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Hardy-Weinberg

Hardy-Weinberg. Allele Frequency. The first equation looks at the percentage of alleles in a population A + a = 1 A is the percentage of dominant alleles a is the percentage of recessive alleles. There are 30 million people in Canada, so how many alleles for eye colour should there be?

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Hardy-Weinberg

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  1. Hardy-Weinberg

  2. Allele Frequency • The first equation looks at the percentage of alleles in a population • A + a = 1 • A is the percentage of dominant alleles • a is the percentage of recessive alleles

  3. There are 30 million people in Canada, so how many alleles for eye colour should there be? • 60 million –two for each person • If 38.9 million of those alleles are brown, what does that mean for the A and a values? • A = 38.9/60 = 0.649 (64.9%) • a = 21.1/60 = 0.351 (35.1%) • So our equation A + a = 1 holds true (0.649 + 0.351 = 1) • % dominant alleles + % recessive alleles = 100% **These are the allele frequencies

  4. Say in New Brunswick (population 750 000), the dominant allele frequency is 71.6% • How many dominant alleles should there be? • 0.716 x 1 500 000 = 1 074 000 • Recessive alleles? • A + a = 1 • a = 1 – A = 1 – 0.716 = 0.284 (28.4%) • 0.284 x 1 500 000 = 426 000

  5. How does this relate to population dynamics? • Probability… • If the dominant allele frequency is 64.9%, then we have a 64.9% chance of selecting that allele • We would have a (0.649)(0.649) of selecting it twice -> (0.649)(0.649) = (0.649)2 = 0.421 • So, what is A2? • You may recognize it as AA -> homozygous dominant

  6. This means two things: • A random individual in Canada has a 42.1% chance of being homozygous dominant for brown eyes • 42.1% of individuals in Canada should be homozygous dominant • This holds true for homozygous recessive • a = 0.351 a2 = (0.351)2 = 0.123 • 12.3% of individuals should be homozygous recessive (blue eyes)

  7. What about the carriers (heterozygotes?) • Aa = (0.649)(0.351) = 0.228 • But, there are two ways to become heterozygous -> your father’s allele is dominant, your mother’s recessive OR you mother’s is dominant, your father’s recessive • So, we multiply by 2 • 2Aa = 2(0.649)(0.351) = 0.456 • 45.6% are heterozygous

  8. These combine into the equation: • A2 + 2Aa + a2 = 1 (look familiar?) • This means percentage of individuals that are homozygous dominant plus heterozygous plus homozygous recessive equals 100% -> everyone! • Please note that you’ll never use this equation as a whole ->you will always merely take out terms from it to do individual calculations • Does it work backwards?

  9. Yes! • If you know the percentage of people that are homozygous recessive (usually the easiest thing to identify), you can make assumptions about the allele frequency • If 15.4% of people are homozygous recessive for PTC paper tasting ability, what percentage are homozygous dominant, and what percentage are carriers?

  10. a2 = 0.154 a = (0.154)0.5 (this means square root of 0.154, but I couldn’t show that on PPT ’07. Thanks Microsoft…) a = 0.392 A + a = 1 A = 1 – a = 1 – 0.392 = 0.608 • With this info, the values are easy to calculate

  11. A2 = (0.609)2 = 0.370 37.0 % are homozygous dominant • 2Aa = 2(0.609)(0.392) = 0.477 47.7 % are carriers • All percentages add up to 100 (ok, 100.1, sig figs…)

  12. Example Batten disease is a rare recessive neurodegenerative disease, affecting 3 out of every 100 000 people in North America. Based on this knowledge, what percentage of people are carriers and could pass it onto their offspring?

  13. Answer • We define the dominant, normal allele as B, and the recessive as b • Since occurrence is 3 out of 100 000, b2=0.00003 • So, frequency of recessive allele is b=√0.00003 = 0.005 • The frequency of the dominant allele is B = 1-b = 1-0.005 = 0.995

  14. The frequency of carriers would be 2Bb=2(0.995)(0.005) = 0.0095 • So, approximately 1% of the population are carriers for this disease

  15. Example (try on your own) It is believed that approximately 4% of Canadians of South American decent are carriers for the recessive condition sickle cell anemia. If 98% of the alleles in this population are dominant, what should the prevalence of sickle cell anemia be?

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