Help
Options
Didn't know it?
click below
click below
Knew it?
click below
click below
Don't Know
Remaining cards (0)
Know
retry
shuffle
restart
0:00
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.
Normal Size Small Size show me how
Normal Size Small Size show me how
Evolutionary Biology
Genetic variation in populations
| Question | Answer |
|---|---|
| The theory of natural selection | 1. Individuals within species are variable 2. Some variations are passed on to offspring 3. In every generation more offspring are produced than can survive 4. Individuals who survive and go on to reproduce are are naturally selected |
| Heritable variation | Natural selection based on heritable variation within populations |
| Heritable variation | Law of segregation: individuals possess two alleles at each gene – one from each parent |
| Heritable variation | Law of independent assortment: genes for separate traits are passed on independently from parents to offspring |
| Heritable variation | Law of dominance: recessive alleles will always be masked by dominance alleles |
| The modern synthesis | •Heritable genetic variation is the principal material for natural selection. •Fixation of beneficial changes by natural selection is the main driving force of evolution |
| The modern synthesis | •Natural selection operates on ‘infinitesimally small’ variations, so evolution is gradual •Species are a central unit of evolution, and speciation a key evolutionary process •The entire evolution of life can be depicted as a single ‘big tree’ |
| Darwin’s four postulates after the modern synthesis | 1. As a result of mutation, gene flow and recombination, individuals within populations are variable for nearly all traits 2. Individuals pass their alleles on to their offspring intact |
| Darwin’s four postulates after the modern synthesis | 3. More offspring are produced than can survive 4. The individuals that survive and reproduce are those with the alleles that best adapt them to their environment |
| How much genetic variation exists within populations? | 1. What fraction of loci are polymorphic? 2. How many alleles are present at each locus? 3. What are the frequencies of the different alleles at loci? |
| Measuring genetic variation in populations | In snow geese (Chen caerulescens) two distinct colour morphs occur, caused by two alleles at a single locu - gel electrophoresis - microsatellite genotyping - DNA sequencing |
| Measuring genetic variation in populations | Measuring genetic variation is much more difficult for continuous traits ie. human height |
| How much genetic variation is in populations? | - nucleotide diversity |
| Case study: scarlet tiger moth (Callimorpha dominula) | • Three colour morphs in Oxfordshire population, differ in white spotting on forewing, and black on red hindwings. • Fisher and Ford wanted to know why these morphs existed at certain proportions, and how these changed over time |
| Genotypes | AA - Dominant homozygote (D) Aa – Heterozygote (H) aa – Recessive homozygote (R) |
| Sample = 1000 | For genotype AA this is 400 / 1000 = 0.4 Aa this is 400 / 1000 = 0.4 aa this is 200 / 1000 = 0.2 |
| Allele frequency calculations | AA - 400, Aa - 400, aa - 200, total = 1000 Allele frequency = proportion of a given allele in the population Each moth has two alleles at the colour locus So there are 1000 x 2 = 2000 alleles in the total sample |
| Calculate the frequency of allele A | Dominants have two copies of allele A, heterozygotes have one, so: The number of copies of A in the sample is: (400 x 2) + (400 x 1) = 1200 The frequency of allele A in the sample is: 1200 / 2000 = 0.6 |
| Formula | All individuals have a genotype, therefore: D + H + R = 1 p and q are traditionally used symbols for dominant and recessive allele frequencies. In a two allele system p + q = 1 |
| Genotype frequencies | D = nD / N = 400 / 1000 = 0.4 H = nH / N = 400 / 1000= 0.4 R = nR / N = 200 / 1000 = 0.2 |
| Allele frequencies | Frequency of A = p = nA / 2N = 0.6 p = D + (H / 2) = 0.6 Frequency of a = q = na / 2N = 0.4 q = (H / 2) + R = 0.4 q = 1 – p = 0.4 |
| Expected genotype frequencies | Consider locus with 2 alleles A and a (frequencies p and q both = 0.5) Under random mating: For an AA homozygote, both sperm and egg have to be A: Probability = p x p = p2 = 0.25 |
| Expected genotype frequencies p2 + 2pq + q2 = 1 | For a heterozygote, we could have Aspermaegg, or aspermAegg: Probability = (p x q) + (p x q) = 2pq = 0.5 For an aa homozygote, both sperm and egg have to be a: Probability = q x q = q2 = 0.25 |
| The Hardy-Weinberg principle | p2 + 2pq + q2 = 1 Where p and q are the frequencies of 2 alleles at a single locus |
| Hardy-Weinberg equilibrium | If observed and expected genotype frequencies are the same, a population is in Hardy-Weinberg equilibrium. Allele frequencies will not change over generations. |
| Hardy-Weinberg equilibrium | This will happen as long as five assumptions are met: a. infinitely large population size b. no mutation c. no selection d. no gene flow e. random mating |
Created by:
reub8n
Popular Biology sets