Genetics: Part II Predicting Offspring

The answer, of course, is no The answer, of course, is no. However, this is a common misconception or misunderstanding about how the numbers work in inheritance. Today we will look at a couple of tools to use, in addition to a Punnett square, to study and predict offspring of specific crosses.

Is a widow’s peak a dominant or recessive trait? b) Key Male Female Affected male Affected female Mating Offspring 1st generation Ff Ff ff Ff 1st generation Ww ww ww Ww 2nd generation 2nd generation FF or Ff ff ff Ff Ff ff Ww ww ww Ww Ww ww 3rd generation 3rd generation ff FF or Ff WW or Ww ww A valuable tool in studying inheritance patterns is a pedigree. A pedigree is a family tree that describes the interrelationships of parents and children across generations Inheritance patterns of particular traits can be traced and described using pedigrees Widow’s peak No widow’s peak Attached earlobe Free earlobe (a) Is a widow’s peak a dominant or recessive trait? b) Is an attached earlobe a dominant or recessive trait?

Interpret this Pedigree Is the trait dominant or recessive? What is the genotype of the first born male in generation II? What is the genotype of the youngest female in generation II? Explain. The “trick” to reading pedigrees is to look for a place where the parents are identical but at least one of their children are different from the parents. In this case, look at the parents in generation III. Both are un-shaded, meaning they do not show the trait in their phenotypes. BUT one child does show it. This means that the parents must have had the allele in question without showing it in their phenotype which means the trait must have been recessive (to “hide” in these two parents). Once that is known, the other questions can be answered. (The reverse situation can be seen in the previous slide in the second generation for widows peak) First born male in generation II is on the left and must be Aa. With the same being true for the first female.

Curriculum Framework 3A EK 3 inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring. a. Rules of probability can be applied to analyze passage of single gene traits from parent to offspring. The chances of producing specific phenotypes can be calculated using the rules of probability. In order for us to determine the “chances” of an organism inheriting a single gene trait, we need to be able to apply the rules of probability.

Probability Mendel’s laws of segregation and independent assortment reflect the rules of probability When tossing a coin, the outcome of one toss has no impact on the outcome of the next toss In the same way, the alleles of one gene segregate into gametes independently of another gene’s alleles

Rule of Addition Rule of addition: Chance that an event can occur 2 or more different ways. Sum of separate probabilities Ex.1/4 Pp +1/4 Pp 1/2 Pp The addition rule states that the probability that any one of two or more exclusive events will occur is calculated by adding together their individual probabilities The rule of addition can be used to figure out the probability that an F2 plant from a monohybrid cross will be heterozygous rather than homozygous

Rule of Multiplication The multiplication rule states that the probability that two or more independent events will occur together is the product of their individual probabilities Chance that 2 or more independent events will occur together Ex. Probability that 2 coins tossed at the same time will land heads up Probability of H x H  HH ½ x ½ = ¼ The multiplication rule (also called the product rule) states that the probability that two or more independent events will occur together is the product of their individual probabilities. Probability in an F1 monohybrid cross can be determined using the multiplication rule. Segregation in a heterozygous plant is like flipping a coin: Each gamete has a 1/2 chance of carrying the dominant allele and a 1/2 chance of carrying the recessive allele.

Rule of Multiplication Cross: GgSs x GgSS What is the probability of producing green, smooth seeds in this cross? Solution Green = 3/4 Smooth = 4/4 3/4 X 4/4 = 12/16 = 3/4 probability of producing green smooth seed Using the multiplication rule can expedite the process of predicting the probability of producing specific phenotypes in the offspring. For example, in peas, green seed color (G) is dominant to yellow seed color (g). Smooth seed coats (S) are dominant to wrinkled seed coats (s). Given that, determine the probability of producing green, smooth seeds in the cross shown here.

From your formula chart: If A and B are mutually exclusive, then P (A or B) = P (A) + P (B) If A and B are independent, then P (A and B) = P(A) X P(B) Ex. Probability of a couple having three girls? Ex. Probability of a couple having three boys? Ex. Probability of having three boys or three girls? The AP Biology Formula chart contains the formula to use when applying the probability rule of addition. The probability of having one girl is ½. The probability of having a second girl is ½. The probability of having the third girl is ½. Each conception is independent of the next. So… ½ x ½ x ½ = 1/8 chance of having three girls. The same would be true of having three boys. However, the odds of producing EITHER three boys or three girls is ¼ because there are two ways to get the result. So the probabilities are added (1/8 + 1/8 = ¼)

In a heterozygous cross YyRr For example: In a heterozygous cross YyRr Probability of YYRR  1/4 (probability of YY)  1/4 (RR)  1/16 Probability of YyRR  1/2 (Yy)  1/4 (RR)  1/8 We can apply the multiplication and addition rules to predict the outcome of crosses involving multiple characters A dihybrid or other multicharacter cross is equivalent to two or more independent monohybrid crosses occurring simultaneously In calculating the chances for various genotypes, each character is considered separately, and then the individual probabilities are multiplied

Cross PpYyRr x PPyyrr ppyyRr 1/4 (probability of pp)  1/2 (yy)  1/2 (Rr)  1/16 ppYyrr  1/16 Ppyyrr  ? PPyyrr  ? ppyyrr  1/16 Chance of at least two recessive traits When considering more than two traits, calculating the probability mathematically is much simpler than drawing a punnett square. Work with a partner to write out how each probability should be calculated?  6/16 or 3/8

Cross PpYyRr x Ppyyrr (Answer) 1/4 (probability of pp)  1/2 (yy)  1/2 (Rr)  1/16 ppYyrr 1/4  1/2  1/2  1/16 Ppyyrr 1/2  1/2  1/2  2/16 PPyyrr 1/4  1/2  1/2  1/16 ppyyrr 1/4  1/2  1/2  1/16 Chance of at least two recessive traits How do your answers compare?  6/16 or 3/8

Complete the genetics card sort matching the term with its definition Practice definitions Genotype Heterozygous Phenotype Allele Monohybrid Test cross Dominant F1 Recessive F2 P1 Homozygous Complete the genetics card sort matching the term with its definition

Practice Now that you have reviewed some basic genetics concepts solidify your skill by completing the set of practice problems available at http://anthro.palomar.edu/practice/mendqui2.htm

Created by: Debra Richards Coordinator of K-12 Science Programs Bryan ISD Bryan, TX