Polygenic Inheritance

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Polygenic Inheritance

• Feb. 13-16, 2007

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Grab your fly vials

• Need to clear P1 parents from Drosophila cultures
• prevents P1 flies from mating with F1 flies that
  will be hatching in a few days
• Take vials outside, release stopper, and make
  sure all adult flies are out of the vial
• replace stopper and return vial to the front
  counter

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Polygenic Inheritance

• more than 1 gene pair influences trait
• polygenes
• contributing or additive alleles
• non-contributing or non-additive alleles
• Many traits show continuous variation
• i.e., human height, IQ
• Traits also influenced by environment

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Continuous vs. Discrete

• Polygenic traits may show continuous or discrete
  phenotypes
• Usually related to number of alleles involved
• May also be affected by environmental factors

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Example

• 4 gene pairs influence rabbit ear length
• each allele contributes equally
• AABBCCDD will be longest
• aabbccdd will be shortest

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Example

• Pure-breeding homozygote with 8cm ears (AABBCCDD)
  mated to pure-breeding homozygote with 4 cm ears
  (aabbccdd)
• F1 generation uniform, 6cm ears (AaBbCcDb)
• F2 has continuum of ears 4-8cm, with most between
  5-7cm

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Estimating number of gene pairs

• Estimated from fraction of F2 with phenotype
  expressed by P1 ancestor
• Estimated through binomial expansion

(p q)n
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Generating Expected Fractions
ps exponent x coefficient number in series

• Use Pascals triangle
• Expand the binomial
• (p q)n where n is number of alleles

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Pascals triangle
1
(p q)0
(p q)1
1
1
(p q)2
2
1
1
(p q)3
3
3
1
1
(p q)4
4
6
1
1
4
(p q)5
5
10
1
5
1
10
(p q)6
6
15
1
15
6
20
1
7
21
1
35
21
35
1
7
(p q)7
(p q)8
8
28
1
70
56
56
8
28
1
(p q)9
9
36
1
126
126
84
36
84
9
1
10
45
1
210
252
120
120
210
45
10
1
(p q)10
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Binomial expansion

• Lets expand the binomial (pq)6 and calculate
  the expected fraction for 6 alleles (3 pairs)

p6 6 p5q 15 p4q2 20 p3q3 15 p2q4 6 pq5
q6
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Binomial expansion

• Lets expand the binomial (pq)6 and calculate
  the expected fraction for 6 alleles (3 pairs)
• Add coefficients
• 161520156164
• Expected fraction is 1/64

p6 6 p5q 15 p4q2 20 p3q3 15 p2q4 6 pq5
q6
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Calculating Expected Fractions
1
(p q)0
(p q)1
1 contributing allele
2 contributing alleles
1
1
0 contributing alleles
AA
Aa
aa
(p q)2
2
1
1
2 polygenes (1 pair)
(p q)3
3
3
1
1
(p q)4
4
6
1
1
4
4 polygenes (2 pair)
0 contributing alleles
2 contributing alleles
4 contributing alleles
1 contributing allele
3 contributing alleles
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Estimating Number of Gene Pairs

• How many offspring of an F2 cross resemble the
  original parent?
• Obtained through ratio of coefficients obtained
  from expansion of the binomial
• 4 genes (2 pair) 1/16
• 6 genes (3 pair) 1/64
• 8 genes (4 pair) 1/256
• 10 genes (5 pairs) 1/1024
• 12 genes (6 pairs) 1/4096 (1/212)
• Hintnotice the expected fraction decreases by
  the reciprocal of 2 raised to the power of the
  number of genes

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Two gene pairs

• (¼)2 1/16
• 1/16 will be AABB
• 1/16 will be aabb

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Estimating Allele Contribution

• Find the number of gene pairs involved
• Find the amount that all gene pairs add
• Distribute that amount over the number of gene
  pairs

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Number of Gene Pairs

• average most extreme phenotypes
• (9 4)/2 6.5
• determine the fraction of these offspring present
• 6.5/1776 0.0037
• find n of (¼)n
• (¼)n 0.0037
• n 4 pairs