Mendelian genetics It’s all about jargon, ratios, and nomenclature.
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Transcript of Mendelian genetics It’s all about jargon, ratios, and nomenclature.
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