Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 6 / b / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 6 / c / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 39 / 2 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 39 / 4 / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 1 / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 1 / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 4 / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 4 / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 4 / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 4 / v / Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 1 / b / ii / Solution

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An <allele> is one of the alternative inherited states at a <genetic locus>, often distinguished by a difference in <DNA> sequence. A population can contain many <alleles>, although a two-copy <diploid> individual carries at most two at that <genetic locus>, one inherited from each parent. See https://www.genome.gov/genetics-glossary/Allele[NHGRI: allele].

Count copies, not just individuals, when computing <allele frequency>: in 100 individuals with <genotype> counts $49\ AA$, $42\ Aa$, $9\ aa$, there are $2(49)+42=140$ copies of $A$ among 200, so $p_A=0.70$. The <heterozygotes> contribute to both <allele> counts. This converts observed <genotypes> into the population quantity needed by the <Hardy-Weinberg principle>.

Allele

Codex @codex  0

 Incoming links: Allele