1. Probability and Pedigree Practice Problems Pick a card

2. Today Practice probability problems Practice probability problems Practice will help in doing these types of questions. Practice will help in doing these types of questions. Any problems you do not have time to do, try doing later. Any problems you do not have time to do, try doing later.

3. Practice 1. Work on the probability problems on pages 1- 4 and 1-5 of the lab manual 1. Work on the probability problems on pages 1- 4 and 1-5 of the lab manual 2. Pedigrees and probability problems 1-28 and 1-30. These you may need to complete at home. 2. Pedigrees and probability problems 1-28 and 1-30. These you may need to complete at home. Please refer to Dr. Colavito’s homepage for solutions. Please refer to Dr. Colavito’s homepage for solutions.

4. Begin with page 1-4 1. 1. – A) Probability of rolling a “5”= 1/6 – B) Probability of rolling an even number: 1/6 + 1/6 + 1/6 = 1/2

5. Problem 1 continued 1c) 1/6 x 1/6 x1/6 = 1/216 1c) 1/6 x 1/6 x1/6 = 1/216 1d) 6x(1/216)=1/36 1d) 6x(1/216)=1/36 1e) 1/216 x 6 = 1/36 1e) 1/216 x 6 = 1/36 1f) 1 (any number) x 5/6 x 4/6 = 20/36= 5/9 1f) 1 (any number) x 5/6 x 4/6 = 20/36= 5/9

6. Problem 2 First identify the probabilities of selecting various colors: First identify the probabilities of selecting various colors: –Jar 1: P (y)= 1/5 P (G) = 4/5 –Jar 2 :P (O) = 2/5 P(G) 3/5 –Jar 3 :P (P) 1/10 P(G) 9/10

7. Problem 2 A. 1/5 x 2/5 x 1/10 = 2/250 = 1/125 A. 1/5 x 2/5 x 1/10 = 2/250 = 1/125 B. 4/5 x 2/5 x 9/10 = 72/250 B. 4/5 x 2/5 x 9/10 = 72/250 C. 1-(probability of no green skittles, see part A.)= 1-1/125=124/125 C. 1-(probability of no green skittles, see part A.)= 1-1/125=124/125

8. Problem 2 D. calculate all possible combinations for each container: D. calculate all possible combinations for each container: (4/5 x 3/5 x 1/10) + (4/5 x 2/5 x 9/10) + 1/5 x 3/5 x 9/10) = 111/250 (4/5 x 3/5 x 1/10) + (4/5 x 2/5 x 9/10) + 1/5 x 3/5 x 9/10) = 111/250

9. Problem 2 E. (2/5 x 1/10 x 4/5) = 8/250= 4/125 E. (2/5 x 1/10 x 4/5) = 8/250= 4/125

10. Problem 3 Autosomal Dominant A. ¾ A. ¾ B. 4!/3!1! (1/4) 3 (3/4) 1 = 3/64 B. 4!/3!1! (1/4) 3 (3/4) 1 = 3/64 C. ¼ x ¾ x/ ¾ = 9/64 C. ¼ x ¾ x/ ¾ = 9/64 D. 1-(probability that all will have straight little fingers)= 1-(1/4) D. 1-(probability that all will have straight little fingers)= 1-(1/4) 3= 63/64

11. Problem 4: Autosomal Recessive A. 2/3 A. 2/3 B. 2/3 B. 2/3 Keep in mind the parents of Jessica and Jerry must be heterozygous for in order to have a sibling that has CF. Keep in mind the parents of Jessica and Jerry must be heterozygous for in order to have a sibling that has CF.

12. Problem 4 C. 2/3 x 2/3 x 1/4 = 1/9 C. 2/3 x 2/3 x 1/4 = 1/9 Remember to refer to your Punnett square and you will see that with heterozygous parents there is ¼ chance of cc. Remember to refer to your Punnett square and you will see that with heterozygous parents there is ¼ chance of cc.

13. Problem 4 D. 3!/2!1! (1/9) 2 (8/9) 1 = 8/243 D. 3!/2!1! (1/9) 2 (8/9) 1 = 8/243 E. 8/9 x 1/9 x 8/9 = 64/729 E. 8/9 x 1/9 x 8/9 = 64/729

14. Problem 5: X linked recessive ½ that Polly has the “d” allele ½ that Polly has the “d” allele ½ x ¼ =1/8 ½ x ¼ =1/8 ½ x 3/8 x 1/8 = 3/128 ½ x 3/8 x 1/8 = 3/128 5!/2!3! (1/8)2 (7/8) 3 = 3430/32768 =0.105 5!/2!3! (1/8)2 (7/8) 3 = 3430/32768 =0.105

15. Pedigrees (page 1-26) A. autosomal dominant A. autosomal dominant B. X linked dominant B. X linked dominant C. X linked recessive C. X linked recessive D. autosomal recessive D. autosomal recessive E. X-linked dominant E. X-linked dominant F. Autosomal recessive F. Autosomal recessive G. Autosomal Dominant G. Autosomal Dominant H. X-linked recessive H. X-linked recessive

16. Answers Check Dr. Colavito’s homepage “study guide” to check your answers. Check Dr. Colavito’s homepage “study guide” to check your answers.