Tutor profile: Darcy-laine M.
Subject: Basic Math
3+5=? 4+8=? 5-2=? 12-1=? 9x3=? 4x6=? 10/2=? 3/1=?
3+5=8 4+8=12 5-2=3 12-1=11 9x3=27 4x6=24 10/2=5 3/1=3
If we know that the percent of homozygous recessive (rr) individuals in a population that is in Hardy Weinberg equilibrium is 64%, then given this, find the frequency of homozygous dominant individuals (RR) in this same population, the frequency of homozygous recessive individuals (rr) in this population, the frequency of the R (dominant) allele, the frequency of the r (recessive) allele, and the frequency of heterozygotes (Rr) within this population.
Hardy Weinberg equation: p^2 + 2pq + q^2=1; p+q=1 p=frequency of dominant allele q=frequency of recessive allele p^2= frequency of homozygous dominant individuals q^2= frequency of homozygous recessive individuals pq= frequency of heterozygote individuals first we have identified that this population is indeed in Hardy Weinberg equilibrium so we can use the Hardy Weinberg equations (which can be found above) since we know that the percent of homozygous recessive individuals is 64% then we know that in order to find the frequency of homozygous recessive individuals we must convert this percentage to a decimal (so we move decimal two places to the left) so the frequency of homozygous recessive individuals is 0.64 (=q^2) if we take the square root of this, we can find the frequency of the recessive allele within this population so the square root of 0.64 is 0.8 so we know that q=0.8 then, now that we have the q value, we can plug this into the Hardy Weinberg equation p+q=1 in order to find our p value (the frequency of the dominant allele) so p+0.8=1 and subtract 0.8 from both sides in order to figure out that p=0.2 (frequency of the dominant allele) Since we have found our p value, we can now find p^2 which will give us our frequency of homozygous dominant individuals within our population. 0.2^2=0.04 so the frequency of homozygous dominant individuals is 0.04. In order to then find our frequency of heterozygote individuals within our population we then must use the 2pq part of the Hardy Weinberg equation so 2(0.2)(0.8)=0.32 so our frequency of heterozygote individuals in this population is 0.32.
Jake is half the age of Jill who is one third the age of Sally who is double the age of John. John is 27, how old is Jake?
By going backwards starting with John who we know is 27, we are told that Sally is double the age of John so therefore she must be 27 x 2=54 years old. Then, we are told that Jill is one third the age of Sally so she is 1/3 x 54=18 years old. Lastly, Jake is half the age of Jill so 1/2 x 18=9 years old, therefore Jake is 9 years old.
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