# Inheritance Patterns and Probability July 2008. Pedigrees

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• Slide 1
• Inheritance Patterns and Probability July 2008
• Slide 2
• Pedigrees
• Slide 3
• 1 2 2 1 3 1 I. II. III. 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)? A.I-1 and I-2 B.I-1, I-2, and II-1 C.I-1, II-1, and II-3 D.I-1, I-2, and II-3 E.I-1, I-2, II-1, and II-3 Dd, DD = normal dd = deaf
• Slide 4
• 1 2 2 1 3 1 I. II. III. If II-2 and II-3 just had another baby boy. What is the chance that he is deaf? A.1/8 B.1/4 C.1/2 D.3/4 E.1 Dd dd
• Slide 5
• 1 2 2 1 3 1 I. II. III. dd 1 2 2 1 3 1 I. II. III. dd Dd Dd or DD Dd Dd or DD Dd family 1 family 2 What are the chances that II-1 from family 1 and II-1 from family 2 will have a deaf child together? A.1/4 B.1/9 C.4/9 D.1/16
• Slide 6
• 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
• Slide 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
• Slide 8
• Based on the pedigree above, which inheritance pattern can be ruled out? A.X-linked dominant B.X-linked recessive C.neither of the above
• Slide 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? A.1/3 B.1/4 C.1/6 D.1/8
• Slide 10
• I. II. III. 1 2 1 3 4 1 2 3 4 5 ? 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
• Slide 11
• Diastrophic dysplasia Autosomal recessive D normal allele d mutant allele Matt Amy RonPeggy GordonPat What is Matts genotype (Matt has diastrophic dysplasia)? A. ddaa B. ddAa C. Ddaa D. DdAa E. ddAA Achondroplasia Autosomal dominant A mutant allele a normal allele
• Slide 12
• Diastrophic dysplasia Autosomal recessive SLC26A2 gene D normal allele d mutant allele Matt Amy RonPeggy GordonPat ddaa What is Rons genotype? A. ddaa B. Ddaa C. DDaa Achondroplasia Autosomal dominant FGFR3 gene A mutant allele a normal allele DDAa (note AA embryos are not viable) Ddaa
• Slide 13
• Diastrophic dysplasia Autosomal recessive SLC26A2 gene D normal allele d mutant allele Matt Amy RonPeggy GordonPat ddaa What is Pats genotype? A. DDAa B. DDAA C. Ddaa D. None of the above Achondroplasia Autosomal dominant FGFR3 gene A mutant allele a normal allele DDAa (note AA embryos are not viable) Ddaa DDaa
• Slide 14
• Diastrophic dysplasia Autosomal recessive SLC26A2 gene D normal allele d mutant allele Matt Amy JeremyZachMollyJacob ddaa What is Zachs genotype? A. Ddaa B. DdAa C. DdAA Achondroplasia Autosomal dominant FGFR3 gene A mutant allele a normal allele DDAa (note AA embryos are not viable)
• Slide 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 ??
• Slide 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 ??
• Slide 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) Rr, RR D) 2/3 Rr, 1/3 RR ??
• Slide 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 ??
• Slide 19
• The next few questions are not about pedigrees, but follow the cootie example
• Slide 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. 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) red with antennae, red without antennae, blue with antennae, blue without antennae X P
• Slide 21
• You allow the F1 generation to mate and produce offspring (F2 generation). What is the probability that an F2 cootie will be red? A) 1/4 B) 1/2 C) 3/4 D) 1 X P RrAa F1
• Slide 22
• Here is the F2 generation (RrAa X RrAa) What is expected number of red cooties with antennae? A) 963 B) 1700 C) 1760 D) 2063 1767 543 598 221 observedexpected (O-E) 2 /E Do the red and antenna gene follow rules of independent assortment?
• Slide 23
• Here is the F2 generation (RrAa X RrAa) What is (O-E) 2 /E for the blue antennaless group? A) 625 B) 25 C) 3.2 D) 0.13 1767--- 543586.7 598586.7 221196 observedexpected (O-E) 2 /E Do the red and antenna gene follow rules of independent assortment?
• Slide 24
• Here is the F2 generation (RrAa X RrAa) 3138 total Do the red and antenna gene follow rules of independent assortment? A)Yes, accept hypothesis differences are likely due to chance B)Yes, accept hypothesis differences are not likely due to chance C)No, reject hypothesis differences are likely due to chance D)No, reject hypothesis differences are not likely due to chance 1767---0.02 543586.73.3 598586.70.21 221196--- observedexpected (O-E) 2 /E
• Slide 25
• Calculating probability of inheritance (monhybrid, dihybrid crosses)
• Slide 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
• Slide 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
• Slide 28
• What is the genotype of the yellow, round parent? Y - Yellow y - Green R - Round r - wrinkled A:YYRR B:YyRR C:YYRr D:YyRr E:Cannot be determined Plant 1: Yellow, round peas Plant 2: Green, wrinkled peas X F1: 1/2 Yellow, round peas 1/2 Yellow, wrinkled peas P:
• Slide 29
• Use Mendels Dihybrid cross results: XP 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
• Slide 30
• Results of the F1 cross Yy X Yy 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 1 2 3 4
• Slide 31
• Genotype and phenotype Phenotypes Genotypes You cross a yellow with a green and see a 50:50 ratio of green and yellow progeny. What is the genotype of the original yellow pea? A)YY B)Yy C)yy D)Need more information
• Slide 32
• Dihybrid cross Mating between individuals that differ in two traits X Round, YellowWrinkled, Green RRYYrryy 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 P F1 RrYy 100% Round, Yellow
• Slide 33
• Dihybrid cross F1 X RrYy RY Ry rY ry RY Ry rY ry F2 Generation RRYYRYRy RrYY RrYy RRYyRRyy RrYy Rryy RrYYRrYy rrYYrrYy RrYy RryyrrYyrryy Question 6: What fraction of the F2 generation is green? A) 1/16 B) 1/2 C) 1/9 D) 1/4
• Slide 34
• What is the phenotype ratio of progeny in F1 generation of the following cross? RrYyrryy X Round, yellowWrinkled, green Round, Yellow Wrinkled, Yellow Round, Green Wrinkled, Green 93319331 31313131 11111111 13391339 A B C D
• Slide 35
• Can you use the outcome to deduce 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
• Slide 36
• Monohybrid cross probability Consider Yy X Yy cross What is the probability of getting a Y from parent 1? A)1/4 B)1/2 C)1 D)1/16
• Slide 37
• Monohybrid cross probability Consider Yy X Yy cross What is the probability of getting a Y from one parent *AND* Y from the other parent (i.e. YY)? A)1/4 B)1/2 C)1 D)1/16

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