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MENDELIAN GENETICS. GARDEN PEAS Advantages seeds easy to obtain characters easy to score crosses easily controlled short generation time large numbers.

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Presentation on theme: "MENDELIAN GENETICS. GARDEN PEAS Advantages seeds easy to obtain characters easy to score crosses easily controlled short generation time large numbers."— Presentation transcript:

1 MENDELIAN GENETICS

2 GARDEN PEAS Advantages seeds easy to obtain characters easy to score crosses easily controlled short generation time large numbers of progeny use of statistics in analysis

3 Unit factors (genes) exist in pairs (alleles) Dominance/Recessiveness Alleles segregate into gametes in equal frequencies Alleles from different gene pairs assort independently into gametes MENDEL'S POSTULATES

4 pure (true) breeding line - a group of genetically identical individuals that always produce offspring of the same phenotype when mated to each other

5 F1 generation - first filial generation; the progeny resulting from the first cross in a series F2 generation - second filial generation; the progeny resulting from a cross of the F1 generation monohybrid cross - a genetic cross between two individuals involving only one character (e.g. AA x aa) in which the parents possess different alleles of the character

6 MONOHYBRID CROSS: PARENTAL AND F1 R_ = Round R is dominant rr = wrinkled r is recessive F1 Gametes either R or r produced in equal frequencies LAW OF EQUAL SEGREGATION Parents Gametes F1 seeds Gametes (pure breeding)

7 CONCEPT OF DOMINANT AND RECESSIVE ALLELES Round x wrinkled F1 all Round (selfed or crossed to each other) F2 3/4 Round; 1/4 wrinkled Mendel concluded that one phenotype of the pea shape character (wrinkled) is hidden in the F1 and reappears in the F2.

8 Seed Parent Rr Pollen Parent Rr X F1F1 1/2 R 1/2 r 1/4 RR 1/4 Rr 1/4 Rr 1/4 rr 1/2 Rr 3/4 Round 1/4 wrinkled Genotypic Ratio = 1: 2: 1 Phenotypic Ratio = 3: 1

9 Mendel explained his data by invoking the concept of genes: - each pea plant has a pair of alleles for one character; one allele is dominant and the other is recessive Mendels First Law - Law of Equal Segregation - different alleles of one gene segregate into different gametes in equal frequencies (i.e. each gamete carries only one allele of a gene) - the union of gametes to make a zygote is random (it doesnt matter which allele is in each gamete) Proof: testcross - crossing a homozygous recessive individual to an individual of unknown genotype to determine the unknown genotype (e.g. R? x rr)

10 ParentsRR x rr F1Rr (all round seeds) F23:1 R_ : rr (phenotypic ratio) Testcross (cross F1 individuals to homozygous recessive) Rr x rr GametesR or r r 1:1 Rr : rr (phenotypic ratio) If F1 were RR, all progeny from the test cross would be Rr (phenotypically round)

11 dihybrid cross - a genetic cross involving two phenotypic characters in which the parents possess different alleles of each character (e.g. round, green x wrinkled, yellow peas)

12 DIHYBRID CROSS R_ Round rr wrinkled Y_ Yellow yy green Parents Gametes F1 Progeny F1 Gametes Mendels Second Law - Independent Assortment - different gene pairs assort independently into gametes (pure breeding)

13 PHENOTYPES OF F2 PROGENY IN DIHYBRID CROSS Phenotypic RatioPhenotypesAlleles Present 9round, yellow R_Y_ 3round, green R_yy 3wrinkled, yellow rrY_ 1wrinkled, green rryy

14 Punnett Square

15 The 9:3:3:1 phenotypic ratio observed in the F2 is created by the random superimposition of two independent 3:1 phenotypic ratios. 3/4 R_3/4 Y_ 1/4 rr1/4 yy 9/16 R_Y_ 3/16 R_yy 3/16 rrY_ 1/16 rryy F1 cross:RrYy x RrYy

16 MENDEL IGNORED FROM Mendel was an unknown researcher Darwins Origin of Species (1859) - continuous not discontinuous variation Cytology of time could not explain hypothesis - chromosomal theory of heredity developed by Theodore Boveri and Walter Sutton in early 1900's clearly showed link between chromosome segregation during meiosis and Mendel's unit factors (genes) Unable to replicate results in all organisms MENDEL REDISCOVERED IN 1900 by Carl Correns, Hugo DeVries, Eric Von Tschermak

17 Using branch (forked) diagrams to determine expected phenotypic ratios from a cross Parental cross RrYy x Rryy (R and Y are dominant alleles) 1/2 Y_3/8 R_Y_ R_3/4 1/2 yy3/8 R_yy 1/2 Y_1/8 rrY_ rr1/4 1/2 yy1/8 rryy calculate frequencies for individual genes separately * because events are independent multiply probabilities

18 In the cross Aa bb CC Dd Ee x Aa Bb Cc dd Ee, what proportion of the progeny will have the genotype AA bb CC dd EE? calculate frequencies for individual genes Aa x Aa1/4 of the progeny will be AA bb x Bb1/2 of the progeny will be bb CC x Cc1/2 of the progeny will be CC Dd x dd1/2 of the progeny will be dd Ee x Ee1/4 of the progeny will be EE * because events are independent multiply probabilities Probability of AA bb CC dd EE individuals = 1/4 x 1/2 x 1/2 x 1/2 x 1/4 = 1/128

19 In the cross AaBbCc x AaBbCC the expected number of gametes = 2 x 2 x 2 = 8 and 2 x 2 x 1 = 4 genotypes = 3 x 3 x 2 = 18 phenotypes = 2 x 2 x 1 = 4 Assumes complete dominance and recessiveness for all gene pairs and independent assortment. Don't blindly use 2 n and 3 n rules described in text; they cant be used in situations such as the one above


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