Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 39
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Outcomes are"mutually exclusive" if they are incompatible in the sense that they cannot occur at the same time. For example,there are four mutually exclusive outcomes of the sex distribution of sibships with three children—namely, theinclusion of 0, 1, 2, or 3 girls. These have probability 1/8, 3/8, 3/8, and 1/8, respectively. The addition rule statesthat the overall probability of any combination of mutually exclusive events is equal to the sum of the probabilitiesof the events taken separately. For example, the probability that a sibship of size 3 contains at least one girlincludes the outcomes 1, 2, and 3 girls, so the overall probability of at least one girl equals 3/8 + 3/8 + 1/8 = 7/8.The multiplication rule of probability deals with outcomes of a genetic cross that are independent.
Any twooutcomes are independent if the knowledge that one outcome is actually realized provides no information aboutwhether the other is realized also. For example, in a sequence of births, the sex of any one child is not affected bythe sex distribution of any children born earlier and has no influence whatsoever on the sex distribution of anysiblings born later.
Each successive birth is independent of all the others. When possible outcomes are independent,the multi-Page 107plication rule states that the probability of any combination of outcomes being realized equals the product of theprobabilities of all of the individual outcomes taken separately. For example, the probability that a sibship of threechildren will consist of three girls equals 1/2 × 1/2 × 1/2, because the probability of each birth resulting in a girl is1/2, and the successive births are independent.Probability calculations in genetics frequently use the addition and multiplication rules together.
For example, theprobability that all three children in a family will be of the same sex uses both the addition and the multiplicationrules. The probability that all three will be girls is (1/2) (1/2) (1/2) = 1/8, and the probability that all three will beboys is also 1/8. Because these outcomes are mutually exclusive (a sibship of size three cannot include three boysand three girls), the probability of either three girls or three boys is the sum of the two probabilities, or 1/8 + 1/8 =1/4. The other possible outcomes for sibships of size three are that two of the children will be girls and the other aboy, and that two will be boys and the other a girl.
For each of these outcomes, three different orders of birth arepossible—for example, GGB, GBG, and BGG—each having a probability of 1/2 ×1/2 × 1/2 = 1/8. The probabilityof two girls and a boy, disregarding birth order, is the sum of the probabilities for the three possible orders, or 3/8;likewise, the probability of two boys and a girl is also 3/8.
Therefore, the distribution of probabilities for the sexratio in families with three children isGGGGGBGBBGBGBGBBGGBBGBBBThe sex ratio information in this display can be obtained more directly by expanding the binomial expression (p +q)n, in which p is the probability of the birth of a girl (1/2), q is the probability of the birth of a boy (1/2), and n isthe number of children. In the present example,in which the red numerals are the possible number of birth orders for each sex distribution. Similarly, the binomialdistribution of probabilities for the sex ratios in families of five children isEach term tells us the probability of a particular combination.
For example, the third term is the probability of threegirls (p3) and two boys (q2) in a family that has five children—namely,There are n + 1 terms in a binomial expansion. The exponents of p decrease by one from n in the first term to 0 inthe last term, and the exponents of q increase by one from 0 in the first term to n in the last term.
The coefficientsgenerated by successive values of n can be arranged in a regular triangle known as Pascal's triangle (Figure 3.19).Note that the horizontal rows of the triangle are symmetrical, and that each number is the sum of the two numberson either side of it in the row above.In general, if the probability of event A is p and that of event B is q, and the two events are independent andmutually exclusive (see Chapter 2), the probability that A will be realized four times and B two times—in aspecific order—is p4q2, by the multiplication rule.
However, suppose that we were interested in the combination ofevents "four of A and two of B," regardless of order. In that case, we multiply theFigure 3.19Pascal's triangle. The numbers are thecoefficients of each term in the expansion of thepolynomial (p + q)n for successive values of nfrom 0 through 6.Page 108probability that the combination 4A: 2B will be realized in any one specific order by the number of possible orders.The number of different combinations of six events, four of one kind and two of another, isThe symbol ! stands for factorial, or the product of all positive integers from 1 through a given number.
Except forn = 0, the formula for factorial isThe case n = 0 is an exception because 0! is defined as equal to 1. The first few factorials areThe factorial formulais the coefficient of the term p4q2 in the expansion of the binomial (p + q)6. Therefore, the probability that event Awill be realized four times and event B two times is 15p4q2.The general rule for repeated trials of event with constant probabilities is as follows:If the probability of event A is p and the probability of the alternative event B is q, the probability that, in ntrials, event A is realized s times and event B is realized t times isin which s + t = n and p + q = 1.
Equation (1) applies even when either s or t equals 0 because O! is defined toequal 1. (Remember also that any number raised to the zero power equals 1; for example, (1/2)0 = 1.) Anyindividual term in the expansion of the binomial (p + q)n is given by Equation (1) for the appropriate values of sand t.It is worth taking a few minutes to consider the meaning of the factorial part of the binomial expansion in Equation(1), which equals n! (s!t!). This ratio enumerates all possible ways in which s elements of one kind and t elementsof another kind can be arranged in order, provided that the s elements and the t elements are not distinguishedamong themselves. A specific example might include s yellow peas and t green peas. Although the yellow peas andthe green peas can be distinguished from each other because they have different colors, the yellow peas are notdistinguishable from one another (because they are all yellow) and the green peas are not distinguishable from oneanother (because they are all green).The reasoning behind the factorial formula begins with the observation that the total number of elements is s + t =n.
Given n elements, each distinct from the next, the number of different ways in which they can be arranged isWhy? Because the first element can be chosen in n ways, and once this is chosen, the next can be chosen in n— 1ways (because only n— 1 are left to choose from), and once the first two are chosen, the third can be chosen inn — 2 ways, and so forth. Finally, once n — 1 elements have been chosen, there is only 1 way to choose the lastelement. The s + t elements can be arranged in n! ways, provided that the elements are all distinguished amongthemselves.
However, applying again the argument we just used, each of the n! particular arrangements mustinclude s! different arrangements of the s elements and t! different arrangements of the t elements, or s! × t!altogether. Dividing n! by s! × t! therefore yields the exact number of ways in which the s elements and the telements can be arranged when the elements of each type are not distinguished among themselves.Let us consider a specific application of Equation (1). in which we calculate the probability that a mating betweentwo heterozygous parents yields exactly thePage 109expected 3 : 1 ratio of the dominant and recessive traits among sibships of a particular size.
The probability p of achild showing the dominant trait is 3/4, and the probability q of a child showing the recessive trait is 1/4. Supposewe wanted to know how often families with eight children would contain exactly six children with the dominantphenotype and two with the recessive phenotype. This is the "expected" Mendelian ratio. In this case, n = 8, s = 6, t= 2, and the probability of this combination of events isThat is, in only 31 percent of the families with eight children would the offspring exhibit the expected 3 : 1phenotypic ratio; the other sibships would deviate in one direction or the other because of chance variation.
Theimportance of this example is in demonstrating that, although a 3 : 1 ratio is the "expected" outcome (and is alsothe single most probable outcome), the majority of the families (69 percent) actually have a distribution ofoffspring different from 3 : 1.Evaluating the Fit of Observed Results to Theoretical ExpectationsGeneticists often need to decide whether an observed ratio is in satisfactory agreement with a theoreticalprediction. Mere inspection of the data is unsatisfactory because different investigators may disagree.
Suppose, forexample, that we crossed a plant having purple flowers with a plant having white flowers and, among the progeny,observed 14 plants with purple flowers and 6 with white flowers. Is this result close enough to be accepted as a 1 :1 ratio? What if we observed 15 plants with purple flowers and 5 with white flowers? Is this result consistent with a1 : 1 ratio? There is bound to be statistical variation in the observed results from one experiment to the next. Who isto say what results are consistent with a particular genetic hypothesis? In this section, we describe a test of whetherobserved results deviate too far from a theoretical expectation. The test iscalled a test for goodness of fit, where the word fit means how closely the observed results "fit," or agree with, theexpected results.The Chi-Square MethodA conventional measure of goodness of fit is a value called chi-square (symbol, X2), which is calculated from thenumber of progeny observed in each of various classes, compared with the number expected in each of the classeson the basis of some genetic hypothesis.
For example, in a cross between plants with purple flowers and those withwhite flowers, we may be interested in testing the hypothesis that the parent with purple flowers is heterozygousfor one pair of alleles determining flower color and that the parent with white flowers is homozygous recessive.Suppose further that we examine 20 progeny plants from the mating and find that 14 are purple and 6 are white.The procedure for testing this genetic hypothesis (or any other genetic hypothesis) by means of the chisquaremethod is as follows:1. State the genetic hypothesis in detail, specifying the genotypes and phenotypes of the parents and the possibleprogeny.
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