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Genetics: Part II Predicting Offspring

Genetics: Part II Predicting Offspring. Key. Affected female. Mating. Male. Affected male. Female. Offspring. 1st generation. Ff. ff. Ff. Ff. 1st generation. Ww. ww. ww. Ww. 2nd generation. 2nd generation. ff. ff. ff. Ff. FF or Ff. Ff. Ww. ww. ww. Ww. Ww. ww.

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Genetics: Part II Predicting Offspring

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  1. Genetics: Part II Predicting Offspring

  2. Key Affected female Mating Male Affectedmale Female Offspring 1stgeneration Ff ff Ff Ff 1stgeneration Ww ww ww Ww 2ndgeneration 2ndgeneration ff ff ff Ff FF or Ff Ff Ww ww ww Ww Ww ww 3rdgeneration 3rdgeneration ff FForFf WWorWw ww No widow’speak Freeearlobe Widow’speak Attachedearlobe b) (a) Is an attached earlobe a dominantor recessive trait? Is a widow’s peak a dominant orrecessive trait?

  3. 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.

  4. Curriculum Framework 3A EK3 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.

  5. Probability

  6. 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. 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

  8. 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

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

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

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

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

  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

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

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