Mendelian genetics

It’s all about jargon, ratios, and nomenclature

fig. 2-3

advantages of peas: self pollination and cross-pollination

- chose characters with only two states, or phenotypes- created pure-bred lines

Mendel’s peas

figs. 2-4 and 2-5

-cross-pollinated parental generation (P) to create first filial (F1) generation-also did reciprocal crosses

first filial (F1) generation were all the same

self-crossed F1 to obtain second filial (F2) generation – ‘missing’ type reappeared – all progeny in 3:1 ratio

deduced presence of dominant and recessive traits

Mendel’s postulates

1. there are hereditary ‘particles’ that determine traits (genes)2. genes are in pairs (alleles)3. members of a gene pair segregate equally into the gametes4. each gamete carries only one allele5. gametes combine at random with respect to allele type

heterozygotes have different alleleshomozygotes have same alleles

a bit of notation…

dominant trait takes capital letter; recessive is lower case A/aletter is determined by phenotype of the dominant trait

yellow peas are dominant, therefore Y/y

a bit of notation…

dominant trait takes capital letter; recessive is lower case A/aletter is determined by phenotype of the dominant trait

yellow peas are dominant, therefore Y/y

Alternative notationwild type trait is +; mutant is - a+/a-

or a’/a

dominant genotype (when we don’t know what the second allele is) can be abbreviated a+/–this includes a+/a+ and a+/a- genotypes

What causes dominance?

Genes code for proteins

Proteins are either structural or functional (enzymes)

An ‘error’ in the genetic code may yield a non-functional enzyme, or no protein, or a less efficient enzyme

Figure 2-6

Punnett square is used to work out progeny of a cross

PGametes

F1

Overall F2 ratio

phenotypic ratio

yellow green Y y

yellow Y/Y Y/y Y

green Y/y y/y y

ratio = 3 yellow : 1 green

genotypic ratio

yellow green Y y

yellow Y/Y Y/y Y

green Y/y y/y y

ratio = 1 YY: 2 Y/y : 1 y/y

note: Y/y = y/Y

Probability rules:

Multiplication rule: the probability of two independent outcomes occurring simultaneously is equal to the product of each of the two outcomes taken separately.

independent probabilities can be multiplied

what is the probability of a litter of three being all male?

Probability rules:

Multiplication rule: the probability of two independent outcomes occurring simultaneously is equal to the product of each of the two outcomes taken separately.

independent probabilities can be multiplied

what is the probability of a litter of three being all male?

probability than any one offspring will be male = ½probability that all three will be male – ½ x ½ x ½ = 1/8

Probability rules:

Addition rule: the overall probability of any combination of mutually exclusive outcomes is equal to the sum of the probabilities of the outcomes taken separately.

mutually exclusive probabilities can be added

What is the probability that a litter of three will have at least one female?

0 females = ½ x ½ x ½ = 1/8 1 female = ½ x ½ x ½ = 1/8 x 3 = 3/82 females = ½ x ½ x ½ = 1/8 x 3 = 3/83 females = ½ x ½ x ½ = 1/8

sum of probabilities containing a female = 7/8

dihybrid cross (more ratios)

F1 – round, yellow X wrinkled, green

= R/R, Y/Y X r/r, y/y

round, yellow R/r, Y/y

dihybrid cross (more ratios)

R/Y R/y r/Y r/y

R/Y

R/y

r/Y

r/y

R/r, Y/y

R/r,Y/y

dihybrid cross (more ratios)

R/Y R/y r/Y r/y

R/Y

R/y

r/Y

r/y

R/r, Y/y

R/r,Y/y

R/R:Y/Y R/R:Y/y R/r:Y/Y R/r:Y/y

R/R:Y/y R/R,y/y R/r:Y/y R/r:y/y

R/r:Y/Y R/r:Y/y r/r:Y/Y r/r:Y/y

R/r: Y/y R/r,y/y r/r:Y/y r/r:y/y

round, yellow

round, green

wrinkled, yellow

wrinkled, green

dihybrid cross (more ratios)

R/Y R/y r/Y r/y

R/Y

R/y

r/Y

r/y

R/r, Y/y

R/r,Y/y

R/R:Y/Y R/R:Y/y R/r:Y/Y R/r:Y/y

R/R:Y/y R/R,y/y R/r:Y/y R/r:y/y

R/r:Y/Y R/r:Y/y r/r:Y/Y r/r:Y/y

R/r: Y/y R/r,y/y r/r:Y/y r/r:y/y

round, yellow

round, green

wrinkled, yellow

wrinkled, green

phenotype ratio: 9 round, yellow 3 round, green 3 wrinkled, yellow1 wrinkled, green

Fig. 2-10

Gene interactions

• one gene may affect several phenotypes (pleiotropy)

- PKU (phenylketonuria): single gene for enzyme phenylalanine hydroxylase) that converts phenylalanine to tyrosine;

- loss of function results in mental retardation, lower pigmentation, additional traits

• several genes may produce the same phenotype

- importance of duplicate genes

incomplete dominance – heterozygote has intermediate phenotype

RR = red flowerRr = pink flower

rr = white flower

usually found when phenotype is a continuous trait (quantitative)e.g., weight, height, fecundity, amount of enzyme produced

alternative is discrete traits e.g., round vs. wrinkled, yellow vs. green peas

Gene interactions

co-dominance – intermediate phenotype is formed when two dominant alleles are present in heterozygote

- characterized by having three phenotypes

genotype blood typeA/A and A/i A produces A antigen

B/B and B/i B produces B antigeni/i O produce neither antigenA/B AB produce both antigens

Gene interactions

Note that terms are somewhat arbitrary: depend on level of analysis

RR = red flowerRr = pink flower

rr = white flower

Incomplete dominance if R allele produces pigment, r does notCo-dominance if R produces red pigment, r produces white pigment

A better example is roan cattle:RR – red hairrr – white hairRr – red AND white hairs

Gene interactions

epistasis – two or more genes interact to form a phenotype

genotype flower color9 W/-, M/- blue 93 W/-, m/m magenta 33 w/w, M/- white1 w/w, m/m white 4

Fig. 4-13

Gene interactions

sickle-cell anemia: -single gene mutation (SNP) -affects hemoglobin configuration, which distorts RBCs -malaria parasite cannot digest altered hemoglobin

genotype phenotypeA/A no anemia; normal blood cellsA/S no anemia; sickle only with low O2

S/S severe or fatal anemia with sickle cells

sickle cellanemia

malaria

Gene interactions

Gene interactions

sickle-cell anemia: -single gene mutation (SNP) -affects hemoglobin configuration, which distorts RBCs -malaria parasite cannot digest altered hemoglobin

genotype phenotypeA/A no anemia; normal blood cellsA/S no anemia; sickle only with low O2

S/S severe or fatal anemia with sickle cells

dominant:reccessive for expression of anemiaincomplete dominance for cell shapecodominance for production of hemoglobin

ratios so far:

1:3 (phenotypic result of a mono-hybrid crossin dominance : recessive traits)

1:2:1 (genotypic result of a mono-hybrid cross) and phenotypic result for incomplete dominant traits

9:3:3:1 (dihybrid cross in dominance : recessive traits)

1:2 if lethal allele is present A/A ¼ A/a 2/4 a/a ¼ - but they are all dead

15:1 (dihybrid cross with duplicate genes)

penetrance = whether genotype is expressed in phenotype due to modifiers, epistatic genes, suppression

expressivity = degree to which genotype is expressed in the phenotype due to other alleles, or environment

Fig. 4-23

was Mendel honest?

all dominant: recessive traits

only two alleles at each gene

no gene interactions

no sex-linked traits

all independent traits (no linkage; pea has 7 chromosomes)

How do we evaluate all this???? (how close must the data be to the ratios?)

Chi square = Χ2 = (observed – expected)2

expectedMonohybrid cross with incomplete dominance:

dev. from exp. (obs-exp)2

phenotype observed expected (obs-exp) expred 19 25 -6 1.44pink 57 50 7 0.98white 24 25 -1 0.04total 100 100 2.46 = Χ2

degrees of freedom = 2 (= N – 1)

p = probability of obtaining the statistic by random chance

Χ2 = 2.46