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 A 2 ?  You may recognize it as AA -> homozygous dominant

6.  This means two things: 1. A random individual in Canada has a 42.1% chance of being homozygous dominant for brown eyes 2. 42.1% of individuals in Canada should be homozygous dominant  This holds true for homozygous recessive a = 0.351 a 2 = (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: A 2 + 2Aa + a 2 = 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.  a 2 = 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.  A 2 = (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, b 2 =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?