Genetics Lecture 3-5 PDF

Teaching Assistant Office hours Monday – Wednesday evenings FEA 141 Genetics TAs: Monday 6:00 – 8:00 PM Hannah Kotek Tuesday 6:00 – 8:00 PM Hannah Kotek Wednesday 6:00 – 8:00 PM...

Teaching Assistant Office hours Monday – Wednesday evenings FEA 141 Genetics TAs: Monday 6:00 – 8:00 PM Hannah Kotek Tuesday 6:00 – 8:00 PM Hannah Kotek Wednesday 6:00 – 8:00 PM Ashley Lee Thursday 6:00 – 8:00 PM Ashley Lee TA for our class: Hannah Kotek In addition to academic questions – 1 Feel free to drop by for study tips, strategies to succeed in Genetics, etc. Lectures 3 - 5 Mendelian analysis & Probabilities Mendelian Analysis Mendel’s peas and crosses Law of Segregation Law of Independent Assortment Probabilities review Product rule Sum rule 2  Practice problems to help you understand this material: Chapter 3: These problems will NOT be turned in. NOTE – A separate problem set will be assigned later, to be turned in for a grade. This will include (~10) problems covering material from several lectures. 3  test crosses used to determine cross with an individual the genotype of a with a known genotype… phenotypically a phenotypically dominant individual recessive individual pp Pp or PP ? x test for hidden recessive genes with homozygous recessive tester strain – phenotype of progeny from test cross indicates the genotype of tested parent 4  test crosses A test cross allows one to determine the alleles carried by the F1 parent, because the test cross parent can only contribute the recessive allele. The phenotypes of the resulting progeny allow you to determine the genotype of the F1 parent. Is this yellow pea result: YY or Yy? 1/2 yellow F1 X cross it with a yy pea 1/2 green (test cross) 5 so the F1 yellow pea was:____  expected results of a test cross Yellow pea is dominant to green pea If a yellow pea is heterozygous: – Yy X yy (tester) – expect the progeny will be _______% 50 Yy (yellow) 50 _______% yy (green) If a yellow pea is homozygous: – YY X yy (tester) – expect the progeny will be 100 _______% Yy (yellow) _______% 0 yy (green) This works for all 7 of 6 Mendel’s chosen traits (and many others).  Mendel followed two genes at once Consider two traits for pea color and shape: – Y (yellow) and y (green) – R (round) and r (wrinkled) PROBLEM: A true breeding plant with round, green seeds is crossed with a true breeding plant with wrinkled, yellow seeds. What are the genotypes and phenotypes of the parental and F1 generations? 7  Tracking two genes – Are two alleles inherited together? pure-breeding parentals F1 are all RrYy (Round, yellow) Self-fertilize F1 observe 9:3:3:1 ratio Note that the round, green and wrinkled, yellow phenotypic combinations observed in the parents did not stay together 8 in the offspring!  Mendel followed two genes at once In a “dihybrid” cross – parents are heterozygous for two traits Each dihybrid plant produces 4 gamete types, equally frequently YR ___, – YyRr (adult) → four gamete types: ___, Yr yr yR or ___ ___, Each gamete has one allele for each trait. 9 Punnett square of a dihybrid cross each F1 produces four different types of gametes in equal proportions these gametes come together randomly to form a zygote 10 a Punnett square of dihybrid cross each F1 produces four different types of gametes in equal proportions these gametes come together randomly to form a zygote CONCLUSIONS: each single trait still gives 3:1 ratio combined, the overall ratio is 9:3:3:1 (the product rule… next lecture) 11  Mendel’s Second Law Law of independent assortment – segregation of alleles of two different genes are independent of one another – for example: no bias toward yR or Yr in gametes – random fertilization of ovules by pollen no bias of gametes for 12 fertilization  dihybrid test cross heterozygous for both genes YyRr - yellow round 31 Yyrr - yellow wrinkled 27 F1 X YyRr yyrr yyRr - green round 26 round wrinkled yyrr - green wrinkled 26 yellow green Gametes: Gametes: YR yr Yr yR yr this is a ratio of 1:1:1:1 13 test cross confirms independent assortment of traits  patterns of segregation one gene (one trait, two phenotypes) – Monohybrid cross – 3:1 (F2) phenotypic ratio – Test cross of F1 – 1:1 phenotypic ratio two genes (two traits, four total phenotypes) – Dihybrid cross – 9:3:3:1 (F2) phenotypic ratio – Test cross of F1 – 1:1:1:1 phenotypic ratio 14  some genetic vocabulary progeny – offspring, children, results of reproduction gametes – sperm and egg; carry half the genetic info of parent; haploid zygote – results from fusion of gametes; diploid haploid – one copy of each chromosome/gene (in animals, only in gametes) diploid – two copies of each chromosome/gene (in animals, all other cells) homozygous – having 2 identical copies of a gene heterozygous – having two different versions of a gene gene – discrete unit of inheritance – determines the expression of proteins – segments of DNA, one gene is part of a chromosome – one chromosome has many genes alleles – different forms of a specific gene genotype – the alleles carried by an individual for a gene or set of genes phenotype – the appearance (expression) of the alleles for a given gene or set of genes mutation – heritable change in a gene resulting in a new allele (most of the allele variation we study in genetics arises from mutation) wild type – standard allele, the most common allele found in the wild variant allele = new allele = mutant allele 15  Probability probability of # of times event is expected to happen an outcome = # of opportunities (trials) sum of all the probabilities of all possible events = 1 60 red gum balls 40 green gum balls if you buy one gum ball, the probability of getting a red one is ____. 0.6 = # of red 60 red = total # of gumballs 100 gumballs 5¢ 16  Product Rule The probability of independent events occurring together is the product of the probabilities of the individual events. p(a and b) = p(a) x p(b) Example - A game of dice: What is the chance of getting a 5?___ 1/6 What is the chance of getting a 5 on the second roll? ___ 1/6 17  Product Rule The probability of getting a 5 on the second die is independent of what the first die shows. What is the probability of getting 5 on BOTH rolls of the dice? The probability of two independent events occurring together: p(a and b) = p(a) x p(b) – one has probability of 1/6 – second has probability of 1/6 18 chance of getting two 5’s = 1/6 X 1/6 = 1/36  Sum Rule The probability of either of two (or more) mutually exclusive events occurring is the sum of their individual probabilities. p(a or b) = p(a) + p(b) probability of getting two 5’s = 1/36 probability of getting two 6’s = 1/36 so, the probability of getting either two 5’s or two 6’s is 1/36 + 1/36 = 1/18 19  Using both the product and the sum rules You have two nickels (to buy two gum balls). What is the probability of getting one green and one red gum ball? This can happen in two 1st 2nd ways: 5¢ 25¢ 5¢ 25¢ green first & then red, 5¢ or 60 red gum ball 1st 2nd 40 green gum balls 5¢ 25¢ 5¢ 25¢ red first & then green. 20  Probability p(green, then red) = p(green) X p(red) → product rule = 0.4 X 0.6 = 0.24 -- or -- p(red, then green) = p(red) X p(green) → product rule = 0.6 X 0.4 = 0.24 Therefore, the probability of getting one red and one green gum ball is = p(green, then red) + p(red, then green) = 0.24 + 0.24 = 0.48 → sum rule 21