Using Genetic Ratios Probability

Using genetic ratios An important part of genetics today is concerned with predicting the types of progeny that emerge from a cross and calculating their expected frequency—in other words, their probability. We have already examined two methods for doing so—Punnett squares and branch diagrams. Punnett squares can be used to show hereditary patterns based on one gene pair, two gene pairs or more. Such squares are a good graphic device for representing progeny, but making them is time consuming. Even the 16-compartment Punnett square takes a long time to write out, but, for a trihybrid, there are 23 , or 8, different gamete types, and the Punnett square has 64 compartments. The branch diagram is easier and is adaptable for phenotypic, genotypic, or gametic proportions, as illustrated for the dihybrid A /a ; B /b . (The dash means that the allele can be present in either form; that is, dominant or recessive.) Note that the “tree” of branches for genotypes is quite unwieldy even in this case, which uses two gene pairs, because there are 32 =9 genotypes. For three gene pairs, there are 33 , or 27, possible genotypes. The application of simple statistical rules is the third method for calculating the probabilities (expected frequencies) of specific phenotypes or genotypes coming from a cross. The two probability rules needed are the product rule and the sum rule, which we will consider in that order. The product rule states that the probability of independent events occurring together is the product of the probabilities of the individual events. The possible outcomes of rolling dice follow the product rule because the outcome on each separate die is independent of the others. As an example, let us consider two dice and calculate the probability of rolling a pair of 4s. The probability of a 4 on one die is 1/6 because the die has six sides and only one side carries the 4. This probability is written as follows: P(of a 4) = 1/6 Therefore, with the use of the product rule, the probability of a 4 appearing on both dice is 1/6 × 1/6 = 1/36 , which is written P(of two 4s) = 1/6 x 1/6 = 1/36 The sum rule states that the probability of either of two mutually exclusive events occurring is the sum of their individual probabilities. In the product rule, the focus is on outcomes A and B. In the sum rule, the focus is on the concept of outcome A or B. Dice can also be used to illustrate the sum rule. We have already calculated that the probability of two 4s is 1/36 and, with the use of the same type of calculation, it is clear that the probability of two 5s will be the same, or 1/36. Now we can calculate the probability of either two 4s or two 5s. Because these outcomes are mutually exclusive, the sum rule can be used to tell us that the answer is 1/36 + 1/36 which is 1/18. This probability can be written as follows: p(two 4s or 5s) = 1/36 + 1/36 = 1/18 “Advanced Genetic Probability Problem Set” Tip: Remember that for any gene or trait, 'homozygous dominant' means AA, 'heterozygous' (also called a carrier) means Aa and 'homozygous recessive' means aa 1. A woman is heterozygous for Tay-Sachs. What are the chances of her giving birth to a child with Tay-Sachs (a homozygous recessive disorder) if the father is normal (homozygous dominant?) What are the chances of giving birth to a carrier like herself? 2. Both you and your sister have attached earlobes, yet your parents have unattached ones. Unattached earlobes are dominant over attached. Give the genotypes of you, your sister, and your parents. 3. Parents who are carriers for have sickle-cell disease produce a child who has sicklecell disease. What are the chances that their next child will have sickle-cell disease? 4. A woman with PKU (kk) marries a man who is genotypically normal (KK). Which answer below best describes the chances that their child will: be genotypically normal (KK), be a carrier for PKU (Kk), or have the disease PKU (kk) ? 5. If both your parents have a continuous hairline (cc), what best describes your possibilities for hairlines? 6. Both Dick and Jane have freckles (dominant) and attached earlobes (recessive). Six of their children do not have freckles. What are the chances that their next child will have freckles and attached earlobes? 7. In a dihybrid cross of a heterozygous pea plant with purple flowers and round seeds (PpRr), what is the probability that a) a white-flowered pea plant with round seeds will be produced?_______________ b) a purple-flowered pea plant with round seeds will be produced? ______________ c) a white-flowered pea plant with wrinkled seeds will be produced? _____________ d) a purple-flowered pea plant with wrinkled seeds will be produced? _____________ e) the genotype PpRr will result? ______________ f) the genotype ppRr will result? ______________ g) the genotype PpRR will result? ______________ h) the genotype ppRr will result? ______________ 8. In a trihybrid cross of a tall, purple-flowered pea plant with round seeds (TtPpRr) with a tall, white-flowered pea plant with wrinkled seeds (Ttpprr), what is the probability: a) that a tall, white-flowered plant with wrinkled seeds will be produced?__________ b) that a short, purple-flowered plant with round seeds will be produced? __________ c) that a short, white-flowered plant with wrinkled seeds will be produced? ________ d) that a tall, purple-flowered plant with round seeds will be produced? ___________ e) that a tall, white-flowered plant with round seeds will be produced?____________ 9. A woman is heterozygous for polydactyly and has 6 fingers on each hand (polydactyly is a dominant trait). What are the chances of her giving birth to a child with polydactyly if the father is normal (homozygous recessive)? 11. Both a husband and wife have normal vision but their son is color blind. What are the likely genotypes of the boy’s parents? Of the boy?