BIOL 239: Mendelian Genetic Textbook PDF

BIOL 239 Mendelian Genetics Textbook 1.1 1.2 1  Lesson-level learning objectives Compare and contrast natural and artificial selection Explain the laws of segregation and independent assortment Explain the molecular mechanisms of dominant and r...

BIOL 239 Mendelian Genetics Textbook 1.1 1.2 1  Lesson-level learning objectives Compare and contrast natural and artificial selection Explain the laws of segregation and independent assortment Explain the molecular mechanisms of dominant and recessive alleles Determine gametes, genotypes and phenotypes of offspring from crosses, using Punnett squares Determine offspring from multihybrid crosses using probabilities Define key terms including phenotype, genotype, gene, allele, gamete, pure breeding, monohybrid, dihybrid, 2 test cross  Genetic traits are passed on from one generation to the next Natural selection – Individuals with certain traits are more likely survive and reproduce in a given environment Artificial selection – Humans choose which individual plants/animals reproduce 3 Images: single rat © Alan Baker / Stockillustrations, 4 rats © Victor Chavez / Getty Images  4 Image: https://doi.org/10.1073/pnas.060433510  5 Image: https://doi.org/10.3389/fevo.2020.00103  Bred for desirable traits (phenotypes) 6 Image: https://www.farmersalmanac.com/what-queen-annes-lace 7 Random DNA mutations cause small phenotypic (observable) changes Humans choose which organisms reproduce Accumulate over time 8  Dogs - domesticated wolves (same species) - human-compatible wolves were successful - self-domestication - lots of genetic variation - more variation in skeletal size and proportion than any other mammal - inbreeding - models for disease 9 Images: https://www.pbssocal.org/redefine/gray-wolves-in-california-a-timeline https://www.snexplores.org/article/turning-wolves-dogs-may-have-occurred-twice https://www.science.org/content/article/urban-foxes-may-be-self-domesticating-our-midst 10  Fox domestication experiment (1950’s, Dmitry Belyayev) Selected for tameness 11 Images: silver fox © Tom Reichner/Shutterstock, domestic fox from https://www.nationalgeographic.com/animals/article/fox-dogs-wild-tame-genetics-study-news 12  Domestication syndrome (animals) floppy ears variations in coat color shorter muzzle smaller tooth size prolonged juvenile behaviour extended breeding cycle hormonal changes 13 Gregor Mendel (1822-1884) Augustinian monk and expert plant breeder (Austria 1800s) Experiments with garden peas 14 (not allowed to use mice)  Garden peas Self-fertile Easy to cross-fertilize Large number of offspring Short growing season 15  Clear alternative forms of particular traits 16 Image: https://opentextbc.ca/biology/chapter/8-1-mendels-experiments/  Pure-breeding lines – Offspring have same traits as parents – Inbred Carefully controlled breeding 17  Mendel’s experiments with garden peas Self- vs Cross-fertilization (cross-pollination) Gametes (reproductive cells) Male ( ): pollen/sperm female ( ): ovules/eggs Each pea is a separate individual 18 Mating of true-breeding parents having antagonistic traits Dominant: yellow, round, purple Recessive: green, wrinkled, white Doesn’t matter which parent is male and which is female 19 Found a consistent pattern of inheritance, from which he based his theories Prior to Mendel: theories that one parent contributes most to an offspring’s inherited features (disproved through reciprocal crosses) parental traits become mixed and forever changed in the offspring (disproved through reappearance of recessive traits) 20  Monohybrid crosses Parental (P) true breeding Matings between individuals that differ in only one trait ALL F1 progeny resembled First filial (F1) one of the parental strains Monohybrids / heterozygotes In F2 progeny, lost trait reappeared Second filial (F2) 21 Phenotype = observable characteristic (largely determined by genotype) -commonly referred to as a trait Genotype = genetic make-up; description of the genetic information carried by an individual (Today we can sequence DNA. Mendel didn’t know DNA existed) 22 Discrete units of inheritance are alleles of genes. Alleles are alternative forms of a single gene. Chromosome 3 from parent 1 homologous same gene different alleles chromosomes -different DNA sequence Chromosome 3 from parent 2 23 Hypothetical example -let’s say each of the following features is determined by a single gene GENE for trait: ALLELE eye pigment (HERC2) brown or blue skin pigment albino or pigmented height tall or short hair texture curly or straight seed texture (peas) smooth or wrinkled Most human traits (including the above) are determined by multiple genes with multiple alleles. 24 Genes/alleles are always designated by italics WW Ww ww dominant recessive Smooth seeds Wrinkled seeds Trait: seed texture 25 In EVERY population, there will ALWAYS be different versions (alleles) of EVERY gene 26 A gene may have several alleles that normally occur in a population = polymorphic but a maximum of two for one gene can exist in a diploid individual e.g., humans Some genes have only one allele that is normally present in a population (other alleles are very rare) = monomorphic 27  Xenopus laevis is tetraploid 28 Image © William Leonard Mendel’s law of segregation: The two alleles for each trait separate (segregate) during gamete formation, then unite at random, one from each parent, at fertilization. -describes how alleles of 1 gene behave 29 Note that the symbols we see given to the gene for seed texture can vary -commonly W (meaning wrinkled), but also R (rough) -does not matter as long as the genotype notation used is defined -you’ll be making up your own genotype code for some of your assignments The symbols don’t really matter, as long as they make sense! The convention is to use a symbol representative of the mutant or recessive trait, but we are not going to adhere to this. 30  The law of segregation YY Y Y yy y y Y Yy y Y y 31  Law of segregation Punnett square monohybrids (heterozygotes) R r W R RR Rr Rr Rr W w W w R r XR r r Rr rr phenotype genotype F2 genotypes: ¼ RR, ½ Rr, ¼ rr F2 phenotypes: ¾ smooth, ¼ wrinkled Each parent carries 2 copies of each gene specific allele exists for each gene 32 each individual receives 1 from each parent Numbers! Punnett square shows the POSSIBILITIES, but there could by 100s of offspring, not just 4......or there could be just 1 offspring R r W R RR Rr F2 genotypes: ¼ RR, ½ Rr, ¼ rr F2 phenotypes: ¾ smooth, ¼ wrinkled r Rr rr 33  You don’t always get the exact expected ratios, especially with smaller numbers of offspring e.g. Family with 4 children, expected ½ male and ½ female 34 Photo © Louise Gleeson Approx. 3:1 phenotypic ratio for simple dominant/recessive traits that result from a cross between monohybrids (one gene) …provided we look at enough offspring R r W R RR Rr r Rr rr F2 genotypes: ¼ RR, ½ Rr, ¼ rr F2 phenotypes: ¾ smooth, ¼ wrinkled 35  Monohybrid crosses Parental (P) true breeding Matings between individuals that differ in only one trait ALL F1 progeny resembled First filial (F1) one of the parental strains Monohybrids / heterozygotes In F2 progeny, lost trait reappeared Second filial (F2) 36 Q How many different gametes and genotypes are produced from the cross Aa x aa? Do a Punnett square A a a a Punnett square uses different gamete possibilities and determines the possible genotypes 37 Q How many different gametes and genotypes are produced from the cross AaBb x Aabb? Do a Punnett square 38 Q How many different gametes and genotypes are produced from the cross AaBb x Aabb? Do a Punnett square AB Ab aB ab Ab ab This is for 2 genes Probability of each square is 1/8 39 How does genotype influence phenotype? YY Yy 40 yy At the DNA level, different alleles differ in nucleotide sequence. This can cause differences in the amino acid sequence or the amount of protein produced. 41 e.g., trait: texture; round (smooth) or wrinkled alleles: round (R) or wrinkled (r) gene: codes for starch-branching enzyme 1 (SBE1) Mutation causes amino acid change that inactivates SBE1 SBE1 Phenotype Genotype Active enzyme RR Soluble starch Insoluble starch Inactive enzyme rr Soluble starch Soluble starch (Enzymes make chemical reactions happen faster) 42 RR large amounts of branched insoluble starch - holds less water rr no branched starch - holds more water - pea loses water and shrivel Rr still produces enzyme - sufficient R allele is dominant over r 43 What if someone gives you a yellow pea and asks you to determine its genotype? What if you’re using plants that are not self-fertile? 44  Test crosses -cross against a (homozygous) recessive phenotype for the trait in question YY yy Yy yy g g G Yy G Yy g yy 45  Mendel’s results reflect basic rules of probability The law of product : The probability of two or more independent events occurring together is the product of the probabilities that each event will occur by itself. AND P of event 1 P of event 2 46  Law of Product AND e.g., coin toss : chance of heads twice in a row = probability of a head and a head 1/2 x 1/2 = 1/4 Chance of cross between heterozygotes (Aa) producing homozygous recessive offspring a and a 1/2 x 1/2 = 1/4 47 THINK IN TERMS OF ALLELES ie GAMETES! Genotypic ratios (AA:Aa:aa) = 1:2:1 Phenotypic ratios: (orange:blue) = 3:1 48 The law of sum : The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities OR 49  Law of Sum OR e.g., dice throw : chance of an even number = probability of a 2 or a 4 or a 6 1/6 + 1/6 + 1/6 = 3/6 = 1/2 chance of cross between heterozygous plants (Aa) producing an orange offspring (orange is dominant) = AA or Aa or aA = (1/2 x 1/2) + [(1/2 x 1/2) + (1/2 x 1/2)] = 1/4 + 1/2 = 3/4 50 Phenotypic ratios: Orange (AA or Aa) = 3/4 3:1 Blue (aa) = 1/4 51 In tutorial 2, you will learn how to use binomial expansion to answer genetic prediction questions. responsible for this on tests/exam 52  Dihybrids individuals that carry different alleles of two genes 53  Dihybrid crosses dihybrids 2n gametes 2 = diploid, n = # genes/traits examined Ratio of yellow (dominant) to green (recessive) = 12:4 or 3:1 Ratio of round (dominant) to wrinkled (recessive) = 12:4 or 3:1 54  Dihybrid crosses 9:3:3:1 phenotypic ratio 55 Warning Do not assume dihybrids are always produced From Parentals who are homozygous dom/dom X rec/rec e.g., GG WW x gg ww can be GG ww x gg WW 56 When dihybrids (2 genes) are crossed and the alleles act in a simple dominant and recessive manner, NEW phenotypic combinations appear, not present in previous generations The two traits act independently Law of segregation describes how different alleles of a single gene behave Law of independent assortment describes how different alleles of different genes behave 57 Law of segregation -each allele of a gene acts independently Ratio of yellow (dominant) to green (recessive) = 12:4 or 3:1 Ratio of round (dominant) to wrinkled (recessive) = 12:4 or 3:1 58  Branching diagrams - genotype YyRr x YyRr - dihybrid cross Trait 1 AND Trait 2 1/4 RR 1/16 YYRR (yellow, round) 1/4 YY 1/2 Rr 2/16 YYRr (yellow, round) 1/4 rr 1/16 YYrr (yellow, wrinkled) 1/4 RR 2/16 YyRR (yellow, round) 1/2 Yy 1/2 Rr 4/16 YyRr (yellow, round) 2/16 Yyrr (yellow, wrinkled) 1/4 rr 1/4 RR 1/16 yyRR (green, round) 1/4 yy 1/2 Rr 2/16 yyRr (green, round) 1/4 rr 1/16 yyrr (green, wrinkled) Phenotypic ratio : 9/16 yellow, round 3/16 yellow, wrinkled 59 3/16 green, round 1/16 green, wrinkled Test crosses and dihybrids -a yellow, round pea Y_R_ ? YY RR YY Rr Yy RR Yy Rr 60 Test crosses and dihybrids -a yellow, round pea Y_R_ ? YY RR YY Rr Yy RR Yy Rr 61 Test cross – use to uncover an unknown genotype, when inheritance pattern and phenotype is known ALWAYS uses a homozygous recessive individual for the genotype in question What’s the genotype? 62  YyRr x YyRr YR Yr yR yr YR Yr yR yr # diff gametes/parent = 2n (2 for 2 diff alleles (diploid) and n for #of genes) =4 63  3 Traits What if we were dealing with 3 or more traits? -how to handle that? 23 = 8 gametes = 8x8 =64 squares -law of segregation allows us to treat alleles of a single gene independently -law of independent assortment allows us to treat different genes separately -lets us apply probability laws 64 Multihybrid crosses Matings between individuals that differ in three or more traits e.g., Aa Bb Cc Dd x Aa Bb Cc Dd Punnett square would be huge as would branch diagram -break each down into the 4 individual gametes and work the probabilities for each combination occurring. -use the genotypic ratios established for heterozygous crosses i.e., 1/4 recessive, 1/2 hybrid, 1/4 dominant Probability of an AA bb Cc Dd offspring: =1/4 x 1/4 x 1/2 x 1/2 =1/64 65  Next: Modifications of Mendelian Ratios Textbook 2.1 2.2 2.3 4.4 (last part only, “Autosomal genes contribute..”) 66