I. Mendel’s postulates
Postulate 1. Unit factors in pairs
•Genetic characters are controlled by unit factors in pairs.
•In other words, genes are present in two associated copies in
diploid organisms.
•For example, DD plants have two alleles for tallness, dd plants
have two alleles for dwarfism.
Postulate 2. Dominance/recessiveness
•In the case of unlike unit factors, one can be dominant and
the other can be recessive.
•In other words, when two different alleles of a gene are
present, one may show its effect while the other may be
masked.
•For example, Dd plants have a tall allele D and a dwarf allele
d, but are phenotypically tall.
Postulate 3. Segregation
•During the formation of gametes, unit factors
segregate randomly.
•In other words, when sperm and eggs are formed,
one of each allelic pair is randomly distributed to to
each gamete.
•For example, a Dd plant makes pollen or eggs, each
randomly receives either the D allele or the d allele.
IV. Independent assortment
Mendel’s Fourth Postulate:
An Interpretation from the Dihybrid cross
• During gamete formation, segregating pairs of unit
factors assort independently.
• In other words, segregation of 2 alleles at one genetic
locus has no effect on the segregation of 2 alleles at
another locus (unless linked).
• For example, the assortment of yellow and green
alleles has no effect on the assortment of round and
wrinkled alleles, and vice versa.
Independent Assortment
• Results in extensive genetic variation
• Number of possible gametes = 2n where n is the
haploid number
• For humans, 223 = 8 million
• Each individual represents one of (8 X 106)2 = 64 X
1012 possible genetic combinations from her parents
Practice: Yellow/Green Peas
•In garden peas, an allele Y for yellow peas are
dominant to an allele y for green peas.
•If we crossed 2 heterozygous pea plants
What would be the expected genotypes of the
F1 generation (progeny)?
Need to set up a punnett square to predict
possible outcomes
Further prediction powers
– In the F2 generation of a monohybrid cross, what is the
expected ratio of Yellow : Green?
– Depends on which of the F1 progeny will be crossed
– Lets review the outcome of the F1
• 50% Heterozygotes
• 25% Homozygotes Recessive
• 25% Homozygotes Dominant
II. Monohybrid Test Cross (Backcross)
• How can you determine genotype from individual
expressing dominant phenotype? - DD or Dd?
• Cross individual with dominant phenotype to a
homozygous recessive individual.
Dihybrid Cross: P1 cross
Since yellow and round are dominant,
Let G = yellow, g = green, W = round, w = wrinkled.
What would the F2 progeny look like?
Dihybrid Cross: F1 cross
III. Dihybrid cross - phenotypes
1.Which traits are dominant?
2. Did the phenotypes of the P1 generation affect the F1 or F2 generations?
Dihybrid Cross: Summary
9
3
3
1
An Easier Way?
• Called the “Fork Line Method”
• This is just another way to be able to predict
genotype and phenotype ratios in dihybrid
problems
• this way you don’t have to write the box
• but it does require you to know the basic ratios
that arise from monohybrids
• based on the idea that: in a dihybrid, the two traits
sort INDEPENDENTLY of one another
• i.e. what happens with one trait is completely unrelated to what
happens with the other trait
PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
Pp x Pp will give:
¼ PP
½ Pp
¼ pp
PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
Pp x Pp will give:
similarly,
Yy x Yy will give:
¼ YY
½ Yy
¼ yy
¼ PP
½ Pp
¼ pp
PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
Yy x Yy will give:
Pp x Pp will give:
¼ YY
½ Yy
¼ yy
¼ YY
¼ PP
½ Pp
¼ pp
multiply fractions
1/16 PPYY
½ Yy
1/8 PPYY
¼ yy
1/16 PPYY
PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
Pp x Pp will give:
¼ YY
¼ PP
½ Pp
¼ pp
multiply fractions
1/16 PPYY
½ Yy
1/8 PPYy
¼ yy
1/16 PPyy
¼ YY
1/8 PpYY
½ Yy
1/4 PpYy
¼ yy
1/8 Ppyy
¼ YY
1/16 ppYY
½ Yy
1/8
¼ yy
1/16 ppyy
ppYy
PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
convert all to
16ths for
consistency
Pp x Pp will give:
1/16 PPYY
1
½ Yy
1/8 PPYy
2
¼ yy
1/16 PPyy
1
¼ YY
1/8 PpYY
2
½ Yy
1/4 PpYy
4
¼ yy
1/8 Ppyy
2
¼ YY
1/16 ppYY
1
½ Yy
1/8
ppYy
2
¼ yy
1/16 ppyy
1
¼ YY
¼ PP
½ Pp
¼ pp
multiply fractions
PpYy x PpYy
so to solve this dihybrid, separate the two traits (since they sort independently):
convert all to
16ths for
consistency
Pp x Pp will give:
1/16 PPYY
1
½ Yy
1/8 PPYy
2
¼ yy
1/16 PPyy
1
¼ YY
1/8 PpYY
2
½ Yy
1/4 PpYy
4
¼ yy
1/8 Ppyy
2
¼ YY
1/16 ppYY
1
½ Yy
1/8
ppYy
2
¼ yy
1/16 ppyy
1
¼ YY
¼ PP
½ Pp
¼ pp
multiply fractions
so – this gives you the same results as the Punnet square – but in some cases might be a faster cleaner way of doing it.
Analysis of dihybrid cross phenotypes
(forked-line/probability method)
Trait 1
Trait 2
Combined traits
Fork Line Practice
• Fruit Fly Genotypes
– Ebony (e)
– Gray (E)
– Long wings (V)
– Vestigial wings (v)
Find the phenotype probabilities of a diheterozygote
female crossed with a diheterozygote male for:
1) Grey and vestigial
2) Ebony and Long
V. Gamete Formation: Trihybrid cross
What size of Punnett square needed for analysis?
Use the Forkline!
• Possible Gametes
– AaBbCc X AaBbCc
Trihybrid Cross - Phenotypes
Forked-line Method
27:9:9:9:3:3:3:1
Product law
• For simultaneous outcomes (AND)
• What is the chance that you will roll snake eyes with
two dice? (1 and 1)
– Chance of rolling 1 with first die = 1/6
– Chance of rolling 1 with second die = 1/6
– Chance of rolling two 1’s = 1/6 X 1/6 = 1/36
• We used product law when calculating probabilities
by the forked-line method.
Sum law
• For outcomes that can occur more than one way
(OR)
• What is the chance that you will roll either a 1 or a 6
with one die?
– Chance of rolling 1 = 1/6
– Chance of rolling 6 = 1/6
– Chance of rolling 1 or 6 = 1/6 + 1/6 = 2/6 = 1/3
VI. Molecular Basis of Mendel’s
Postulates: Chromosome behavior
• 1879: Walter Flemming discovers chromosomes in
living cells.
• 1900: DeVries, Correns, and Tschermak repeat,
rediscover Mendel.
• 1902: Sutton and Boveri and others link behavior of
chromosomes to Mendelian segregation and
independent assortment; propose chromosomal
theory of heredity.
Theodor Boveri
• Hypothesis: Removal of
chromosomes should
result in some change
to organism.
• Did this in sea urchin
embryos, saw abnormal
embryos.
1862-1915
Walter Sutton (1877-1916)
• Studied grasshopper testis and ovary cells, observed
that haploid numbers of chromosomes mixed to form
diploid set.
• Noted that this chromosome behavior correlated with
Mendel’s genes, and proposed that genes are carried
on the chromosomes.
Correlation Between Unit Factors
and Genes on Chromosomes
• Unit factors in pairs ~ genes on homologous
chromosomes in pairs
• Segregation of unit factors during gamete formation ~
genes on homologs segregate during meiosis
• Independent assortment of segregating unit factors ~
genes on nonhomologous chromosomes assort
independently
• Stronger evidence for the chromosomal theory of heredity
came from experiments of T.H. Morgan and others with
fruit flies from 1909 onwards.
VII. Laws of probability help explain
genetic events
Genetic ratios are most properly expressed as probabilities:
ex. 3/4 tall: 1/4 dwarf
The probability of each zygote having the genetic potential
for becoming tall is 3/4, etc..
Probabilities range from 0 (an event is certain NOT to
happen), to 1.0 (an event is certain to happen).
How do we calculate the probability of 2 or more
events happening at the same time?