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 I A I B ii IAIA IBIB i i

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 I A I B ii IAIA IBIB iIAiIAiIBiIBi iIAiIAiIBiIBi

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 I A I B ii IAIA IBIB iIAiIAiIBiIBi iIAiIAiIBiIBi 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 I A I B ii IAIA IBIB iIAiIAiIBiIBi iIAiIAiIBiIBi 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: Pp PPPPp p pp Ii IIIIi i 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: YellowI___ PurpleiiP_ Rediipp Pp PPPPp p pp Ii IIIIi i 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: YellowI___1 x ¾ = ¾ = 12/16 PurpleiiP_¾ x ¼ = 3/16 Rediipp¼ x ¼ = 1/16 Pp PPPPp p pp Ii IIIIi i 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: YellowI___1 x ¾ = ¾ = 12/16 PurpleiiP_¾ x ¼ = 3/16 Rediipp¼ x ¼ = 1/16 Pp PPPPp p pp Ii IIIIi i 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: YellowI___1 x ¾ = ¾ = 12/16 PurpleiiP_¾ x ¼ = 3/16 Rediipp¼ x ¼ = 1/16 Pp PPPPp p pp Ii IIIIi i 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 F 1 plants? B) How many phenotypic classes will there be in the F 2 ? c) What is the probability of seeing a 20cm plant in the F 2 generation?

A) If a tall plant ( AABBCC ) is crossed with a base-height plant ( aabbcc ), what is the height of the F 1 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 F 2 ?  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 F 2 generation?  Phenotype: 20cm  Genotype: 2 dominant alleles AaBbCc x AaBbCc AAbbcc ¼ x ¼ x ¼ = 1/64 AaBbcc 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 = 1/64 aaBbCc 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

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