Mendelian Genetics Concept 2: Analyzing the effects of complex genetic crosses such as incomplete/co- dominance, multiple alleles, pleiotropy, epistasis, polygenics, and lethal alleles.
Dec 18, 2015
Mendelian GeneticsConcept 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, 13Genetics Problems: 1, 2, 6, 12, 13
Learning Goal: How are traits
passed down from
parents to offspring?
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
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 IB
ii
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 IB
i IAi IBii IAi 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
IA IB
i IAi IBii IAi 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
IA IB
i IAi IBii IAi IBi
Predicted Phenotypic Ratio:
1 Type A: 1 Type B
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:
P pP PP Ppp Pp pp
I iI II Iii Ii ii
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
P pP PP Ppp Pp pp
I iI II Iii Ii ii
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___ 1 x ¾ = ¾ = 12/16Purple iiP_ ¾ x ¼ = 3/16Red iipp ¼ x ¼ = 1/16
P pP PP Ppp Pp pp
I iI II Iii Ii ii
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___ 1 x ¾ = ¾ = 12/16Purple iiP_ ¾ x ¼ = 3/16Red iipp ¼ x ¼ = 1/16
P pP PP Ppp Pp pp
I iI II Iii Ii ii
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___ 1 x ¾ = ¾ = 12/16Purple iiP_ ¾ x ¼ = 3/16Red iipp ¼ x ¼ = 1/16
P pP PP Ppp Pp pp
I iI II Iii Ii ii
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 cm5: 35 cm4: 30 cm3: 25 cm2: 20 cm1: 15 cm0: 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 ¼ x ¼ x ¼ = 1/64AaBbcc ½ x ½ x ¼ = 1/16 = 4/64 AabbCc ½ x ¼ x ½ = 1/16 = 4/64 aaBBcc ¼ x ¼ x ¼ = 1/64aaBbCc ¼ x ½ x ½ = 1/16 = 4/64 aabbCC 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