# Probability and the Idea of Chance

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1 Probability and the Idea of Chance Instructions Activity 1. The Idea of Chance Consider a simple demonstration of the operation of chance (i.e., probability) in the tossing of coins. It is usually impossible to measure or control the many factors involved when a coin is tossed and allowed to return to the resting position. The end result is said to occur by chance. The coin returns to a resting position with one or the other of its faces upturned, that is, heads or tails. When a single coin is tossed many times, it is expected that about half of the tosses will result in heads facing up and the other half in tails facing up. Procedure 1. Toss a single coin 24 times. Record the results in Table 1 on the Lab Report. 2. Calculate the expected number of heads and tails and determine the deviations (O E) between observed and expected. Be sure to indicate whether each deviation is a positive or a negative number. Place the sum of the deviations at the bottom of the Deviations column. Answer Question 1 on the Lab Report. Activity 2. Independent Events Occurring Simultaneously The results obtained from a series of tosses vary with the number of coins tossed at the same time. The expected results can be determined on the basis of the following rule: The probability of two or more unrelated events occurring at the same time is the product of their individual probabilities. Procedure 1. Toss two coins together 36 times. Record the results in Table 2. In calculating the expected number in each category, keep in mind that, when two or more coins are tossed together, each coin acts independently of the other and has an equal chance of falling heads or tails. Answer Question 2 on the Lab Report. Note that this situation is similar to what we see in a simple cross. When an individual with the genotype Aa produces gametes, the probability is that ½ of the gametes will contain the A allele and ½ will contain a. When an Aa female is crossed with an Aa male and progeny are produced, the probability is ¼ that an A egg and an A sperm will come together to produce an AA offspring. Similarly, the probability is ½ for Aa and ¼ for aa progeny. In studying this cross, you are considering a situation in which independent events (the union of different kinds of gametes) occur simultaneously. Thus, there is a basic probability principle that Mendel s first law is based upon. 2. The same law of probability applied to the experiment with two coins can be used to calculate the expected ratios when three, four, or more coins are tossed together. For example, calculate the expected results from tossing three coins together. Let H represent the individual coins that fall heads and let T indicate tails for the same coins. Complete Table 3 showing the various possible combinations and the chances of their occurring. The first combination of three heads, which can occur only 33

2 one way, that is, when all three coins show heads, is worked out as an example. Finally, toss three coins 64 times and record the frequency of occurrence of each of the classes. Calculate the expected numbers and deviations. Answer Question 3 on the Lab Report. 4. Now, calculate the expected results for tossing four coins together. Complete Table 4, showing the probability of the different combinations. (Note: by now, you shouldn t have to toss coins anymore to see the patterns!) Activity 3. Probability and Genetic Counseling In many genetic counseling situations, the counselor will prepare a pedigree for the family or families seeking advice. As much as possible, the counselor will determine the phenotype and the genotype for each person in the pedigree with respect to the trait in question (e.g., albinism). The counselor can then apply probability principles to determine the probability that a child having a particular abnormality will be produced among the offspring of a certain marriage. To illustrate the application of these concepts, consider the pedigree shown in Figure 4. (For the purposes of this problem, your instructor will indicate which members of the pedigree express the genetic abnormality.) Unless there is evidence to the contrary, assume that individuals who have married into this family do not carry the recessive gene for the trait. Fig. 4. Human pedigree showing four generations. Circles represent females; squares represent males. (A larger version of this figure is provided on p. 4.) For a specific example, you might assume that four members of this pedigree are albinos: the woman in the first generation, her second daughter (the mother of individuals 5, 6, 7, and 8), and individuals 4 and 11 in the third generation. Now, assume that you are a genetic counselor and that individuals 6 and 12 in the third generation of 34

3 this pedigree come to you and ask, What is the probability that if we marry and have a family, an albino child will be born to us? The counselor must determine the probability that individuals 6 and 12 are heterozygous carriers of the recessive gene for albinism. The counselor must also consider the probability of two heterozygous carriers producing a homozygous recessive child. First of all, the mother of individual 6 is an albino (cc), which means that 6 must be (probability = 1 or 100%) a heterozygote. The father of individual 12 must be heterozygous (Cc) since his mother is an albino. Although individual 12 is not an albino, he has a ½ chance of being heterozygous, depending on which allele (C or c) he inherited from his father. Finally, two heterozygotes (Cc x Cc) have a ¼ chance of producing an albino (cc) child. The genetic counselor may advise individuals 6 and 12 that if they were to have children, the probability of their having an albino child will be 1 x ½ x ¼ = ⅛. This is the probability that the three independent events will occur simultaneously. A slightly more confusing situation arises in determining the probability that an unaffected child of two knownheterozygous parents is itself heterozygous. For example, in the pedigree shown, we know with certainty that the parents of individual 10 must be heterozygous. The question, then, is What is the probability that one of their phenotypically normal children is heterozygous? We know that in the mating Cc x Cc the expected ratio of offspring is 1 CC : 2 Cc : 1 cc, that is, among the expected three normal 1 CC + 2 Cc offspring, two may be expected to be heterozygous. Thus, the probability is 2/3 that the normal offspring of heterozygous parents will themselves be heterozygous. Answer Question 4 on the Lab Report. 35

4 36

5 Probability and the Idea of Chance Lab Report Name Activity 1. The Idea of Chance Heads (H) Tails (T) Table 1. Results of Tossing a Coin 24 Times Results Observed (O) Expected (E) Deviation (O E) Totals Getting a boy or a girl is a biological situation similar to what happens when you flip a coin. If you randomly select 100 families having only one child, in how many of these families might you expect the child to be a boy? A girl? Activity 2. Independent Events Occurring Simultaneously Table 2. Results of Tossing Two Coins 36 Times Results Combinations Observed (O) Expected (E) Deviation (O E) Heads on both coins HH Heads on one, tails on the other HT, TH Tails on both coins TT Totals

6 3. If, for example, two coins are tossed together, then the chance for each falling with heads up is ½. Likewise, the chance for tails is ½. Therefore, the chance that both will be heads is. The chance for the first coin to fall heads and the second coin tails is also, but one head and one tail can be obtained in another way: The first coin could be tails and the second heads. Therefore, the total probability of obtaining a head on one coin and a tail on the other is. The chance for both coins to fall tails up is the same as for both to be heads, that is,. To summarize, two coins tossed together many times are expected to fall heads, heads about of the time; heads, tails (and vice versa) about of the time; and tails, tails about of the time. Stated as a ratio instead of as fractions, the expected result is : :. (You can use numbers or percents.) Table 3. Expected Results from Tossing Three Coins Together Classes Combinations Probability of Each Class Occurring Observed (O) Expected (E) Deviation (O E) 3 heads HHH ½ x ½ x ½ = ⅛ 8 2 heads; 1 tail HHT, HTH, THH 1 head; 2 tails 3 tails Totals A similar situation would be observed if you were to study families consisting of three children. If you were to select randomly 160 families each of which had three children, in how many would the three children be expected to be all boys? Two boys and a girl? One boy and two girls? Three girls? 38

7 Table 4. Expected Results from Tossing Four Coins Together Classes Combinations Probability of Each Class Occurring 4 heads HHHH 3 heads : 1 tail HHHT, HHTH, HTHH, THHH 2 heads : 2 tails 1 head : 3 tails 4 tails Activity 3. Probability and Genetic Counseling 4. Continue to assume that the same four individuals in the pedigree are albinos, and calculate the probability of the albino trait appearing in the offspring if the following cousins and second cousins should marry and reproduce (yuck): a. 4 x 5 b. 6 x 10 c. 7 x 14 d. 2 x 10 e. 16 x 15 f. 16 x 17 g. 2 x 12 39

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