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Inheritance Patterns and Probability

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100 diapositives · 1.9K Views · Sep 09, 2014

Inheritance Patterns and Probability. July 2008. Pedigrees. I. 1. 2. Dd, DD = normal dd = deaf. II. 1. 2. 3. III. 1.

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  1. Inheritance Patterns and Probability July 2008

  2. Pedigrees

  3. I. 1 2 Dd, DD = normal dd = deaf II. 1 2 3 III. 1 This pedigree shows a family with a form of deafness that is inherited in a recessive manner. Members of the family with filled symbols are deaf. Which members of this family are definitely heterozygous (Dd)? I-1 and I-2 I-1, I-2, and II-1 I-1, II-1, and II-3 I-1, I-2, and II-3 I-1, I-2, II-1, and II-3

  4. I. 1 2 II. Dd dd 1 2 3 dd III. 1 If II-2 and II-3 just had another baby boy. What is the chance that he is deaf? 1/8 1/4 1/2 3/4 1

  5. family 2 family 1 Dd Dd I. Dd Dd I. 1 2 1 2 Dd or DD II. dd Dd Dd or DD II. Dd dd 1 2 3 1 2 3 dd III. dd III. 1 1 What are the chances that II-1 from family 1 and II-1 from family 2 will have a deaf child together? 1/4 1/9 4/9 1/16

  6. Based on the pedigree above, which inheritance pattern can be ruled out? Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive None of the above

  7. Based on the pedigree above, which inheritance pattern can be ruled out? • A. Autosomal dominant • B. Autosomal recessive • C. X-linked dominant • D. X-linked recessive • E. None of the above

  8. Based on the pedigree above, which inheritance pattern can be ruled out? X-linked dominant X-linked recessive neither of the above

  9. ? Phenylketonuria (PKU) is an inherited disorder that can lead to mental retardation if left untreated. PKU is inherited in an recessive manner. What is the chance that the boy marked with a “?” in the pedigree will have PKU? 1/3 1/4 1/6 1/8

  10. I. 1 4 2 3 II. 1 3 4 5 2 ? III. 1 You would like to use mitochondrial DNA to try to determine if III-1 is a member of the family shown in this pedigree. II-2 and II-3 are dead as indicated with a slash and you are unable to collect mitochondrial DNA from them. If III-1 is a member of this family his mitochondrial DNA should match: A) I-1 and II-1 only B) I-1, I-2 and II-1 only C) I-1, I-3, II-1, and II-4 only D) I-3 and II-4 only E) I-3, II-4, and II-5 only

  11. Diastrophic dysplasia Autosomal recessive “D” normal allele “d” mutant allele Achondroplasia Autosomal dominant “A” mutant allele “a” normal allele Ron Peggy Gordon Pat Matt Amy What is Matt’s genotype (Matt has diastrophic dysplasia)? A. ddaa B.ddAa C. Ddaa D.DdAa E. ddAA

  12. Ddaa Ddaa Diastrophic dysplasia Autosomal recessive SLC26A2 gene “D” normal allele “d” mutant allele Achondroplasia Autosomal dominant FGFR3 gene “A” mutant allele “a” normal allele Ron Peggy Gordon Pat Matt Amy ddaa DDAa (note AA embryos are not viable) What is Ron’s genotype? A.ddaa B.Ddaa C.DDaa

  13. DDaa DDaa Diastrophic dysplasia Autosomal recessive SLC26A2 gene “D” normal allele “d” mutant allele Achondroplasia Autosomal dominant FGFR3 gene “A” mutant allele “a” normal allele Ddaa Ddaa Ron Peggy Gordon Pat Matt Amy ddaa DDAa (note AA embryos are not viable) What is Pat’s genotype? A.DDAa B.DDAA C. Ddaa D. None of the above

  14. Diastrophic dysplasia Autosomal recessive SLC26A2 gene “D” normal allele “d” mutant allele Achondroplasia Autosomal dominant FGFR3 gene “A” mutant allele “a” normal allele Matt Amy ddaa DDAa (note AA embryos are not viable) Jeremy Zach Molly Jacob What is Zach’s genotype? A. Ddaa B.DdAa C.DdAA

  15. Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele. ? ? What is the phenotype of the twins’ father? A) RR B) Rr C) rr D) red

  16. Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele. ? ? What is the genotype of the twins’ father? A) RR B) Rr C) rr D) 1/2 Rr, 1/2 RR

  17. Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele. ? ? What is the genotype of the twins' mother? A) RR B) Rr C) ½ Rr, ¼ RR D) 2/3 Rr, 1/3 RR

  18. Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele. ? ? What is the probability that the first twin born will have blue cootie disease? A) 1/4 B) 1/3 C) 1/6 D) 0

  19. The next few questions are not about pedigrees, but follow the cootie example

  20. Antennaless is an autosomal recessive disorder that leads to cooties without antenna. A is the dominant WT allele and a is the mutant recessive allele. A true-breeding WT red cootie mates with a true-breeding blue antennaless cootie. X P What is the phenotype of the F1 generation? A) All red with antennae B) All red but half with antennae and half without C) 9 red antennae: 3 red no antennae: 3 blue antennae: 1 blue no antennae D) ¼ red with antennae, ¼ red without antennae, ¼ blue with antennae, ¼ blue without antennae

  21. You allow the F1 generation to mate and produce offspring (F2 generation). X P F1 RrAa What is the probability that an F2 cootie will be red? A) 1/4 B) 1/2 C) 3/4 D) 1

  22. Here is the F2 generation (RrAa X RrAa) observed expected (O-E)2/E Do the red and antenna gene follow rules of independent assortment? What is expected number of red cooties with antennae? A) 963 B) 1700 C) 1760 D) 2063

  23. Here is the F2 generation (RrAa X RrAa) observed expected (O-E)2/E Do the red and antenna gene follow rules of independent assortment? What is (O-E)2/E for the blue antennaless group? A) 625 B) 25 C) 3.2 D) 0.13

  24. Here is the F2 generation (RrAa X RrAa) 3138 total observed expected (O-E)2/E • Do the red and antenna gene follow rules of independent assortment? • Yes, accept hypothesis – differences are likely due to chance • Yes, accept hypothesis – differences are not likely due to chance • No, reject hypothesis – differences are likely due to chance • No, reject hypothesis – differences are not likely due to chance

  25. Calculating probability of inheritance (monhybrid, dihybrid crosses)

  26. Results of the F1 cross Yy X Yy What is the phenotype of the circled green pea? A) YY B) Yy C) yy D) green E) need more information

  27. Results of the F1 cross Yy X Yy What is the genotype of the circled yellow pea? A) YY B) Yy C) yy D) yellow E) need more information

  28. Plant 1: Yellow, round peas Plant 2: Green, wrinkled peas X P: F1: 1/2 Yellow, round peas 1/2 Yellow, wrinkled peas Y - Yellow y - Green R - Round r - wrinkled What is the genotype of the yellow, round parent? A: YYRR B: YyRR C: YYRr D: YyRr E: Cannot be determined

  29. Use Mendel’s Dihybrid cross results: P X F1 F2 315 101 108 32 Given this data, what do you think the ratio of offspring is? A: 3:1 B: 1:2:1 C: 9:3:3:1 D: 2:1

  30. Results of the F1 cross Yy X Yy 1 2 3 4 The test cross that would most clearly distinguish the genotype of the circled yellow pea is: A) Yellow pea 1 X Yellow pea 2 B) Yellow pea 2 X Yellow pea 3 C) Yellow pea 2 X Green pea 4 D) You would need to do all of the above crosses

  31. Genotype and phenotype Phenotypes Genotypes You cross a yellow with a green and see a 50:50 ratio of greenand yellow progeny. What is the genotype of theoriginal yellow pea? YY Yy yy Need more information

  32. Dihybrid cross Mating between individuals that differ in two traits Round, Yellow Wrinkled, Green X P RRYY rryy F1 100% Round, Yellow RrYy What are the possible gametes produced by the F1 peas? A) rryy, RrYy, RRYY B) R, r, Y, y C) Rr, Yy, RR, rr, YY, yy D) RY, Ry, rY, ry

  33. RrYY RRYY RYRy RrYy RRYy RRyy Rryy RrYy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy F2 Generation Dihybrid cross F1 X RrYy Question 6: What fraction of the F2 generation is green? A) 1/16 B) 1/2 C) 1/9 D) 1/4 RY Ry rY ry RY Ry rY RrYy ry

  34. What is the phenotype ratio of progeny in F1 generation of the following cross? Round, yellow Wrinkled, green X RrYy rryy A B C D 9 3 3 1 3 1 3 1 1 1 1 1 1 3 3 9 Round, Yellow Wrinkled, Yellow Round, Green Wrinkled, Green

  35. Can you use the outcome todeduce the parental genotype? Suppose you cross a yellow and green and get 50% yellow and 50% green? What are the parental genotypes? A) YY X yy B) Yy x yy C) yy x yy D) Yy x Yy

  36. Monohybrid cross probability Consider Yy X Yy cross What is the probability ofgetting a Y from parent 1? 1/4 1/2 1 1/16

  37. Monohybrid cross probability Consider Yy X Yy cross What is the probability ofgetting a Y from oneparent *AND* Y from theother parent (i.e. YY)? 1/4 1/2 1 1/16

  38. Monohybrid cross probability Consider Yy X Yy cross What is the probability ofbeing Yellow (i.e. YY ORYy)? 1/4 1/2 3/4 1

  39. Consider the following cross: • AaBBCcddEe X aabbCCDdEe • What is the probability their first offspring will be • aaBbCCDdee? • 1/8 • 1/16 • 1/32 • 1/64 • Cannot be determined

  40. What is the probability of rolling a two OR a three with one role of a six-sided die? A)1/3 B)1/2 C)1/6 D)1/36 E)1/64

  41. A male smurf has an dominant X-linked disorderthat causes red skin. He marries smurfette (who is normal blue). What are the possible phenotypes of their male children?A) all blue skin B) all red skin C) patches of red and blue skin D) more than one of the above

  42. A male smurf has an dominant X-linked disorder that causes red skin. He marries smurfette (who is normal blue). What are the possible phenotypes of their female children?A) all blue skin B) all red skin C) patches of red and blue skin D) more than one of the above

  43. Statistical Analysis of Crosses

  44. Use Mendel’s Dihybrid cross results: P X Total seeds observed = 556 F1 F2 315 101 108 32 3. Calculate Expected (e) numbers for each class if hypothesis correct How many seeds would you expect to be green and round (to the nearest whole number)? A: 100 B: 104 C: 105 D: 108

  45. Calculate Expected (e) numbers for each class if hypothesis correct Calculate X2 = ∑ (o -e)2/e Always use real numbers, not % or fraction ∑ means ’Sum of all classes’ Use a table: Observed expected (o-e)2/e 315 312 (315-312)2/312=0.029 101 104 0.087 108 104 0.154 32 35 0.257 X2 = 0.527 5. Calculate Degree of Freedom A: 1 B: 4 C: 3

  46. 6. Look up probability (p) for X2 at a given df in the table A: Accept the hypothesis B: Reject the hypothesis

  47. P X F1 F2 315 101 108 32 How many degrees of freedom are there in the F2 generation of the following cross? A) 1 B) 2 C) 3 D) 4 E) 5

  48. What if his results had been 5120 yellow and 2903 green? Could Mendel still accept his hypothesis? X2 = 535 A) Accept the hypothesis B) Reject the hypothesis

  49. What does P = 0.005mean for the 28:20 ratio? 28:20 is likely to be 3:1 28:20 is NOT likely to be 3:1 28:20 is not statistically “significant” and so cannot be used to assess 3:1 ratio This experiment is totally flawed and cannot be interpretted

  50. Exceptions to Mendel’s Laws (maternal, cytoplasmic/mitochondrial, sex-limited, co-dominance, incomplete dominance, lethal, epistatsis, heterozygous advantage, imprinting)