1. Genetics: Part II Predicting Offspring

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3. Key Male Female Affected male Affected female Mating Offspring 1st generation 2nd generation 3rd generation 1st generation 2nd generation 3rd generation Is a widow’s peak a dominant or recessive trait? (a)Is an attached earlobe a dominant or recessive trait? b) Widow’s peak No widow’s peak Attached earlobe Free earlobe FF or Ff WW or Ww Ww ww Ww ww Ff ff FF or Ff ff

4. 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.

5. 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.

6. Probability 6

7. 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 7

8. 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

9. 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

10. 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? 10

11. Probability of YYRR Probability of YyRR 1 / 4 (probability of YY) 1 / 2 (Yy) 1 / 4 (RR) 1 / 16 1/81/8       In a heterozygous cross YyRr For example:

12. Chance of at least two recessive traits ppyyRr ppYyrr Ppyyrr PPyyrr ppyyrr 1 / 4 (probability of pp)  1 / 2 (yy)  1 / 2 (Rr)  1 / 16  ? ?  ?  1 / 16  6 / 16 or 3 / 8 Cross PpYyRr x Ppyyrr

13. Chance of at least two recessive traits ppyyRr ppYyrr Ppyyrr PPyyrr ppyyrr 1 / 4 (probability of pp)  1 / 2 (yy)  1 / 2 (Rr) 1/4  1/2  1/21/4  1/2  1/2 1/2  1/2  1/21/2  1/2  1/2 1/4  1/2  1/21/4  1/2  1/2 1/4  1/2  1/21/4  1/2  1/2  1 / 16  2 / 16  1 / 16  6 / 16 or 3 / 8 Cross PpYyRr x Ppyyrr (Answer)

14. 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 14