1. Eye color is determined by multiple alleles at a single gene locus. There are multiple versions (alleles) of the eye color gene - blue, green, and brown. 2. Polygenic inheritance is when multiple genes contribute to a single trait. For traits determined by polygenic inheritance, there are many possible combinations of alleles across several gene loci that result in a range of phenotypes. 3. Both multiple allele inheritance and polygenic inheritance can contribute to natural variation in human eye color. Multiple alleles at a single gene locus explain some of the variation, but many genes likely have small effects, making polygenic inheritance important as well.

1. MULTIPLE ALLELES AND POLYGENIC INHERITANCE Biology 3 Kent Kawashima

4. Character:eye color Trait: blue Genotype: AAAATATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTATCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCTGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTCGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCTGATCA… Phenotype: Megan Fox’s blue eyes

5. Human have 23 pairs of chromosomes. one set comes from the mother one set comes from the father

6. A gene occurs in pairs in diploid cells. eye color gene pair The combination of what alleles you have in your gene pair is what is called your genotype. Homologous chromosomes AA, BB, Cc, RrQq, IAIA… These letters are just representations for your genotype. They also simply represent your alleles.

7. The eye color gene located in chromosome 19… gene for eye color blue version, green version, brown version blue green brown …can contain any of the versions for that gene. These versions are called your alleles

8. “blue” allele “green” allele “brown” allele …AAAATATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTATCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCTGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTCGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCT… …AAATTATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTCTCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCAGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTTGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCT… …AAACTATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTGTCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCCGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTAGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCT… versions of a gene occur because of small changes in the DNA (in pink)

9. Since your eye color gene occurs as a pair, you may have two similar alleles green green Genotype: Homozygous for the green allele

10. …or you may have two different alleles blue green Genotype: Heterozygous

11. …or you may have two different alleles green brown Genotype: Heterozygous

12. …or you may have two different alleles brown blue Genotype: Heterozygous

14. More than two versions of the gene exists…This is called “Multiple Allele Inheritance”

15. gene for eye color gene for eye color gene for eye color gene for eye color What if many genes have an effect to one character? This is called “Polygenic Inheritance”

16. In Polygenic Inheritance… gene for eye color + add more of something! gene for eye color 0 no effect. gene for eye color gene for eye color alleles that either contribute to a trait or not many genes

17. Example. For each + allele you have, it tells your cells to add more pigment to your iris. The 0 alleles however do not add nor subtract pigment, they do not contribute to your phenotype. gene for eye color + + gene for eye color + 0 gene for eye color + 0 gene for eye color + +

18. + allele is called a contributory allele + 0 0 allele is called a non-contributory allele

19. Compare. + 0 + 0 + 0 + 0 + 0 + 0 + + 0 0 0 + + + 0 0 0 + + 0 + 0 0 + 0 + + 0 + 0 0 + 0 + 6 +, 2 0 2 +, 6 0 4 +, 4 0 8 + 8 0

20. Compare. + 0 + 0 + 0 + 0 + 0 + 0 + + 0 0 0 + + + 0 0 0 + + 0 + 0 0 + 0 + + 0 + 0 0 + 0 + gray dark brown green blue light brown

21. Arranged from least to most melanin produced + 0 + 0 + 0 + 0 + 0 + 0 + + 0 0 0 + + + 0 0 0 + + 0 + 0 0 + 0 + + 0 + 0 0 + 0 + gray dark brown green blue light brown

23. There is no dominance, an allele either contributes to a phenotype or not

27. Shortcuts for Polygenic Inheritance Genetics

28. If you cross… AA x aa AABB x aabb Aa x Aa AaBb x AaBb 1 AABB 4 AaBB, aABB, AABb, AAbB 6 AaBb, aAbB, AabB, aABb, aaBB, AAbb 4aAbb, Aabb, aabB, aaBb 1 aabb all possible genotypes 1 AA 2 Aa, aA 1 AA relative phenotype portions expected

29. If you cross… AA x aa P Aa x Aa F1 Number of gene pairs: n = 1 Number of phenotypes in F2: 21 + 1 = 3 Number for genotypes regardless of position in the gene pair: 31 = 3 (two big A, one big A, no big A) 1 AA 2 Aa, aA 1 aa F2

30. If you cross… AABB x aabb Number of gene pairs: n = 2 Number of phenotypes in F2: 22 + 1 = 5 Number for genotypes regardless of position in the gene pair: 32 = 9 2 big A, 2 big B 2 big A, 1 big B 2 big A, 0 big B 1 big A, 2 big B 1 big A, 1 big B 1big A, 0 big B 0 big A, 2 big B 0 big A, 1 big B 0 big A, 0 big B P AaBb x AaBb F1 1 AABB 4 AaBB, aABB, AABb, AAbB 6 AaBb, aAbB, AabB, aABb, aaBB, AAbb 4aAbb, Aabb, aabB, aaBb 1 aabb regardless of position in the gene pair F2

31. Use Pascal’s Triangle to get the coeefficientsof the binomial expansion Pascal’s triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 (A + a)2n binomial equation

32. for 1gene pair, n=1 (A + a)2 = 1A2 + 2Aa + 1a2 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 Use Pascal’s triangle for coefficients

33. for 2gene pairs, n=2 (A + a)4 = 1A4 + 4A3a + 6A2a2 + 4Aa3 + 1a4 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 Use Pascal’s triangle for coefficients

34. for 3gene pair, n=3 (A + a)6 = 1A6 + 6A5a + 15A4a2 + 30A3a3 + 15A2a4 + 6Aa5 + 1a6 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 Use Pascal’s triangle for coefficients

35. for n number of gene pairs, (A + a)2n = (2nC0)A2na0+ (2nC1)A2n-1a1+ (2nC2)A2n-2a2+ … + (2nC2n)A0a2n 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 2n C x, x=0 -> 2n OR Use Pascal’s triangle for coefficients combination

36. for 2gene pairs, n=2 number of contributory alleles means 1/16 has 4 contributory alleles number of non-contributory alleles (A + a)4 = 1A4 + 4A3a + 6A2a2 + 4Aa3 + 1a4 Coefficients show the relative portion of the generation showing that polygenic genotype means 4/16 has just 3 contributory alleles