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Mendel: father of genetics Quick review of terminology

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1 Inheritance of Single-Gene Differences – discovered by Gregor Mendel!!!
Mendel: father of genetics Quick review of terminology Mendel’s Empirical approach Monohybrid cross Mendel’s Postulates to explain his data Mendel’s First “law” equal segregation Punnent Square Dihybrid cross Mendel’s Second “law” independent assortment The branch diagram & probabilities Using the testcross

2 Who was this “Father of Genetics”?
I. Gregor Johann Mendel Who 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 gametes Future 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 color Each 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 phenotype Recessive 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 organisms Self Cross = cross to oneself (plants, fungi) Haploid Cross = simplest, each gene present in 1 copy only (fungi) Diploid Cross = each gene present in 2 copies Homozygote cross (AA x AA), aka pure-breeding Heterozygote cross (Aa x Aa) Testcross = cross with a known homozygote recessive Backcross = hybrid offspring are crossed with one of the parents Reciprocal 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 design chose a good “model” organism: Pisum sativum restricted his examination to one or very few pairs of contrasting traits in each experiment took meticulous notes with accurate quantitative records

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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 characteristics P – parental generation F1 – First filial generation F2 – Second filial generation

11 Mendel’s results from the monohybrid crosses
Parental F1 F2 F2 ratio Round x wrinkled All round 5474 round wrinkled 2.96:1 Yellow x green seeds All yellow 6022 yellow 2001 green 3.01:1 Purple x white All purple 705 purple 224 white 3.15:1 Inflated x pinched All inflated 882 inflated pinched 2.95:1 Green x yellow pods All green 428 green 152 yellow 2.82:1 Axial x terminal All axial 651 axial 207 terminal 3.14:1 Long x short All long 787 long 277 short 2.84:1

12 A. Mendel’s Postulates to explain his data
the existence of unit “factors” – particulate theory of inheritance Traits inherited as discrete units that remain unchanged as they pass from parent to offspring genes are in pairs, thus 2 phenotypes must be determined by 2 different alleles of 1 gene When 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 1:1 segregation in a heterozygote 5) random fertilization – gametes are brought together for fertilization in a random manner

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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.

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19 C. Using Punnett Squares in Genetic Crosses
Punnett squares used for monohybrid crosses Considers only genes of interest List sperm genotypes across top List egg genotypes down side Fill in boxes with zygote genotypes

20 Pollen of Heterozygous Plant
Making a Punnett Square: Heterozygous X Heterozygous P p Eggs of Heterozygous Plant P P P P p Pollen of Heterozygous Plant p P p p p Frequencies Phenotypes Genotypes PP p P P p p 2 1 1 Purple White 3 (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

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24 F1 cross: GgWw x GgWw

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 formation The alleles assort independently Possible gametes produced from meiosis

26 (Hair color) & (Hair length)
Black/Brown Short/Long P: Black, short x Brown, long F1: all black, short F2: Black, short x Black, short: BbSs x BbSs

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28 Review Mendel’s Laws

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 performed Accuracy of prediction depends on sample size

30 Probability Rules Product rule: the probability of independent events occurring together is the product of the probabilities of the individual events Pr(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 dependent Prc = 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 = 9/16 What 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/r F2? 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 1 The 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 trait Genetic ratios – expressed as probabilities Based on the product rule of probability Pr(A) x Pr(B) = Pr(A and B)

36 Binomial expansion Used to predict the probability of an unordered combination of events each event possesses one of two mutually exclusive characteristics, eg. curly hair or straight hair the outcome for any one event is independent of the outcome for any other event Example: 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 events from 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-x Step 1. calculate individual probabilities (p & q) p = ¼, q = ¼ Step 2. determine # of events in category x and the total # of events x = 2 n = 5 Step 3. substitute values for p, q, x in the equation 5! 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 girl b. that their first child will be a taster girl c. that two out of three children will be nontasters 1/8 (½ x ¼) (½ x ¾) 3/8 14%

39 Non-Mendelian ratios - Interactions between the alleles of one gene
Inheritance of Gene Differences – non-Mendelian geneic interactions, part 2 Non-Mendelian ratios - Interactions between the alleles of one gene Interactions between the alleles of more than one gene Gene interaction Epistasis

40 A. Incomplete Dominance
Two alleles (heterozygote) produce an intermediate phenotype At the molecular level, the mutant allele results in a reduced amount of functional protein 2 doses = 100% 1 dose = 50% 0 dose = None

41 incomplete dominance F1 hybrids have an appearance somewhere in between the phenotypes of the two parental varieties F1 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 alleles Some genes are found in three or more alleles that are different from each other e.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 c ch = himalayan coat, color in only certain parts of body; dominant to c c = albino, no color; recessive to all other alleles A 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
AB i/i O Specific type of multiple alleles, when two alleles are equally expressed in the heterozygous individual e.g. ABO blood group

46 C. Lethal alleles Allele in an essential gene that has the potential of causing the death of an organism. Age of onset Conditional lethal alleles Semilethal alleles

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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 genes Examples: 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 trait 1 trait: comb 9/16 Walnut 3/16 Rose 3/16 Pea 1/16 Single RP Rp rP rp RRPP RRPp RrPP RrPp RRpp Rrpp rrPP rrPp rrpp

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 pair Epistatic = 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. Enzyme C Enzyme P Colorless colorless Purple Pigment precursor intermediate Enzyme 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 White F1: all purple, CcPp F2: 9 purple, 7 white CCpp x ccPP CP Cp cP cp CCPP CCPp CcPP CcPp CCpp Ccpp ccPP ccPp ccpp C-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+/mm F1: w+/w;m+/m (all blue) F2: 9:3:4 9/16 Blue: w+/-; m+/- 3/16 Magenta: w+/-; m/m 4/16 White: w/w; m+/- & w/w; m/m Two 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/E B/B, e/e b/b, E/E The progeny of a dihybrid cross 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- ee Precursor Molecule 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.


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