Lec10 Genetics Recap

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Lec10 Genetics Recap

  1. 1. MULTIPLE ALLELES AND POLYGENIC INHERITANCE<br />Biology 3<br />Kent Kawashima<br />
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  4. 4. Character:eye color<br />Trait: blue<br />Genotype: AAAATATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTATCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCTGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTCGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCTGATCA…<br />Phenotype: Megan Fox’s blue eyes<br />
  5. 5. Human have 23 pairs of chromosomes.<br />one set comes from the mother<br />one set comes from the father<br />
  6. 6. A gene occurs in pairs in diploid cells.<br />eye color gene pair<br />The combination of what alleles you have in your gene pair is what is called your genotype.<br />Homologous chromosomes<br />AA, BB, Cc, RrQq, IAIA… <br />These letters are just representations for your genotype. They also simply represent your alleles.<br />
  7. 7. The eye color gene located in chromosome 19…<br />gene for eye color<br />blue version, green version,<br /> brown version<br />blue<br />green<br />brown<br />…can contain any of the versions for that gene.<br />These versions are called your alleles<br />
  8. 8. “blue” allele<br />“green” allele<br />“brown” allele<br />…AAAATATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTATCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCTGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTCGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCT…<br />…AAATTATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTCTCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCAGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTTGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCT…<br />…AAACTATTATATATCCCCGTAGCTACGATCTATCTATCATCTATCTACTATCTATCTATCTATCTGTCTACTATCTATCCGGCGTATCATCGGGGGGGGGGGTACTAGCTAGCTAGCTATCTACGTAGTCTACTGATCTAGCTAGTCATGCTGATCATCGTAGCTGAGCTACTAGTCGATCGTAGCTAGTCATCTAGCTGAGCTAGCCGATCGATCTAGCTAGCTGATCGTAGCGATCGTAGCTAGCTAGTAGATGCTAGCTGATCGTAGCATGCTAGCTGATGCATGCTAGCTAGCTAGCT…<br />versions of a gene occur because of small changes in the DNA (in pink)<br />
  9. 9. Since your eye color gene occurs as a pair, you may have two similar alleles<br />green<br />green<br />Genotype: Homozygous for the green allele<br />
  10. 10. …or you may have two different alleles<br />blue<br />green<br />Genotype: Heterozygous<br />
  11. 11. …or you may have two different alleles<br />green<br />brown<br />Genotype: Heterozygous<br />
  12. 12. …or you may have two different alleles<br />brown<br />blue<br />Genotype: Heterozygous<br />
  13. 13. gene for eye color<br />blue<br />green<br />brown<br /><ul><li>Only one gene controls the nature of the character…
  14. 14. More than two versions of the gene exists…</li></ul>This is called “Multiple Allele Inheritance” <br />
  15. 15. gene for eye color<br />gene for eye color<br />gene for eye color<br />gene for eye color<br />What if many genes have an effect to one character?<br />This is called “Polygenic Inheritance” <br />
  16. 16. In Polygenic Inheritance…<br />gene for eye color<br />+<br />add more of something!<br />gene for eye color<br />0<br />no effect.<br />gene for eye color<br />gene for eye color<br />alleles that either contribute to a trait or not<br />many genes<br />
  17. 17. Example.<br />For each + allele you have, it tells your cells to add more pigment to your iris.<br />The 0 alleles however do not add nor subtract pigment, they do not contribute to your phenotype.<br />gene for eye color<br />+<br />+<br />gene for eye color<br />+<br />0<br />gene for eye color<br />+<br />0<br />gene for eye color<br />+<br />+<br />
  18. 18. + allele is called a contributory allele<br />+<br />0<br />0 allele is called a non-contributory allele<br />
  19. 19. Compare.<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />+<br />0<br />0<br />0<br />+<br />+<br />+<br />0<br />0<br />0<br />+<br />+<br />0<br />+<br />0<br />0<br />+<br />0<br />+<br />+<br />0<br />+<br />0<br />0<br />+<br />0<br />+<br />6 +, 2 0<br />2 +, 6 0<br />4 +, 4 0<br />8 +<br />8 0<br />
  20. 20. Compare.<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />+<br />0<br />0<br />0<br />+<br />+<br />+<br />0<br />0<br />0<br />+<br />+<br />0<br />+<br />0<br />0<br />+<br />0<br />+<br />+<br />0<br />+<br />0<br />0<br />+<br />0<br />+<br />gray<br />dark brown<br />green<br />blue<br />light brown<br />
  21. 21. Arranged from least to most melanin produced<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />0<br />+<br />+<br />0<br />0<br />0<br />+<br />+<br />+<br />0<br />0<br />0<br />+<br />+<br />0<br />+<br />0<br />0<br />+<br />0<br />+<br />+<br />0<br />+<br />0<br />0<br />+<br />0<br />+<br />gray<br />dark brown<br />green<br />blue<br />light brown<br />
  22. 22. In polygenic inheritance…<br /><ul><li>Many genes contribute to affect a singular character
  23. 23. There is no dominance, an allele either contributes to a phenotype or not
  24. 24. There is an equal effect for each contributory allele</li></li></ul><li>In polygenic inheritance…<br /><ul><li>Polygenes have two types of alleles, a contributory allele that adds to the phenotype, and a non-contributory allele that has no effect on the phenotype</li></ul>A(BIG LETTER) - contibutory allele<br />a(small letter) - non-contributory allele<br />
  25. 25. In polygenic inheritance…<br /><ul><li>Phenotypes show a great number of variations
  26. 26. Polygenic inheritance is usually the mode of inheritance for traits that are continuous – weight, height and skin tone</li></li></ul><li>So…<br />Is iris color inherited through multiple alleles or polygenes?<br />
  27. 27. Shortcuts for Polygenic Inheritance Genetics<br />
  28. 28. If you cross…<br />AA x aa<br />AABB x aabb<br />Aa x Aa<br />AaBb x AaBb<br />1 AABB<br />4 AaBB, aABB, AABb, AAbB<br />6 AaBb, aAbB, AabB, aABb, aaBB, AAbb<br />4aAbb, Aabb, aabB, aaBb<br />1 aabb<br />all possible genotypes<br />1 AA<br />2 Aa, aA<br />1 AA<br />relative phenotype portions expected <br />
  29. 29. If you cross…<br />AA x aa<br />P<br />Aa x Aa<br />F1<br />Number of gene pairs: n = 1<br />Number of phenotypes in F2: 21 + 1 = 3<br />Number for genotypes regardless of position in the gene pair: 31 = 3<br />(two big A, one big A, no big A)<br />1 AA<br />2 Aa, aA<br />1 aa<br />F2<br />
  30. 30. If you cross…<br />AABB x aabb<br />Number of gene pairs: n = 2<br />Number of phenotypes in F2: 22 + 1 = 5<br />Number for genotypes regardless of position in the gene pair: 32 = 9<br />2 big A, 2 big B<br />2 big A, 1 big B<br />2 big A, 0 big B<br />1 big A, 2 big B<br />1 big A, 1 big B<br />1big A, 0 big B<br />0 big A, 2 big B<br />0 big A, 1 big B<br />0 big A, 0 big B<br />P<br />AaBb x AaBb<br />F1<br />1 AABB<br />4 AaBB, aABB, AABb, AAbB<br />6 AaBb, aAbB, AabB, aABb, aaBB, AAbb<br />4aAbb, Aabb, aabB, aaBb<br />1 aabb<br />regardless of position in the gene pair<br />F2<br />
  31. 31. Use Pascal’s Triangle to get the coeefficientsof the binomial expansion <br />Pascal’s triangle<br />1<br />1 1<br />1 2 1<br />1 3 3 1<br />1 4 6 4 1<br />1 5 10 10 5 1<br />1 6 15 20 15 6 1<br />1 7 21 35 35 21 7 1<br />1 8 28 56 70 56 28 8 1<br />(A + a)2n<br />binomial equation<br />
  32. 32. for 1gene pair, n=1 <br />(A + a)2 = 1A2 + 2Aa + 1a2<br />1<br />1 1<br />1 2 1<br />1 3 3 1<br />1 4 6 4 1<br />1 5 10 10 5 1<br />1 6 15 20 15 6 1<br />1 7 21 35 35 21 7 1<br />1 8 28 56 70 56 28 8 1<br />Use Pascal’s triangle for coefficients<br />
  33. 33. for 2gene pairs, n=2 <br />(A + a)4 = 1A4 + 4A3a + 6A2a2 + 4Aa3 + 1a4<br />1<br />1 1<br />1 2 1<br />1 3 3 1<br />1 4 6 4 1<br />1 5 10 10 5 1<br />1 6 15 20 15 6 1<br />1 7 21 35 35 21 7 1<br />1 8 28 56 70 56 28 8 1<br />Use Pascal’s triangle for coefficients<br />
  34. 34. for 3gene pair, n=3 <br />(A + a)6 = 1A6 + 6A5a + 15A4a2 + 30A3a3 + 15A2a4 + 6Aa5 + 1a6<br />1<br />1 1<br />1 2 1<br />1 3 3 1<br />1 4 6 4 1<br />1 5 10 10 5 1<br />1 6 15 20 15 6 1<br />1 7 21 35 35 21 7 1<br />1 8 28 56 70 56 28 8 1<br />Use Pascal’s triangle for coefficients<br />
  35. 35. for n number of gene pairs, <br />(A + a)2n = (2nC0)A2na0+ (2nC1)A2n-1a1+ (2nC2)A2n-2a2+ … + (2nC2n)A0a2n<br />1<br />1 1<br />1 2 1<br />1 3 3 1<br />1 4 6 4 1<br />1 5 10 10 5 1<br />1 6 15 20 15 6 1<br />1 7 21 35 35 21 7 1<br />1 8 28 56 70 56 28 8 1<br />2n C x, x=0 -> 2n<br />OR<br />Use Pascal’s triangle for coefficients<br />combination<br />
  36. 36. for 2gene pairs, n=2 <br />number of contributory alleles<br />means 1/16 has 4 contributory alleles<br />number of non-contributory alleles<br />(A + a)4 = 1A4 + 4A3a + 6A2a2 + 4Aa3 + 1a4<br />Coefficients show the relative portion of the generation showing that polygenic genotype<br />means 4/16 has just 3 contributory alleles<br />

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