Advanced Probability and Dihybrid Crosses

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. 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 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 The cross below of RrYy x RrYy is illustrated with two punnet squares and a branch diagram. The probability of each genotype is calculated using the product and sum rules. 1. 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? 2. 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? ______________ 3. 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?____________ 4. Suppose a white, straight haired guinea pig mates with a brown, curly-haired animal. All five babies in their first litter have brown fur, but three are curly and two have straight hair. The second litter consists of six more brown offspring, where two are curly and four are straight haired. a) Assuming curly is dominant to straight, what are the genotypes of the parents and the offspring? b) What is the probability of getting two female guinea pigs with straight hair in a row? 5. . About 70% of Americans get a bitter taste from the substance called phenylthiocarbamide (PTC). It is tasteless to the rest. The "taster" allele is dominant to non-taster. Also, normal skin pigmentation is dominant to albino. A normally pigmented woman who is taste-blind for PTC has an albino-taster father. She marries an albino man who is a taster, though the man's mother is a non-taster. Show the expected offspring of this couple.