Mendelian genetics
It’s all about jargon, ratios, and nomenclature
Mendel’s peas
fig. 2-3
advantages of peas: self pollination and cross-pollination
- chose characters with only two states, or phenotypes
- created pure-bred lines
-cross-pollinated parental generation (P) to create first filial (F1)
generation
-also did reciprocal crosses
first filial (F1) generation were all the same
figs. 2-4 and 2-5
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 gametes
4. each gamete carries only one allele
5. gametes combine at random with respect to allele type
heterozygotes have different alleles
homozygotes have same alleles
a bit of notation…
dominant trait takes capital letter; recessive is lower case A/a
letter 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/a
letter is determined by phenotype of the dominant trait
yellow peas are dominant, therefore Y/y
Alternative notation
wild type trait is +; mutant is or
a+/aa’/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
Punnett square is used to work out progeny of a cross
P
Gametes
F1
Figure 2-6
Overall
F2 ratio
genotypic ratio
phenotypic ratio
yellow
Y
green
y
yellow
Y
Y/Y
Y/y
green
y
Y/y
y/y
ratio = 3 yellow : 1 green
yellow
Y
green
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 = ½
probability 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 = 3/8
2 females
= ½ x ½ x ½ = 1/8 x 3 = 3/8
3 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
round, yellow
R/r, Y/y
r/r, y/y
dihybrid cross (more ratios)
R/r, Y/y
R/Y
R/Y
R/r,
Y/y
R/y
r/Y
r/y
R/y
r/Y
r/y
dihybrid cross (more ratios)
R/r, Y/y
R/Y
R/Y
R/r,
Y/y
R/y
r/Y
r/y
R/R:Y/Y R/R:Y/y R/r:Y/Y R/r:Y/y
R/y R/R:Y/y R/R,y/y R/r:Y/y R/r:y/y
r/Y
R/r:Y/Y R/r:Y/y r/r:Y/Y r/r:Y/y
round, yellow
round, green
wrinkled, yellow
r/y
R/r: Y/y R/r,y/y
r/r:Y/y
r/r:y/y
wrinkled, green
dihybrid cross (more ratios)
R/r, Y/y
R/Y
R/Y
R/r,
Y/y
R/y
r/Y
r/y
R/R:Y/Y R/R:Y/y R/r:Y/Y R/r:Y/y
R/y R/R:Y/y R/R,y/y R/r:Y/y R/r:y/y
r/Y
R/r:Y/Y R/r:Y/y r/r:Y/Y r/r:Y/y
round, yellow
round, green
wrinkled, yellow
r/y
R/r: Y/y R/r,y/y
r/r:Y/y
r/r:y/y
wrinkled, green
phenotype ratio: 9 round, yellow
3 round, green
3 wrinkled, yellow
1 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
Gene interactions
incomplete dominance – heterozygote has intermediate phenotype
RR = red flower
Rr = 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
A/A and A/i
B/B and B/i
i/i
A/B
blood type
A
produces A antigen
B
produces B antigen
O
produce neither antigen
AB produce both antigens
Gene interactions
Note that terms are somewhat arbitrary: depend on level of analysis
RR = red flower
Rr = pink flower
rr = white flower
Incomplete dominance if R allele produces pigment, r does not
Co-dominance if R produces red pigment, r produces white pigment
A better example is roan cattle:
RR – red hair
rr – white hair
Rr – red AND white hairs
Gene interactions
epistasis – two or more genes interact
to form a phenotype
9
3
3
1
genotype
W/-, M/W/-, m/m
w/w, M/w/w, m/m
flower color
blue
9
magenta 3
white
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
A/A
A/S
S/S
malaria
phenotype
no anemia; normal blood cells
no anemia; sickle only with low O2
severe or fatal anemia with sickle cells
sickle cell
anemia
Gene interactions
sickle-cell anemia:
-single gene mutation (SNP)
-affects hemoglobin configuration, which distorts RBCs
-malaria parasite cannot digest altered hemoglobin
genotype
A/A
A/S
S/S
phenotype
no anemia; normal blood cells
no anemia; sickle only with low O2
severe or fatal anemia with sickle cells
dominant:reccessive for expression of anemia
incomplete dominance for cell shape
codominance for production of hemoglobin
ratios so far:
1:3 (phenotypic result of a mono-hybrid cross
in 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
expected
Monohybrid cross with incomplete dominance:
dev. from exp.
phenotype observed
expected
(obs-exp)
red
19
25
-6
pink
57
50
7
white
24
25
-1
total
100
100
degrees of freedom = 2 (= N – 1)
(obs-exp)2
exp
1.44
0.98
0.04
2.46 = Χ2
p = probability of obtaining the statistic by random chance
Χ2 = 2.46