Use of binomial expansion and probability in solving genetics problems PDF

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how to solve genetic problems from binomials...

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BIO 325 Use of binomial expansion and probability in solving genetics problems When multiple independent events are analyzed, there can be several different sequences in which the events are ordered with each sequence representing a set. For example, in a family with three children where both parents are heterozygous for albinism (an autosomal recessive condition), what is the probability that all three children will exhibit albinism? Considering A as the allele for normal pigmentation, and a as the allele for albinism, in an Aa x Aa cross the probability of having a child with albinism (aa) is ¼ and the probability of having a child with normal pigmentation (A-) is ¾. In this family, the probability that all three children will have albinism (aa) is ¼ x ¼ x ¼ = 1/64. As seen, there is only one sequence and the product rule is sufficient to get the combined probability. On the other hand, in this family if you are required to determine the probability where one child has albinism and the remaining two show normal pigmentation, there are three possible sequences: Sequence 1: First child with albinism (1/4) x second child normal (3/4) x third child normal (3/4) = 9/64 Sequence 2: First child normal (3/4) x second child with albinism (1/4) x third child normal (3/4) = 9/64 Sequence 3: First child normal (3/4) x second child normal (3/4) x third child with albinism (1/4) = 9/64 Since there are three possible sequences, and any one would satisfy the requirement of having one child with albinism and two normal children, it can be sequence 1 or sequence 2 or sequence 3, and in this case we apply the sum rule to add the probabilities of all the sequences (sequence 1 probability + sequence 2 probability + sequence 3 probability) which is 9/64 + 9/64 + 9/64 = 27/64 to give the combined probability. Using the same trait, where both parents are heterozygous, what is the probability that in a family with five children, two children exhibit albinism and three are normal? Once again there can be a number of sequences in which the children with albinism and the normal children can occur, and although it may be possible to write the different sequences as shown in the previous example, determine the probability of each sequence and add them up, it will take much more time. However, application of the binomial expansion makes it easier and faster to solve

these types of problems (and not just genetics problems!). The binomial takes the form (p + q)n, where p represents the probability of occurrence of one event, q represents the probability of occurrence of the alternative event, and n represents the number of times the event occurs. For determining the probability of having two children with albinism and three normal children in a family of five children, where both parents are heterozygous, binomial expansion is applied as follows: p = probability of a child having albinism (1/4) q = probability of having a child with normal pigmentation (3/4) n = total number of children (5) The binomial in this case is (p + q)5, and the expansion is: (p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5 Each of the terms in the expansion provides the probability for one particular combination of phenotypes (albinism and normal pigmentation) in the children. The first term of the expansion (p5) represents the probability of having five children, all exhibiting albinism, where p is the probability of a child having albinism. The second term of the expansion (5p4q) represents the probability of having four children with albinism and one child with normal pigmentation. The third term of the expansion (10p3q2) represents the probability of having three children with albinism and two children with normal pigmentation. The fourth term of the expansion (10p2q3) represents the probability of having two children with albinism and three children with normal pigmentation, and this is what you are interested in. To calculate the probability, insert the values for p (which is 1/4) and q (which is 3/4) in this term and you have your answer. 10p2q3 = 10 (1/4)2 x (3/4)3 = 270/1024 = 0.26 Using the same expansion, you can calculate the probability of any combination of events as long as you are able to identify the correct term. In this example, the next term (5pq4) will give you the probability of having one child with albinism and four children with normal pigmentation, and the last term (q5) will give you the probability of having all five children with normal pigmentation.

How to expand a binomial? In general, the expansion of any binomial (p + q)n consists of a series of n + 1 terms. In the example used earlier, n = 5, so the number of terms are 5 + 1 = 6 and they are p5, 5p4q, 10p3q2, 10p2q3, 5pq4, and q5. To write the terms, first figure out their exponents. The exponent of p in the first term always begins with the power to which the binomial is raised, or n. In our example, n = 5, so our first term is p5. The exponent of p decreases by one in each successive term; so the exponent of p is 4 in the second term (p4), 3 in the third term (p3), and so forth. The exponent of q is 0 (no q) in the first term and it increases by 1 in each successive term, increasing from 0 to 5 in our example. Next, determine the coefficient of each term. The coefficient of the first term is always 1; so, in our example, the first term is 1p5, or just p5. The coefficient of the second term is always the same as the power to which the binomial is raised; in our example, this coefficient is 5 and the term is 5p4q. For the coefficient of the third term, look at the preceding term; multiply the coefficient of the preceding term (5 in our example) by the exponent of p in that term (4) and then divide by the number of that term (second term, or 2). So the coefficient of the third term in our example is (5 x 4)/2 = 10 and the term is 10p3q2. Follow this logic for each successive term. A different method to determine the probability of any combination of events An even simpler way to determine the probability of any particular combination of events is to use the formula: n! P= ps qt s! t! where P represents the overall probability of event X with probability p occurring s times and event Y with probability q occurring t times. In the example used in the binomial expansion, event X may be treated as the occurrence of a child with albinism (p = ¼) and event Y as the occurrence of a child with normal pigmentation (q = ¾); s would be the number of children with albinism (2) and t would be the number of children with normal pigmentation (3). The ! symbol stands for factorial, and it means the product of all the integers from n to 1. In this example, n = 5; so n! = 5 x 4 x 3 x 2 x 1. Applying this formula to obtain the probability of having two children with albinism and three children with normal pigmentation in a family of five children, gives the following:

5! P=

(1/4)2 (3/4)3

2! 3!

5x4x3x2x1 (1/4)2 (3/4)3 = 0.26

= 2x1x3x2x1

This value is the same as that obtained with the binomial expansion! (From: Genetics: A Conceptual Approach, by Benjamin A. Pierce. Fifth Edition. Freeman)...