Mendelian Genetics
Concept 2: Analyzing the effects of complex genetic crosses such
as incomplete/co- dominance, multiple alleles, pleiotropy,
epistasis, polygenics, and lethal alleles.
We just need to practice…
 Try… On Worksheets:
Multiple Choice: 1, 2, 3, 5, 7, 8, 10, 13
Genetics Problems: 1, 2, 6, 12, 13
Try This!
A brown eyed male has 5 children
with a blue eyed female. 4 of the
children have brown eyes and 1 child
has blue eyes.
If brown eyes are dominant to blue
eyes, what is the genotype of each
parent?
Degrees of Dominance
 Complete Dominance
 Homozygous Dominant and Heterozygous are
indistinguishable
 Monohybrid Phenotypic Ratio – 3:1
 Incomplete Dominance
 Heterozygotes display a blended phenotype
 Monohybrid Phenotypic Ratio – 1:2:1
 Codominance
 Heterozygotes display both traits (separate, distinguishable)
 Monohybrid Phenotypic Ratio - 1:2:1
Degrees of Dominance
 Not cut and dry…
 Organism
 Biochemical
 Molecular
Multiple Alleles
 Many alleles for one gene
 Blood Type
 Example: If a male with blood type AB has children with a
female with blood type O, what is the predicted
phenotypic ratio of their children?
Multiple Alleles
 Example: If a male with blood type AB has children with a
female with blood type O, what is the predicted
phenotypic ratio of their children?
Type AB male x Type O Female
IAIB
ii
IA
i
i
IB
Multiple Alleles
 Example: If a male with blood type AB has children with a
female with blood type O, what is the predicted
phenotypic ratio of their children?
Type AB male x Type O Female
IAIB
ii
i
i
IA
IAi
IAi
IB
IBi
IBi
Multiple Alleles
 Example: If a male with blood type AB has children with a
female with blood type O, what is the predicted
phenotypic ratio of their children?
Type AB male x Type O Female
IAIB
ii
i
i
IA
IAi
IAi
IB
IBi
IBi
Predicted Phenotypic Ratio:
Multiple Alleles
 Example: If a male with blood type AB has children with a
female with blood type O, what is the predicted
phenotypic ratio of their children?
Type AB male x Type O Female
IAIB
ii
i
i
IA
IAi
IAi
IB
IBi
IBi
Predicted Phenotypic Ratio:
1 Type A: 1 Type B
Pleiotropy
 One gene that has multiple effects on phenotypic
characters
 Sickle cell
 Cystic fibrosis
Epistasis
 One gene having an effect over another gene
 Mouse coat colour
 Example: In corn, a dominant allele I inhibits kernel
colour, while the recessive allele i permits colour when
homozygous. At a different locus, the dominant allele P
causes purple kernel colour, while the homozygous
genotype pp causes red kernels. If plants heterozygous
at both loci are crossed, what will be the phenotypic
ratio of the offspring?
Epistasis
 Example: In corn, a dominant allele I inhibits kernel colour,
while the recessive allele i permits colour when homozygous.
At a different locus, the dominant allele P causes purple
kernel colour, while the homozygous genotype pp causes
red kernels. If plants heterozygous at both loci are crossed,
what will be the phenotypic ratio of the offspring? (no
colour=yellow)
IiPp x IiPp
Phenotypes:
I
i
I
II
Ii
i
Ii
ii
P
p
P PP Pp
p Pp pp
Epistasis
 Example: In corn, a dominant allele I inhibits kernel colour,
while the recessive allele i permits colour when homozygous.
At a different locus, the dominant allele P causes purple
kernel colour, while the homozygous genotype pp causes
red kernels. If plants heterozygous at both loci are crossed,
what will be the phenotypic ratio of the offspring? (no
colour=yellow)
IiPp x IiPp
Phenotypes:
Yellow I___
Purple iiP_
Red
iipp
I
i
I
II
Ii
i
Ii
ii
P
p
P PP Pp
p Pp pp
Epistasis
 Example: In corn, a dominant allele I inhibits kernel colour,
while the recessive allele i permits colour when homozygous.
At a different locus, the dominant allele P causes purple
kernel colour, while the homozygous genotype pp causes
red kernels. If plants heterozygous at both loci are crossed,
what will be the phenotypic ratio of the offspring? (no
colour=yellow)
IiPp x IiPp
Phenotypes:
Yellow I___
Purple iiP_
Red
iipp
I
i
I
II
Ii
1 x ¾ = ¾ = 12/16
¾ x ¼ = 3/16
¼ x ¼ = 1/16
i
Ii
ii
P
p
P PP Pp
p Pp pp
Epistasis
 Example: In corn, a dominant allele I inhibits kernel colour,
while the recessive allele i permits colour when homozygous.
At a different locus, the dominant allele P causes purple
kernel colour, while the homozygous genotype pp causes
red kernels. If plants heterozygous at both loci are crossed,
what will be the phenotypic ratio of the offspring? (no
colour=yellow)
IiPp x IiPp
Phenotypes:
Yellow I___
Purple iiP_
Red
iipp
I
i
I
II
Ii
1 x ¾ = ¾ = 12/16
¾ x ¼ = 3/16
¼ x ¼ = 1/16
i
Ii
ii
P
p
P PP Pp
p Pp pp
Predicted Phenotypic Ratio:
Epistasis
 Example: In corn, a dominant allele I inhibits kernel colour,
while the recessive allele i permits colour when homozygous.
At a different locus, the dominant allele P causes purple
kernel colour, while the homozygous genotype pp causes
red kernels. If plants heterozygous at both loci are crossed,
what will be the phenotypic ratio of the offspring? (no
colour=yellow)
IiPp x IiPp
Phenotypes:
Yellow I___
Purple iiP_
Red
iipp
I
i
I
II
Ii
1 x ¾ = ¾ = 12/16
¾ x ¼ = 3/16
¼ x ¼ = 1/16
i
Ii
ii
P
p
P PP Pp
p Pp pp
Predicted Phenotypic Ratio:
12 Yellow: 3 Purple: 1 Red
Polygenic Inheritance
 Two or more genes affecting one phenotypic character
 Quantitative characters – continuum rather than “either/or”
 Skin colour in humans
 Spike weed height
 Number of phenotypic classes possible = #alleles+1
Try This!
 The height of spike weed is a result of polygenetic
inheritance involving three genes, each of which can
contribute an additional 5cm to the base height of the
plant, which is 10cm. The tallest plant (AABBCC) can
reach a height of 40cm.
A) If a tall plant (AABBCC) is crossed with a base-height
plant (aabbcc), what is the height of the F1 plants?
B) How many phenotypic classes will there be in the F2?
c) What is the probability of seeing a 20cm plant in the F2
generation?
A) If a tall plant (AABBCC) is crossed
with a base-height plant (aabbcc),
what is the height of the F1 plants?
 The parental cross produced 25cm tall F1 plants, all
AaBbCc plants with 3 units of 5 cm added to the base
height of 10cm.
B) How many phenotypic classes
will there be in the F2?
 General Rule: number of phenotypic classes resulting
from a cross of heterozygotes equals the number of
alleles involved plus one.
 (AaBbCc ) 6 alleles + 1 = 7
 6 dominant alleles: 40 cm
5: 35 cm
4: 30 cm
3: 25 cm
2: 20 cm
1: 15 cm
0: 10 cm
c) What is the probability of seeing a
20cm plant in the F2 generation?
 Phenotype: 20cm
 Genotype: 2 dominant alleles
AaBbCc x AaBbCc
AAbbcc
AaBbcc
AabbCc
aaBBcc
aaBbCc
aabbCC
¼ x ¼ x ¼ = 1/64
½ x ½ x ¼ = 1/16 = 4/64
½ x ¼ x ½ = 1/16 = 4/64
¼ x ¼ x ¼ = 1/64
¼ x ½ x ½ = 1/16 = 4/64
¼ x ¼ x ¼ = 1/64
Different ways of getting the same thing… add!
15/64
Lethal Alleles
 Recessive lethal allele
 2:1 ratio if lethal before birth
 Cystic fibrosis
 Dominant lethal allele
 Must show effects after reproductive age
 Why?
 Huntington’s Disease
Environmental Effects
 Hydrangeas
 Plant a tin can next to them and see what happens!
Don’t forget about pedigrees!
We just need to practice…
 On Worksheets for Concept 2:
Multiple Choice: 4, 6, 9, 11, 12, 14, 15, 16
Genetics Problems: 3, 4, 5, 7, 8, 9, 10, 11, 14