p=f(A1)= (2×total number of individuals) • (2×count of A2A2)+(A2 A1)+(A2 A3) • q=f(A2)= (2×total number of individuals) • (2×count of A3A3)+(A3 A1)+(A3 A2) • r=f(A3)= (2×total number of individuals)

  12. Genetic structure of populations • Allelic frequencies at an X-linked locus: • (2×XAXA females)+(XA Xa females)+(XA Y males) • p=f(XA)= (2×number of females)+(number of males) • (2×XaXa females)+(XA Xa females)+(Xa Y males) • q=f(Xa)= (2×number of females)+(number of males) • If the numbers of males and females are equal: • 2 1 1 • p=f(XA)= — [f(XAXA) + — f(XAXa)] + — f(XAY) 3 2 3 • 2 1 1 • q=f(Xa)= — [f(XaXa) + — f(XaXA)] + — f(XaY) 3 2 3

  13. The Hardy-Weinberg Law • The Hardy-Weinberg Law says that: • In an infinitely large, randomly mating population, free from mutation, migration, and natural selection, • the frequencies of the alleles do noy change over time, • genotypic frequencies remain in the proportions p2 (frequency of AA), 2pq (frequency of Aa) and q2 (frequency of aa), • Assumptions: • If population size is limited: chance deviations cause genetic drift • Random mating not necessary for all traits: humans mate preferentially for height, IQ, skin colour, etc. but random for blood type. p2 + 2pq + q2 = 1

  14. The Hardy-Weinberg Law • When is a population in genetic equilibrium? • The allele frequencies will not change from generation to generation • The genotypic frequencies will be in the proportions p2, 2pq and q2 after one generation of random mating • Why? • Each generation of zygotes produces A and a in proportions p and q. • If p=0.6 and q=0.4 then AA=0.36 (p2), Aa=2×0.24 (2pq), and aa=0.16 (q2) • Total=1.00

  15. The Hardy-Weinberg Law • Example: • In the US the frequency of tyrosinase-negative albinism is 1 in 40,000, or 0.000025. • The trait is recessive, so the genotype is aa. • So q2 equals 0.000025, thus q=0.005 and p=0.995. • The heterozygote frequency is therefore 2pq=2×0.995×0.005=0.00995 (almost 1%!)

  16. The Hardy-Weinberg Law • Extension for more than 2 alleles. • 2 alleles: p2 + 2pq + q2 = (p + q)2 • 3 alleles: (p + q + r)2 = p2 + q2 + r2 + 2pq + 2pr + 2qr • 4 alleles: (p + q + r + s)2 • etc. • Example: allozym-polymorphism in Mytilus edulis • Allele frequency •  • LAP98p=0.52LAP96q=0.31LAP94r=0.17 •  • LAP=leucine aminopeptidase • Assignment: calculate the genotypic frequencies

  17. The Hardy-Weinberg Law • Testing for Hardy-Weinberg equilibrium using 2 • Example tiger moth:observed genotypic frequencies:f(BB)= 452/497=0.909 f(Bb)= 43/497= 0.087 f(bb)= 2/497= 0.004Allelic frequencies:p=f(B)=0.953q=f(b)=0.047 • The number of degrees of freedom is 1, because p and q are not independent, so the correction for continuity is applied • (0.6-0.5)2 (1.5-0.5)2 (0.9-0.5)2 • 2 =  +  +  = 0.168 • 451.4 44.5 1.1 • The change P for this result is >0.5, so there is no significant deviation from the Hardy-Weinberg equilibrium.

  18. The Hardy-Weinberg Law • The Hardy-Weinberg Law can be used to estimate allelic frequencies • Example: • In Arizona (US) live the Hopi Indians. • In this tribe 26 cases of albinism wereobserved in a population of 6000. • So q2 = 26/6000 = 0.0043 and q= 0.065. • Therefore p=0.935. • The frequency of heterozygotes is 2pq = 2×0.935×0.065 = 0.122, thus one out of eight Hopi’s carries an allele for • albinism. • Is this calculation allowed? Photograph taken in 1900

  19. Genetic variation • The genetic structure of populations varies in space and time. • In mussel we observe cline:change of allelic frequencies alonga geographical transect. • In humans 15% of the variationis found between different popula-tions, 85% is shared across populations.

  20. Genetic variation • How is genetic variation measured? • The genetic basis of most traits is too complex to assign genotypes to individuals. Only a few traits behave in a Mendelian fashion. • Cytological variation (chromosome morphology and banding pattern) was observed in salivary glands of fruit flies. • Use of starch gel electrophoresis to study protein polymorphism made it possible to determine genotypes of many individuals at many loci. • By measuring variation at the DNA- or RNA-level.

  21. Genetic variation • Allozyme electrophoresis • proportion of polymorphic loci = frequency of loci with more than one allele in a population • heterozygosity HI=proportion of an individal’s loci that are heterozygous • allozymes=alleles for different enzymes • isozymes=enzymes of like structure and function coded by different loci

  22. Genetic variation • Electrophoretic pattern • Monomer11 homozygous12 heterozygous22 homozygous • Neutral mutation model states that observed variation in enzymes is neutral to natural selection. • Study assignments: “Measuring Genetic Variability in Natural Population by Allozyme Electrophoresis” See Blackboard External Links • “Flash animation Measuring Genetic Variation” • See Blackboard Course Documents Week 2 b) dimer 11 homozygous 12 heterozygous 22 homozygous c) tetramer 11 homozygous 22 heterozygous 22 homozygous

  23. Genetic variation • Measuring geneticvariation at the DNA level by Restriction Fragment Length Polymorphism (RFLP) • Amplify DNA-sequence of interest of a number of individuals with PCR • Use restriction enzymes to cut amplified fragments into smaller fragments • Separate restriction fragments on agarose gel • RFLP’s are inherited like alleles for other traits

  24. Genetic variation • Individual nucleotide heterozygosity HI in eukaryotes varies from 0.001 to 0.02 • In the human genome nucleotide heterozygosity HI is estimated 0.0008, which means that an individual is heterozygous at about 1 in every 1250 nucleotides • The highest diversity occurs at sites that do not change the amino acid sequence, which are called synonymous changes • Non-synonymous changes may affect a protein’s function and are subjected to natural selection and have been eleminated from the population • In many regions of the DNA there are short, identical segments, called Short Tandem Repeats (STR) or microsatellites, the number of which varies from person to person. They are used in forensics, parent testing, etc. • Exercises: 22.1-22.14 with a *

  25. Forces that change gene frequencies • The simplest way to calculate the population heterozygosity HT for a single locus is as: • HT = 1 – pi2 • i  • where pi is the frequency of the i-th of k alleles. [Note that p1, p2, p3 etc. may correspond to what you would normally think of as p, q, r, s etc.]. • If we want the gene diversity over several loci we need double summation and subscripting as follows: • HT = 1 – pij2 • ij • 2 alleles: if p=q=0,5 then HT = 1–(0,25+0,25) = 0,5 • 10 alleles: if p1=p2=p3 … p10=0,1 then HT = 1–(10×0,01) = 0,9

  26. Forces that change gene frequencies • For many populations the conditions for Hardy-Weinberg equilibrium do not hold, because: • Mutation occurs • They are small • Natural selection favours/dismisses alleles • Mating may be non-random • Migration takes place

  27. Forces that change gene frequencies • Mutations are heritable random changes in the DNA of a locus • New alleles occur only as a result of mutation • Mutations provide raw genetic material on which natural selection acts • Kinds of mutations:- base substitutions: transitions A«»G and C«»T are more common than transversionsA«»T, G«»T, C«»G and A«»C - Deletions and insertions - Chromosomal mutations (inversions and translocations) • see Flash animation Translocations on Blackboard

  28. Forces that change gene frequencies • Estimates of mutation rates • Using sequence analysis • 1/1.000.000.000 mutations per base pair per cell division • Varies 1000-fold within genome: high at mutational hotspots • Using single base mutations with phenotypic effect • 1/100.000 per gamete (humans, mice, Drosophila) • By keeping DNA non-expressed for many generations • Mutation rate = 1 per zygote, mostly mildly deleterious and recessive • Using amino acid chang • 4.2 amino acid altering mutations per ind. per generation • 1.6 of these are deleterious • Mutation rates in males > females

  29. Forces that change gene frequencies • A mutation can be neutral, detrimental, or beneficial, which depends on the specific environment e.g. DDT-resistance in insects • Forward mutation rate u is higher than reverse mutation rate v • Forward mutation from A to a lowers p and the reverse mutation form a to A increases p. At equilibrium: • v u • p =  and q =  • u + vu + v

  30. Forces that change gene frequencies • How fast do mutations change allele frequencies? • Example: • p=0.9 and q=0.1 • u=5×10–5 and v=2×10–5 • After one generation: • p=vq–up=(2×10–5×0.1)–(5×10–5×0.9)=–0.000043 • At equilibrium p=0.286 • This means that a change from 0.50 to 0.49 takes 680 generations • A change from 0.30 to 0.29 takes 17,200 generations!

  31. Forces that change gene frequencies • Random genetic drift

  32. Forces that change gene frequencies • Random genetic drift

  33. Forces that change gene frequencies • Random genetic drift is caused by sampling error when gametes are drawn randomly from a small population. • Example: • A small population with 12 individuals p=0.5 and q=0.5 has genotypes 3AA, 6Aa and 3 aa. Suppose an accident kills 25% of the population, which all happen to be aa. • This chance is ....... • In the next generation p becomes ...... • and q becomes .......

  34. Forces that change gene frequencies • Bottlenecks and founder effects • When populations remain small over many generations drift plays an important role in allelic frequencies. Scattered populations in isolated habitats each undergo drift. • Founder effect occurs when a population is established with a small number of breeding individuals. • Example:Tristan da Cunha was settled in 1817 by William Glass and wife. A few more settlers joined them. In 1855 population size was 100 with 26% of all genes from the Glasses. • In 1961 after a volcanic eruption the whole population of 300 was evacuated to England. 14% of their genes were still Glass’s.

  35. Forces that change gene frequencies • Bottleneck effect occurs after a drastic reduction in population size. • Example:On Tristan da Cunha in 1857 to population dropped from 103 to 33. In 1885 15 men drownded, many left the island and the population decreased further from 106 to 59. Many alleles were lost as a result. • Main effect of genetic drift is fixation: only one allele is left.

  36. Forces that change gene frequencies • Genetic drift is random, individual populations do not change in the same direction. This divergence increases over generations. • This is the basis of the Neutral theory of molecular evolution: a new mutation that is neutral with respect to natural selectionwill most likely be lost as a result of genetic drift. Occasionally the mutation will drift to fixation. As mutation is a recurring event, a gene will accumulate differences over time by chance alone.In this way the genes of two related lineages can be compared and used to estimate the date since they last shared a common ancestor.Following this method humans and chimpanzees shared a common ancestor 6 million years ago.What is your opinion about this method?

  37. Forces that change gene frequencies • Can we predict Genetic drift? • Depends on effective population size = number of individuals that contribute gametes to the next generation • 4×Nf×Nm pqpq • Ne =  , variance =  , standard error s=  • Nf+Nm 2Ne 2Ne • In which Nf=number of females, Nm= number of males • Example: • In a population of 70 females where only 2 males fertilise all the females, the females each contribute 1/2×1/70= 0.007 and each male 1/2×1/2= 0.25 of the gametes • Ne = (4×70×2)/(70+2) = 7.8, which means that drift occurs in about the same rate as in a population with 4 females and 4 males! • If p=0.8 then sp= √pq/2Ne = 0.1 √

  38. Forces that change gene frequencies • Genghis Khan a Prolific Lover, DNA Data Implies • Hillary Mayellfor National Geographic NewsFebruary 14, 2003 • Genghis Khan, the fearsome Mongolian warrior of the 13th century, may have done more than rule the largest empire in the world; according to a recently published genetic study, he may have helped populate it too. • An international group of geneticists studying Y-chromosome data have found that nearly 8 percent of the men living in the region of the former Mongol empire carry y-chromosomes that are nearly identical. That translates to 0.5 percent of the male population in the world, or roughly 16 million descendants living today.

  39. Forces that change gene frequencies • The amount of drift is inversely proportional to population size. Example: If the initial allele frequency p0=0.3 and N=10 then it takes 2.4×10 generations on average for the allele to be lost.

  40. Forces that change gene frequencies • Balance between mutation and random genetic drift. Mutation introduces new variation, genetic drift removes it. • The Infinite alleles model states that every mutation generates a novel allele. In a large gene of 10,000 base pairs the chance that two mutations generate the same allele is very small. (300 amino acid protein ‡ 900 nucleotides ‡ 4900 = 10542 possible sequences) • In this situation the forces of mutation and genetic drift balance each other. The chance of drawing two alleles and having them be different is • 4Ne×u • HT =  • 1+4Ne×u • HT is equal to the heterozygosity in the total population.

  41. Forces that change gene frequencies • Migration • Immigration introduces new alleles into populations: gene flow • Gene flow changes allelic frequencies in the recipient populationExample:px = f(A)x = 0.8 in population x and py = f(A)y = 0.3 in population yIn each generation some individuals migrate from x to y. After migration population y consists of two groups of individuals: a proportion m of migrants with px = 0.8 and 1–m residents with py = 0.3.The frequency of A in y after migration is p’y = mpx + (1–m)py • p= m(px–py) • Through exchange of alleles different populations remain similar, divergence is reduced • Example: Monarch butterfly

  42. Forces that change gene frequencies • Natural selection • In contrast to mutation, genetic drift and migration, natural selection results in adaptation. Adaptation is the proces by which traits evolve that make organisms more suited to their environment. • Natural selection is the dominant force in evolution. The concept comes from Darwin (who had no clue about transmission genetics). • Natural selection can be defined as differential reproduction of genotypes. It simply means that individuals with certain alleles produce more offspring than others. Therefore those alleles increase in frequency in the next generation. • Example Industrial melanism, described around 1900.

  43. Forces that change gene frequencies • Estimation of fitness must be done with great care!Example:Starlings have an optimal number of eggs they could raise to mature offspring. Laying more eggs, they produced fewer succesful chicks.So, assigning higher fitness values to birds that laid mote eggs would be incorrect. • At times natural selection can elimate genetic variation, and at other times it maintains variation. It can change allelic frequencies or prevent them from changing. It can produce genetic divergence between populations or maintain genetic uniformity.

  44. Forces that change gene frequencies • Natural selection is measured in terms of Darwinian fitness W, which is defined as the relative reproductive ability of a genotype. • The genotype that produces the most offspring has fitness W=1. • Selection coefficient s is the the relative intensity of selection against a genotype = 1–W • Example:

  45. Forces that change gene frequencies • Possible outcomes of natural selection • W11= W12= W22. No selection, allelic frequencies do not change. • W11= W12<1 and W22= 1. Selection operates against the dominant allele. • W11= W12= 1 and W22<1. Selection operates against the recessive allele. • W11< W12< 1 and W22=1. Selection operates without effects of dominance, the heterozygote has intermediate fitness. • W11 and W22<1 and W12= 1. Selection is favouring the heterozygote. • W12< W11 and W22= 1. Selection operates is favouring both homozygotes.

  46. Forces that change gene frequencies

  47. Forces that change gene frequencies Effectiveness of selection against a recessive lethal genotype at different initial allelic frequencies Fitnesses of the genotypes AA, Aa and aa are 1, 0.5, and 0.5 for the dominant case, 1, 0.75 and 0.5 for the additive case, and 1, 1 and 0.5 for the recessive case

  48. Forces that change gene frequencies • Heterozygote superiority = heterosis = overdominance • In this case both alleles are maintained in the population at an equilibrium, depending on the relative fitnesses of the homozygotes.If the selection coefficient of AA is s and the selection coefficient of aa is t, then at equilibrium: • peq=f(A)=t/(t+s) and qeq=f(a)=s/(t+s) • Example:Sickle-cell anemia is common in areas with malaria.Hb-A/Hb-A hasnormal bloodHb-A/Hb-S hassickle-cell traitHb-S/Hb-S has sickle-cell disease • Heterozygotes are relatively resistent to the malaria parasite.

  49. Forces that change gene frequencies • Balance between mutation and natural selection • Natural selection reduces the frequency of deleterious alleles:q=–spq2 • Mutation produces new alleles that can be harmful: q=up • At equilibrium both forces are in balance: spq2 =up • So at equilibrium: qeq=(u/s) • If the recessive homozygote is lethal: qeq=u • Example:mutation rate u=10–6 and s=0.1at equilibrium the frequency of the allele will be qeq=u/s=0.0032 • Most deleterious alleles remain within the population at low frequency because of equilibrium between mutation and natural selection

  50. Forces that change gene frequencies • Assortative mating • Many populations do not mate randomly for some traits • Positive assortative mating occurs when similar phenotypes mate preferentially • Negative assortative mating occurs when dissimilar phenotypes mate preferentially • Assortative mating does not alter the allelic frequencies, but it may influence the genotypic frequencies

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