Gene$cs: Part II Predic$ng Offspring APGRU5L2

Transcription

1 Gene$cs: Part II Predic$ng Offspring APGRU5L2

2 The answer, of course, is no. However, this is a common misconception or misunderstanding about how the numbers work in inheritance. 2

3 Key Male Female Affected male Affected female Mating Offspring 1st generation 2nd generation 3rd generation Ww ww ww Ww Ww ww ww Ww Ww WW or Ww ww ww 1st generation 2nd generation FF or Ff 3rd generation Ff Ff ff Ff ff ff Ff ff Ff FF or Ff ff 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?

4 A pedigree 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 4

5 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.

6 ANS TO PEDIGREE QUESTIONS 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. 6

7 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.

8 Probability 8

9 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 9

10 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 à1/2 Pp 10

11 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 F 2 plant from a monohybrid cross will be heterozygous rather than homozygous 11

12 Rule of Multiplication 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 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 à HH ½ x ½ = ¼

13 multiplication rule 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: Cross: GgSs x GgSS 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 13

14 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

15 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? 15

16 RULE OF ADDITION 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) 16

17 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

18 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 18

19 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? Cross PpYyRr x Ppyyrr ppyyrr ppyyrr Ppyyrr PPyyrr ppyyrr 1 / 4 (probability of pp) 1 / 2 (yy) 1 / 2 (Rr) = 1 / 16 = 1 / 16 =? =? = 1 / 16 Chance of at least two recessive traits = 6 / 16 or 3 / 8

20 Cross PpYyRr x Ppyyrr (Answer) ppyyrr ppyyrr Ppyyrr PPyyrr ppyyrr 1 / 4 (probability of pp) 1 / 2 (yy) 1 / 2 (Rr) 1 / 4 1 / 2 1 / 2 1 / 2 1 / 2 1 / 2 1 / 4 1 / 2 1 / 2 1 / 4 1 / 2 1 / 2 Chance of at least two recessive traits = 1 / 16 = 1 / 16 = 2 / 16 = 1 / 16 = 1 / 16 = 6 / 16 or 3 / 8

21 Practice Now that you have reviewed some basic genetics concepts solidify your skill by completing the set of practice problems available at 21

22 Dihybrid crosses combine 2 traits on separate chromosomes Law of Independent Assortment says genes on different chromosomes assort independently into gametes So SsBb can form all these gametes: SB, Sb, sb, sb These gamete possibilities go on the outside of the Punnett! 22

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24 Write down these answers silently and independently 1. Draw the dihybrid cross for the mating of two individuals, both with the genotype GgYy. 24

25 Check yo self!

26 OBJECTIVES Define and give examples of epistasis and polygenic inheritance. Define linked genes and explain why they don t follow the Law of Independent Assortment. 26

27 In epistasis, one gene alters the expression of another gene Epistasis = like a light switch to a power outlet (the power outlet doesn t work unless you flip the light switch) EX: In mice, one gene determines whether pigment will be deposited in the hair, another gene determines the pigment color. 27

28 An example of epistasis BbCc BbCc Gene B: B = black b = brown Eggs 1 / 4 BC Sperm 1 / 4 1 / 4 1 / 4 1 / 4 BC bc Bc bc BBCC BbCC BBCc BbCc Gene C: C = pigment c = no pigment 1 / 4 1 / 4 bc Bc BbCC bbcc BbCc bbcc BBCc BbCc BBcc Bbcc 1 / 4 Gene C affects Gene B! bc BbCc bbcc Bbcc 9 : 3 : 4 bbcc

29 Polygenic inheritance occurs in genes with a spectrum of phenotypes Polygenic Inheritance = additive effect of two or more genes on a single phenotype EX: Skin color - a couple with dark skin pigmentation can have a lighter skinned child (or vice versa) because many genes interact 29

30 Linked genes are located on the same chromosome Chromosome 1 Linked Not Linked Chromosome 2 Linked genes are inherited together, so they do NOT independently assort

31 Exit Ticket 1. What is the probability that a couple has 3 children, all of them are boys?

32 EXIT TICKET Review 1) Probabilities: More than one way of getting 3 boys? NO! So we use the MULTIPLICATION RULE Multiply the probability of each event together ½ * ½ * ½ = 1/8

33 Exit Ticket 2. Describe the Law of Segregation.

34 EXIT TICKET Review 2) Law of Segregation One allele for each gene will be segregated or separated into different gametes Tt Parent Cell T Gametes t

35 Exit Ticket 3. A black wolf and a white wolf can have black and white spotted offspring. Draw a Punnett Square crossing a black wolf with a black and white spotted wolf. What type of dominance gives us spots?

36 EXIT TICKET Review 3) Black wolf x Spotted wolf What type of dominance gives us spots? Co-dominance Write out genotypes Black wolf: BB B Spotted: BW Do Punnett W B BB BW B BB BW

37 Exit Ticket 1. Name all the gametes that a tulip with the genotype AaTt could make. 2. When one gene controls another gene, what is this called?