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Inheritance Patterns and Probability July 2008 Pedigrees 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)? 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 I. 1 II. 1 III. 2 Dd dd 2 dd 3 1 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 family 2 family 1 I. I. Dd Dd 1 II. Dd or DD 1 Dd 1 2 Dd dd 2 dd III. Dd 3 1 II.Dd or DD III. 1 2 dd Dd 2 dd 3 1 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 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 Based on the pedigree above, which inheritance pattern can be ruled out? A. B. C. D. E. Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive None of the above 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 ? 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 I. II. III. 1 1 2 4 3 3 2 4 5 ? 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 Diastrophic dysplasia Autosomal recessive Achondroplasia Autosomal dominant “A” mutant allele “D” normal allele “d” mutant allele Ron Peggy Gordon Matt Pat “a” normal allele Amy What is Matt’s genotype (Matt has diastrophic dysplasia)? A. ddaa B. ddAa C. Ddaa D. DdAa E. ddAA Diastrophic dysplasia Autosomal recessive SLC26A2 gene Achondroplasia Autosomal dominant FGFR3 gene “D” normal allele “A” mutant allele “d” mutant allele “a” normal allele Ddaa Ron Peggy Ddaa Matt ddaa What is Ron’s genotype? A. ddaa B. Ddaa C. DDaa Gordon Pat Amy DDAa (note AA embryos are not viable) Diastrophic dysplasia Autosomal recessive SLC26A2 gene Achondroplasia Autosomal dominant FGFR3 gene “D” normal allele “A” mutant allele “d” mutant allele “a” normal allele Ddaa Ron Peggy Ddaa Matt ddaa What is Pat’s genotype? A. DDAa B. DDAA C. Ddaa D. None of the above DDaa Gordon Pat DDaa Amy DDAa (note AA embryos are not viable) Diastrophic dysplasia Autosomal recessive SLC26A2 gene Achondroplasia Autosomal dominant FGFR3 gene “D” normal allele “A” mutant allele “d” mutant allele Matt ddaa What is Zach’s genotype? A. Ddaa B. DdAa C. DdAA Jeremy “a” normal allele Amy DDAa (note AA embryos are not viable) Zach Molly Jacob 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 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 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 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 The next few questions are not about pedigrees, but follow the cootie example 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. P X 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 You allow the F1 generation to mate and produce offspring (F2 generation). P X F1 RrAa What is the probability that an F2 cootie will be red? A) 1/4 B) 1/2 C) 3/4 D) 1 Here is the F2 generation (RrAa X RrAa) observed expected (O-E)2/E 1767 543 598 221 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 Here is the F2 generation (RrAa X RrAa) observed expected 1767 --- 543 586.7 598 586.7 221 196 (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 Here is the F2 generation (RrAa X RrAa) 3138 total observed expected 1767 --- 0.02 543 586.7 3.3 598 586.7 0.21 221 196 --- (O-E)2/E 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 Calculating probability of inheritance (monhybrid, dihybrid crosses) 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 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 Plant 1: Yellow, round peas X Plant 2: Green, wrinkled peas P: F1: 1/2 Yellow, round peas Y - Yellow y - Green R - Round r - wrinkled 1/2 Yellow, wrinkled peas What is the genotype of the yellow, round parent? A: B: C: D: E: YYRR YyRR YYRr YyRr Cannot be determined 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 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 Genotype and phenotype 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) B) C) D) YY Yy yy Need more information Phenotypes Dihybrid cross Mating between individuals that differ in two traits Round, Yellow P Wrinkled, Green X 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 Dihybrid cross F1 RrYy X RY Ry rY ry RY RRYY RYRy RrYY RrYy Ry RRYy RRyy RrYy Rryy RrYy rY ry RrYY RrYy rrYY rrYy RrYy rrYy rryy Rryy F2 Generation Question 6: What fraction of the F2 generation is green? A) 1/16 B) 1/2 C) 1/9 D) 1/4 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 Round, Yellow 9 3 1 1 Wrinkled, Yellow 3 1 1 3 Round, Green 3 3 1 3 1 1 1 9 Wrinkled, Green 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 Monohybrid cross probability • Consider Yy X Yy cross What is the probability of getting a Y from parent 1? A) B) C) D) 1/4 1/2 1 1/16 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) B) C) D) 1/4 1/2 1 1/16 Monohybrid cross probability • Consider Yy X Yy cross What is the probability of being Yellow (i.e. YY OR Yy)? A) B) C) D) 1/4 1/2 3/4 1 Consider the following cross: AaBBCcddEe X aabbCCDdEe What is the probability their first offspring will be aaBbCCDdee? A)1/8 B)1/16 C)1/32 D)1/64 E)Cannot be determined 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 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 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 Statistical Analysis of Crosses Use Mendel’s Dihybrid cross results: P X F1 F2 315 101 108 32 Total seeds observed = 556 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 3. Calculate Expected (e) numbers for each class if hypothesis correct 4. 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 6. Look up probability (p) for X2 at a given df in the table A: Accept the hypothesisB: Reject the hypothesis How many degrees of freedom are there in the F2 generation of the following cross? P X F1 F2 315 101 108 32 A) 1 B) 2 C) 3 D) 4 E) 5 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 What does P = 0.005 mean for the 28:20 ratio? A) 28:20 is likely to be 3:1 B) 28:20 is NOT likely to be 3:1 C) 28:20 is not statistically “significant” and so cannot be used to assess 3:1 ratio D) This experiment is totally flawed and cannot be interpretted Exceptions to Mendel’s Laws (maternal, cytoplasmic/mitochondrial, sexlimited, co-dominance, incomplete dominance, lethal, epistatsis, heterozygous advantage, imprinting) Maternal A maternal effect gene exists in a dominant N (normal) allele and a recessive n (mutant) allele. What would be the ratios of genotypes and phenotypes for the following cross? nn female X NN male A) all Nn, all normal B) all Nn, all mutant C) all nn, all mutant D) all NN, all normal A maternal effect gene exists in a dominant N (normal) allele and a recessive n (mutant) allele. What would be the ratios of genotypes and phenotypes for the following cross? NN female X nn male A) all Nn, all normal B) all Nn, all mutant C) all nn, all mutant D) all NN, all normal Worm Mel2 gene products are deposited into the egg by the mother and are required from embryonic development. Mutations in the mel2 gene are recessive and cause maternal effect embryonic lethality. In a cross between mel2 heterozygotes, what percent of embryos will die? A) 100% B) 50% C) 25% D) 0% Zebrafish Ack15 gene products are deposited into the egg by the mother and are required from embryonic development. Mutations in the ack15 gene are recessive and cause maternal effect embryonic lethality. In a cross between ack15 homozygous mutant female and a heterozygous male, what percent of embryos will die? A) 100% B) 50% C) 25% D) 0% You are studying Drosophila mulleri and you discover a maternal effect gene you call nanu. nanu mRNA is localized to the anterior end of the embryo and promotes the formation of anterior structures. Mutations in the nanu gene are recessive. In a cross between two nanu heterozygotes, how many of the embryos will have defects in their anterior structures? A. 100% B. 50% C. 25% D. 0% You are studying Drosophila mulleri and you discover a maternal effect gene you call nanu. nanu mRNA is localized to the anterior end of the embryo and promotes the formation of anterior structures. Mutations in the nanu gene are recessive. In a cross between a nanu/nanu mutant female and a +/+ male, how many of the embryos will have defects in their anterior structures? A. 100% B. 50% C. 25% D. 0% Cytoplasmic/mitochondrial Sex Limited Co-dominance/incomplete dominance Variable Expression Conditional You cross a true-breeding white buffalo to a true breeding black buffalo. All the F1 are brown. P F1 An F1 brown buffalo is crossed to the white parent. If they have 4 offspring, how many do you predict will be white? A. 0 B. 1 C. 2 D. 4 E. Not enough information A true-breeding albino buffalo is crossed to a true-breeding black buffalo and all of the progeny are brown. Crossing the brown buffalos to each other yields an approximate ratio of 1 albino: 2 brown: 1 black. The alleles for buffalo color show: A) complete dominance B) partial dominance C) co-dominance CU has asked the MCDB2150 class to help it with the breeding of Ralphie buffalo. You do the following cross: X Long haired Ralphie buffalo Short haired Ralphie buffalo Ralphie buffalo with a mix of short and long hairs You try to establish a true breeding herd of Ralphie buffalo with a mix of short and long hair using the F1 Ralphies but you are unsuccessful. Which mode of interaction between alleles is a possible reason for your lack of success? A. Codominance B. Incomplete dominance C. Complete dominance D. Recessive epistasis The blood type alleles in humans show example(s) of: A) co-dominance B) complete dominance C) multiple alleles D) two of the above E) all of the above A man with blood type A and a woman with blood type B have a child with blood type O. This couple can also have children with which blood types? A: O only B: AB and O C: A, B and O D: A, B, AB and O Charlie Chaplin (multiple alleles) • Charlie was blood type O • Girlfriend was blood type A • Her (out-of-wedlock child) B FACTS: I locus (blood group) has 3 alleles A = genotype IAIA or Iai -> Ab to B B = genotype IBIB or Ibi -> Ab to A AB = genotype IAIB -> No Ab O = genotype ii -> Ab to both A and B Mr. Chaplin’s case • • A) B) C) Given IA and IB are dominant to I His girlfriend sued for paternity who won? Girlfriend won - baby COULD be his Chaplin won - baby COULD NOT be his Hung jury, can’t tell from the facts Only 66% of women with a heterozygous BRCA1 mutation get breast cancer by age 55 and most do so in only one breast. BRCA1 mutant allele… A: shows incomplete penetrance B: shows variable expressivity C: both Lethal Lethal Alleles All the sneetches want their children to have stars on their bellies. The combination of alleles that makes a black, starless sneetch is lethal. If a true breeding black star bellied sneetch mates with a true breading yellow sneetch, what is the probability that their first child will have a star? A) 1 B) 1/2 C) 1/4 D) 3/16 YYss X yySS YY, Yy = yellow yy = black SS, Ss = star ss = no star Lethal Alleles All the sneetches want their children to have stars on their bellies. The combination of alleles that makes a black, starless sneetch is lethal. If two heterozygous yellow star-bellied sneetches mate, what is the likelihood that their first child will not have a star? YySs X YySs A) 1 B) 1/4 C) 1/5 D) 3/16 YS Ys yS YS YYSS YYSs YySS Ys YYSs YYss YySs yS YySs YySs yySS ys YySs Yyss yySs ys YySs Yyss yySs yyss Sue SLC26A gene DdAa DA Zach DdAa SLC26A gene FGFR3 gene ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8 /00015218-E719-12F1-88F40C01AC1BF814.jpg FGFR3 gene Da dA da DA DDAA DDAa DdAA DdAa Da DDAa DDaa DdAa Ddaa dA DdAA DdAa ddAA ddAa da DdAa Ddaa ddAa ddaa What is the chance that Zach and Sue will have a child with diastrophic dysplasia and achondroplasia? A. 1/6 B. 1/8 C. 3/16 D. 9/16 E. None of the above Sue SLC26A gene DdAa DA Zach DdAa SLC26A gene FGFR3 gene ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8 /00015218-E719-12F1-88F40C01AC1BF814.jpg FGFR3 gene Da dA da DA DDAA DDAa DdAA DdAa Da DDAa DDaa DdAa Ddaa dA DdAA DdAa ddAA ddAa da DdAa Ddaa ddAa ddaa What is the chance that Zach and Sue will have a child with only diastrophic dysplasia? A. 1/3 B. 1/6 C. 1/12 D. 1/16 Epistasis P Brown X Yellow F1 Black F2 Brown: Black: Yellow Partial Dominance Model 1: BB -> Brown 2: Bb -> Black 1: bb -> Yellow Recessive epistasis Model 9 B_E_ -> Black 3 bbE_ -> Brown 4 _ _ee -> Yellow Crossing F2 yellows to brown parent gave a mix of brown, yellow and black. Which model does this support? A: partial dominance B: recessive epistasis Epistasis and Labrador retriever coat color The B locus determines if pigment can be produced “B” codes for black pigment “b” codes for brown pigment The E locus determines if the pigment can be deposited in the hair shaft “E” allows dark pigment (black or brown) to be deposited “e” prevents dark pigment from being deposited (the dogs are yellow) What is the phenotype of a BbEe lab? A. Black B. Brown C. Yellow The B locus determines if pigment can be produced “B” codes for black pigment “b” codes for brown pigment The E locus determines if the pigment can be deposited in the hair shaft “E” allows dark pigment (black or brown) to be deposited “e” prevents dark pigment from being deposited (the dogs are yellow) What is the phenotype of a bbEe lab? A. Black B. Brown C. Yellow The B locus determines if pigment can be produced “B” codes for black pigment “b” codes for brown pigment The E locus determines if the pigment can be deposited in the hair shaft “E” allows dark pigment (black or brown) to be deposited “e” prevents dark pigment from being deposited (the dogs are yellow) What is the phenotype of a bbee lab? A. Black B. Brown C. Yellow The B locus determines if pigment can be produced “B” codes for black pigment “b” codes for brown pigment The E locus determines if the pigment can be deposited in the hair shaft “E” allows dark pigment (black or brown) to be deposited “e” prevents dark pigment from being deposited (the dogs are yellow) What is the phenotype of a BBee lab? A. Black B. Brown C. Yellow You cross a hairless mouse aaBB to a mouse with curly hair AAbb. All of the F1s have straight hair. You cross two of the F1 mice together. In the F2 generation: 18 mice have straight hair, 8 mice are hairless, and 6 have curly hair. What is the phenotype of the aabb mice in the F2 generation? A. hairless B. curly hair C. straight hair hairless mouse aaBB X curly hair AAbb F1s have straight hair AaBb F2 9 A- B4 aa -3 A- bb 18 straight hair mice A-B8 hairless mice aaB- & aabb 6 curly hair mice A-bb What is the order of function? A. A, then B B. B, then A C. A and B act simultaneously D. Not enough data Recessive epistasis Model 9 B_E_ -> Black 3 bbE_ -> Brown 4 _ _ee -> Yellow What is the order of function? A. B, then E B. E, then B C. E and B act simultaneously D. Not enough data Mutant strain A: intermediate 2 builds up Mutant strain B: intermediate 3 builds up Mutant strain C: intermediate 1 builds up If gene A is epistatic to gene B which intermediate will build up in a AB double mutant? A) intermediate 1 B) intermediate 2 C) intermediate 3 Epistasis AA – fluffy hair aa – bald BB – red hair pigment bb – no red pigment (blue hair) X AaBb AaBb What are the possible gametes produced by each chuzzel? A) Aa, Bb B) AaBb C) A, a, B, b D) AB, Ab, aB, ab 1.You cross a true-breeding white buffalo to a true-breeding tan buffalo in the P generation 2. You find out that all of the F1 buffalos are gold 3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 47 33 The T locus determines if pigment can be produced: “T” codes for gold pigment “t” codes for tan pigment The A locus determines if pigment can be deposited into the hair shaft: “A” allows pigment (gold or tan) to be deposited into the hair shaft “a” prevents pigment from being deposited into the hair shaft (the buffalo are white) What is the genotype of the tan buffalo in the P generation? A. AAtt B. Aatt C. Either AAtt or Aatt 1.You cross a true-breeding white buffalo to a true-breeding tan buffalo in the P generation aaTT AAtt 2. You find out that all of the F1 buffalos are gold AaTt 3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 47 33 The T locus determines if pigment can be produced: “T” codes for gold pigment “t” codes for tan pigment The A locus determines if pigment can be deposited into the hair shaft: “A” allows pigment (gold or tan) to be deposited into the hair shaft “a” prevents pigment from being deposited into the hair shaft (the buffalo are white) What is the genotype of the white buffalo in the P generation? A. aatt B. aaTT C. aaTt X 2. You find out that all of the F1 buffalos are gold AaTt AaTt 3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 47 33 Punnett square for cross of two gold buffalo AaTt x AaTt: AT At aT at AT At aT at AATT AATt AaTT AaTt AATt AaTT AaTt AAtt AaTt Aatt AaTt aaTT aaTt Aatt aaTt aatt How many of the genotypes in the Punnett square will result in a gold buffalo? A. 9 B. 4 C. 3 D. 2 E. 1 X 2. You find out that all of the F1 buffalos are gold AaTt AaTt 3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 AT At aT at 47 33 AT At aT at AATT AATt AaTT AaTt AATt AaTT AaTt AAtt AaTt Aatt AaTt aaTT aaTt Aatt aaTt aatt How many of the genotypes in the Punnett square will result in a white buffalo? A. 9 B. 4 C. 3 D. 2 E. 1 X 2. You find out that all of the F1 buffalos are gold AaTt AaTt 3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 AT At aT at 47 33 AT At aT at AATT AATt AaTT AaTt AATt AaTT AaTt AAtt AaTt Aatt AaTt aaTT aaTt Aatt aaTt aatt How many of the genotypes in the Punnett square will result in a tan buffalo? A. 9 B. 4 C. 3 D. 2 E. 1 3. You cross two of the F1 gold buffalo together and in the F2 generation get: 91 AT At 47 33 AT At aT at AATT AATt AaTT AaTt AATt AAtt AaTt Aatt Approximately how many true-breeding tan buffalo are in the F2 generation in your herd? A. 1 B. 3 aT AaTT AaTt aaTT aaTt C. 11 D. 22 at AaTt Aatt aaTt aatt E. 33 Heterozygous Advantage A man who does not have sickle cell anemia and has no history of it in his family (assume he is not a carrier) marries a woman who has sickle cell anemia. They have a son. This family is planning to travel to the Solomon Islands. Which family member(s) should take Lariam, a very expensive drug that prevents malaria? A. Father B. Mother C. Son D. Father and the son E. Everyone Imprinting A’a’ Aa 1 2 3 4 A’a Aa’ aa’ AA’ A mutation (a) occurs on an imprinted gene (A). The maternal copy of the gene is methylated and not expressed. ‘ denotes the alleles inherited from the father. Which of the offspring will be affected? A) 1 and 3 B) 2 and 3 C) 3 only D) none of the offspring will be affected Prader-Willi if the deletion is on the chromosome inherited from mom. Angelman syndrome if the deletion is on the chromosome inherited from dad. A is the normal chromosome 15 a contains the deletion A is from mom A’ is from dad A’a’ Aa 1 2 3 A’a AA’ aa’ Which disorder does the mother have? A) None B) PWS C) AS D) need more information Prader-Willi if the deletion is on the chromosome inherited from mom. Angelman syndrome if the deletion is on the chromosome inherited from dad. A is the normal chromosome 15 a contains the deletion A is from mom A’ is from dad A’a’ Aa 1 2 3 A’a AA’ aa’ Which disorder does the offspring 1 have? A) None B) PWS C) AS D) need more information Prader-Willi if the deletion is on the chromosome inherited from mom. Angelman syndrome if the deletion is on the chromosome inherited from dad. A is the normal chromosome 15 a contains the deletion A is from mom A’ is from dad A’a’ Aa 1 2 3 A’a AA’ aa’ Which disorder does the offspring 3 have? A) both syndromes B) PWS C) AS D) need more information Prader-Willi if the deletion is on the chromosome inherited from mom. Angelman syndrome if the deletion is on the chromosome inherited from dad. An individual with AS married a normal individual and produced an offspring with PWS. What is the gender of the parent with AS? A) male B) female C) need more information What is the gender of the child with PWS? A) male B) female C) need more information Dominance vs. recessive Chuzzle Population •A chuzzle populations contains 640 red chuzzels and 320 green chuzzles. •Chuzzles are not choosy about their mates. Either color will mate with the other at equal frequencies. •When red chuzzles mate all the pups are red. When red and green chuzzles mate some pups are red and some are green. •There is no advantage (for mating or survival) based on color. Which trait is dominant? A) red B) green