Presentation on theme: "Mendel: father of genetics Quick review of terminology"— Presentation transcript:
1 Inheritance of Single-Gene Differences – discovered by Gregor Mendel!!! Mendel: father of geneticsQuick review of terminologyMendel’s Empirical approachMonohybrid crossMendel’s Postulates to explain his dataMendel’s First “law” equal segregationPunnent SquareDihybrid crossMendel’s Second “law” independent assortmentThe branch diagram & probabilitiesUsing the testcross
2 Who was this “Father of Genetics”? I. Gregor Johann MendelWho was this “Father of Genetics”?
3 Transmission genetics – link between meiosis & Mendel’s postulates Mendel determined the transmission of discrete units (genes located on chromosomes) from parent to offspring, predicting the formation of gametesFuture cytological studies suggested a correlation exists between the behavior of chromosomes during meiosis and the transmission of traits
4 A. Terminology review Genes come in different forms = ALLELES i.e. there may be a single gene for flower color but several alleles, each producing a different colorEach individual has 2 alleles per gene (1 derived from mother, 1 from father)Phenotype = expressed form of a character (what an individual looks like)Genotype = specific set of alleles carried by an individual (the actual genetic composition)Homozygous = the alleles of a gene are identical (AA)Heterozygous = the alleles of a gene are different (Aa)Dominant allele = an allele that expresses its phenotypic effect even when heterozygous… therefore AA and Aa have the same phenotypeRecessive allele = An allele whose phenotypic effect is not expressed in a heterozygote… therefore (a) can only be expressed when the individual is homozygous – (aa).
5 Terminology cont. - Genetic Crosses Controlled mating of two specific organismsSelf Cross = cross to oneself (plants, fungi)Haploid Cross = simplest, each gene present in 1 copy only (fungi)Diploid Cross = each gene present in 2 copiesHomozygote cross (AA x AA), aka pure-breedingHeterozygote cross (Aa x Aa)Testcross = cross with a known homozygote recessiveBackcross = hybrid offspring are crossed with one of the parentsReciprocal Cross = in an initial cross, if the female parent has the mutant condition & the male parent has the wild type condition - The reciprocal cross is the reverse (female is wild type & male is mutant)
6 B. Mendel’s success with the empirical approach Came up with an elegant model of experimental designchose a good “model” organism: Pisum sativumrestricted his examination to one or very few pairs of contrasting traits in each experimenttook meticulous notes with accurate quantitative records
8 Mendel’s Empirical approach By using controlled crosses, Mendel designed experiments to determine the quantitative relationships from which laws could be discovered
9 Looked at contrasting characteristics of the garden pea -seed coat, seed color, petal color, pod shape, pod color, stem size, axial/terminal flowers.
10 II. The Monohybrid cross Hybridization = when two plants of the same species but with different characteristics are crossed (mated) to each other.Mono = dealing with one pair of contrasting characteristicsP – parental generationF1 – First filial generationF2 – Second filial generation
11 Mendel’s results from the monohybrid crosses ParentalF1F2F2 ratioRound x wrinkledAll round5474 round wrinkled2.96:1Yellow x green seedsAll yellow6022 yellow 2001 green3.01:1Purple x whiteAll purple705 purple 224 white3.15:1Inflated x pinchedAll inflated882 inflated pinched2.95:1Green x yellow podsAll green428 green 152 yellow2.82:1Axial x terminalAll axial651 axial 207 terminal3.14:1Long x shortAll long787 long 277 short2.84:1
12 A. Mendel’s Postulates to explain his data the existence of unit “factors” – particulate theory of inheritanceTraits inherited as discrete units that remain unchanged as they pass from parent to offspringgenes are in pairs, thus 2 phenotypes must be determined by 2 different alleles of 1 geneWhen two unlike unit factors responsible for a single character are present in a single individual, one unit factor is dominant to the other (Dominanance/Recessiveness)the principle of segregation, genetic units segregate from each other
13 Points 4 & 5:4) gametic content – the F2 3:1 ratio is based on a 1:1segregation in a heterozygote5) random fertilization – gametes are brought togetherfor fertilization in a random manner
15 B. Mendel’s Law of equal segregation: Equal Segregation = The two members of a gene pair segregate from each other into the gametes; so half the gametes carry one member of the pair and the other half of the gametes carry the other member of the pair.
19 C. Using Punnett Squares in Genetic Crosses Punnett squares used for monohybrid crossesConsiders only genes of interestList sperm genotypes across topList egg genotypes down sideFill in boxes with zygote genotypes
20 Pollen of Heterozygous Plant Making a Punnett Square: Heterozygous X HeterozygousPpEggs of Heterozygous PlantPPPPpPollen of Heterozygous PlantpPpppFrequenciesPhenotypesGenotypesPPpPPpp211PurpleWhite3 (75%)1(25%)
21 D. Using the testcross to determine if the parent is heterozygous The organism of the dominant phenotype is crossed to a known homozygous recessive individual
22 III. Mendel’s Dihybrid Cross Follows the inheritance of two different traits within the same individual.i.e. Yellow, Round x Green Wrinkled
25 A. Mendel’s (postulate) Second Law of independent assortment: Independent Assortment = two different genes on different chromosomes will randomly assort their alleles during gamete formationThe alleles assort independentlyPossible gametes producedfrom meiosis
26 (Hair color) & (Hair length) Black/Brown Short/LongP: Black, short x Brown, longF1: all black, shortF2: Black, short x Black, short:BbSs x BbSs
29 B. Probability and statistics in genetics How can we calculate the expected ratios of the phenotypes/genotypes for progeny?How can we determine if our results are significantly different from what we would expect under Mendelian principles?Pr(A) = Probability of an event A, number between 0 and 1 that measures the likelihood that A will occur when the experiment is performedAccuracy of prediction depends on sample size
30 Probability RulesProduct rule: the probability of independent events occurring together is the product of the probabilities of the individual eventsPr(A) x Pr(B) = Pr(A and B)Sum rule: probability of either of two mutually exclusive events occurring is the sum of their individual probabilities.Pr(A) + Pr(B) = Pr(A or B)Conditional probability: the probability that one outcome will occur, given the specific condition upon which the outcome is dependentPrc = Pr(a)/Pr(b)
31 e.g. Product rule in practice If you self cross an F1 dihybrid yellow, round pea plant-What proportion of offspring will be yellow and round?Probability of producing yellow peas: ¾ (Y/Y or Y/y)Probability of producing round peas: ¾ (R/R or R/r)Therefore, Yellow-Round offspring:¾ x ¾ = 9/16What if you crossed pure breeding tall plants with purple flowers that make yellow round peas with short plants with pure breeding white flowers that make green wrinkled peas?
32 P: T/T;P/P;Y/Y;R/R x t/t;p/p;y/y;r/r F1: T/t;P/p;Y/y;R/rF2?What is the probability of having tall plants with purple flowers that make yellow peas? (T/-;P/-;Y/-;R/-)¾ x ¾ x ¾ x ¾ = 81/256
33 Properties of probabilities The probability of an event always takes on a value between 0 and 1The probability of two events occurring together is equal to Pr(A) x Pr(B)If two events A and B are mutually exclusive, then the probability the either A or B occurs is equal to Pr(A) + Pr(B)
34 A pure breeding black guinea pig is crossed with a pure breeding tan guinea pig. If black is dominant to tan, what will the genotype and phenotype of the F1 be? Give proportions.For the above, what would the genotypes and phenotypes of offspring from an F1 x F1 cross be? Give proportions.
35 The forked-line method, or branch diagram Calculate the probability of obtaining an aa; B-; C- zygote from the cross Aa; Bb; Cc X Aa; Bb; Cc.Much simpler than using the Punnent square for looking at more than one traitGenetic ratios – expressed as probabilitiesBased on the product rule of probabilityPr(A) x Pr(B) = Pr(A and B)
36 Binomial expansionUsed to predict the probability of an unordered combination of eventseach event possesses one of two mutually exclusive characteristics, eg. curly hair or straight hairthe outcome for any one event is independent of the outcome for any other eventExample: from a cross between two tall plants, Tt x Tt, what is the probability of having 2 dwarf plants out of five offspring?
37 Step 1. calculate individual probabilities (p & q) p = ¼, q = ¼ Equation to determine the probability of unordered eventsfrom a cross between two tall plants, Tt x Tt, what is the probability of having 2 dwarf plants out of five offspring?Pr(x successes in n trials) = n!(n-x)! x!pxqn-xStep 1. calculate individual probabilities (p & q)p = ¼, q = ¼Step 2. determine # of events in category x and the total # of eventsx = 2n = 5Step 3. substitute values for p, q, x in the equation5!3!2!x=120/12 x (0.0625)( )= or 0.97%
38 The ability to taste phenylthiocarbamide is an autosomal dominant phenotype, and the inability to taste it is recessive. If a taster woman with a nontaster father marries a taster man who, in a previous marriage had a nontaster daughter, what is the probability:a. that their first child will be a nontaster girlb. that their first child will be a taster girlc. that two out of three children will be nontasters1/8(½ x ¼)(½ x ¾)3/814%
39 Non-Mendelian ratios - Interactions between the alleles of one gene Inheritance of Gene Differences – non-Mendelian geneic interactions, part 2Non-Mendelian ratios - Interactions between the alleles of one geneInteractions between the alleles of more than one geneGene interactionEpistasis
40 A. Incomplete Dominance Two alleles (heterozygote) produce an intermediate phenotypeAt the molecular level, the mutant allele results in a reduced amount of functional protein2 doses = 100%1 dose = 50%0 dose = None
41 incomplete dominanceF1 hybrids have an appearance somewhere in between the phenotypes of the two parental varietiesF1 is pink, an intermediate color between white and red, F2 1:2:1
42 Example: Tay-Sachs disease – Homozygous recessive individuals are severely affected (death by age 3), Heterozygotes express only about 50% of hexosaminidase enzyme for lipid metabolism. Slightly affected.The closer we look, the more we find that heterozygotes are different from homozygous dominant individuals.
43 B. Multiple allelesSome genes are found in three or more alleles that are different from each othere.g. white clover, coat color in rabbits
44 C = full coat color; dominant to all other alleles cch = chinchilla coat, a partial defect in pigmentation;dominant to ch and cch = himalayan coat, color in only certain parts of body;dominant to cc = albino, no color; recessive to all other allelesA rabbit with chinchilla fur is mated to a himalayan. Some of their F1 offspring have himalyan fur, some have chinchilla fur and some are albino. Name the genotypes of the parents and the genotypic ratios of the F1 offspring.
45 Codominance Genotype Blood Type IA/IA or IA/i A IB/IB or IB/i B IA/IB ABi/iOSpecific type of multiple alleles, when two alleles are equally expressed in the heterozygous individuale.g. ABO blood group
46 C. Lethal allelesAllele in an essential gene that has the potential of causing the death of an organism.Age of onsetConditional lethal allelesSemilethal alleles
48 II. Interactions between the alleles of more than one gene Genes interact in concert with other genes and with the environment to influence a particular characteristic.Final product
49 I Interactions between genes produce many different phenotypes Most traits can be affected by the contributions of two or more genesExamples: morphological characteristics - Height, weight, growth rate, pigmentation
50 One trait, involving between two genes Bateson & Punnett 1906 studied comb morphology in chickens – found a departure from expected ratio for a single trait1 trait: comb9/16 Walnut3/16 Rose3/16 Pea1/16 SingleRPRprPrpRRPPRRPpRrPPRrPpRRppRrpprrPPrrPprrpp
51 9:3:3:1 with four different phenotypes R-P- (walnut), R-pp (rose), rrP- (pea), rrpp (single)
52 A. Epistatic interactions Expression of one gene or gene pair masks or modifies the expression of another gene or gene pairEpistatic = gene product that masks another gene. Hypostatic = the second gene product being masked by another gene.Often arise because two or more different proteins participate in an enzymatic pathway leading to the formation of a single product.EnzymeCEnzymePColorless colorless Purple Pigmentprecursor intermediateEnzyme C needed to convert the precursor into the intermediate, Enzyme P converts the colorless intermediate into purple pigment
53 Sweet Peas – flower color P: White x WhiteF1: all purple, CcPpF2: 9 purple, 7 whiteCCpp x ccPPCPCpcPcpCCPPCCPpCcPPCcPpCCppCcppccPPccPpccppC-P- (purple), cc or pp masks C or P (producing white flowers)homozygosity for the white allele at one gene masks the purple producing allele of another gene [EPISTASIS]
54 2 genes: bw+ & st+, both necessary for red eye (wild type eye color in Drosophila)
55 F1: w+/w;m+/m (all blue) F2: 9:3:4 P: w/w;m+m+ x w+w+/mmF1: w+/w;m+/m (all blue)F2: 9:3:49/16 Blue: w+/-; m+/-3/16 Magenta: w+/-; m/m4/16 White: w/w; m+/- & w/w; m/mTwo genes encode enzymes catalyzing successive steps in the synthesis of blue pigment (pathway).If the first step in the pathway is blocked due to a homozygous mutant (w/w) or both steps are blocked, then the flower will be white.If the second step in the pathway is blocked due to a homozygous mutant (m/m) the flower will be magenta.
56 B/B, E/EB/B, e/eb/b, E/EThe progeny of a dihybridcross would be 9:3:4, black,brown, golden.B- (black)bb (Brown)E- (color deposition)
57 BE Be bE be BBEE BBEe BbEE BbEe BBee Bbee bbEe bbee BbEe x BbEe B bb E-eePrecursorMolecule Black brown color deposited(golden)golden
58 Plants of the mustard family were crossed Plants of the mustard family were crossed. When a true-breeding plant with triangular seeds is crossed with a plant with ovate seeds, the F1 generation has triangular seeds. When the F1 is self-fertilized, the result is a 15:1 ratio (triangular:ovate). Explain this ratio.This is a 2 gene interaction. The presence of one dominant allele in either gene results in a triangular seed.