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Problem 1 Consider the following pedigree for a rare human muscle disease A. What unusual feature distinguishes this pedigree? B. Where in the cell do you think the mutant DNA resides that is responsible for the phenotype

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**Answer Problem 1 A. What unusual feature distinguishes this pedigree?**

The pattern of clearly shows maternal inheritance. However males and females are affected equally. Where in the cell do you think the mutant DNA resides that is responsible for the phenotype Mostly likely, the mutant DNA is mitochondrial.

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Problem 2 The following pedigree shows the recurrence of a rare neurological disease (large black symbols) and spontaneous fetal abortion (small black symbols) in one family. (Slashes mean that the individual is deceased.) Provide and explanation for this pedigree in regard to cytoplasmic segregation of defective mitochondria.

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Answer Problem 2 Provide and explanation for this pedigree in regard to cytoplasmic segregation of defective mitochondria. Each cell has many mitochondria, each with numerous genomes. Cytoplasmic segregation of mitochondria mixtures is routinely found within the same cell. The best explanation for this pedigree is that the mother in generation I experienced a mutation in a single cell that was a progenitor of her egg cells (primordial germ cell). By chance alone, the two males with the disorder in the second generation were from egg cells that had experienced a great deal of cytoplasmic segregation prior to fertilization, whereas the two females received mixture. The spontaneous abortions that occurred for the first woman in generation II were the result of extensive cytoplasmic segregation in her primordial germ cells: aberrant mitochondria were retained. The spontaneous abortions of the second woman in generation II also came from such cells. The normal children of this woman were the result of extensive segregation in the opposite direction: normal mitochondria were retained. The affected children of this woman were from egg cells that had undergone less cytoplasmic segregation by the time of fertilization so that they developed to term but still suffered from the disease.

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Problem 3 A study made in 1958 in the mining town of Ashibetsu in the Hokkaido province of Japan revealed the frequencies of MN blood type genotypes (for individuals and for married couples) shown in the following table to the right A. Is the population in Hardy-Weinberg equilibrium with respect to the MN blood types. B. Is mating random with respect to MN blood types Genotype Number of individuals or couples Individuals LM/LM 406 LM/LN 744 LN/LN 332 Total 1482 Couples LM/LMXLM/LM 58 LM/LMX LM/LN 202 LM/LNX LM/LN 190 LM/LMXLN/LN 88 LM/LNXLN/LN 162 LN/LNXLN/LN 41 741

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Answer Problem 3 A. Is the population in Hardy-Weinberg equilibrium with respect to the MN blood types. If the population is in equilibrium then the Hardy-Weinberg equation is true and p2+2pq+q2=1 First you need to calculate the frequency of the genotypes which are p=fA/A+1/2fA/a=frequency of A q=fa/a+1/2fA/a=frequency of a LM=p=406/1482+1/2(744/1482)=0.52 LN=q=(332+1/2(744))/1482)=0.48 Two be in equilibrium the genotypes should be distributed LM/LM=p2(1482)=0.52x0.52(1482)=401 LM/LN=2pq(1482)=2x0.52X0.48X1482=740 LN/LN=q2(1482)=0.48X0.48X1482=341 This compares well with the data, so the population is in equilibrium Genotype Number of individuals or couples Individuals LM/LM 406 LM/LN 744 LN/LN 332 Total 1482 Couples LM/LMXLM/LM 58 LM/LMX LM/LN 202 LM/LNX LM/LN 190 LM/LMXLN/LN 88 LM/LNXLN/LN 162 LN/LNXLN/LN 41 741

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**Answer Problem 3 B. Is mating random with respect to MN blood types**

If the mating is random with respect to blood type, then the following frequency of mating should occur LM/LMXLM/LM=p2xp2x741=0.524x741=54 LM/LMX LM/LN or LM/LNX LM/LM =(2)xp2x2pqx741=2x0.522X2x0.52X0.48=200 LM/LNX LM/LN=2pqx2pq741=4x0.522x0.482x741=185 LM/LMXLN/LN or LN/LNXLM/LM =(2)p2xq2x741=2x0.52x0.52x0.48x0.48x741=92 LM/LNXLN/LN or LN/LNXLM/LN =(2)2pqxq2x741=2x2x0.52x0.48x0.48x0.48x741=170 LN/LNXLN/LN=q2xq2x741=0.484x741=39 These number compare nicely with the actual data therefore the matings are random Genotype Number of individuals or couples Individuals LM/LM 406 LM/LN 744 LN/LN 332 Total 1482 Couples LM/LMXLM/LM 58 LM/LMX LM/LN 202 LM/LNX LM/LN 190 LM/LMXLN/LN 88 LM/LNXLN/LN 162 LN/LNXLN/LN 41 741

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Problem 4 Colorblindness results from a sex-linked recessive allele. One in every 10 males is colorblind. a. What proportion of all women are colorblind b. How many colorblind men are there for each colorblind woman c. In what proportion of marriages would colorblindness affect half the children of each sex? d. In what proportion of marriages would all children be normal? e. In a population that is not in equilibrium, the frequency of the allele for colorblindness is 0.2 in woman and 0.6 in men. After one generation of random mating, what proportion of the female progeny will be color blind? What proportion of the male progeny? f. What will the allelic frequencies be in the male and in the female progeny in part e?

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**Answer Problem 4 ½ colorblind X Xcb Y X Xcb Y no colorblind children**

a. What proportion of all women are colorblind Assuming population in Hardy-Weinberg equilibrium and that the allelic frequency is the same in both sexes, we can directly calculate the frequency of colorblindness allele as q=0.1 (Because this trait is sex linked, q is equal to the frequency of affected males.) For females to be colorblind they must be homozygous for the allele, so their frequency is q2=0.01 b. How many colorblind men are there for each colorblind woman q/q2=10 to 1=frequency in men/frequency in females c. In what proportion of marriages would colorblindness affect half the children of each sex? For the condition to be true the mothers must be heterozygous and the fathers must be color blind. The frequency of heterozygous woman is 2pq and the frequency of affected males if q. Therefore the frequency of a random marriage would be 2pqxq=2X0.9x0.1x0.1=0.018 In what proportion of marriages would all children be normal? All children will be phenotypically normal only if the mother is homozygous for the noncolorblind allele. The fathers genotype does not matter. Therefore the frequency would be p2=.9x.9=0.81 ½ colorblind Xcb Y X X/Xcb Xcb/Xcb X/Y Xcb/Y Xcb Y X X/Xcb X/Y no colorblind children

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**Answer Problem 4 Fathers genotype 0.4X 0.6Xcb Y Mother 0.8X 0.2Xcb**

e. In a population that is not in equilibrium, the frequency of the allele for colorblindness is 0.2 in woman and 0.6 in men. After one generation of random mating, what proportion of the female progeny will be color blind? What proportion of the male progeny? easiest to draw a diagram showing possiblities So the frequency in females will be 0.12 and the frequency in males will be 0.2 f. What will the allelic frequencies be in the male and in the female progeny in part e? From the analysis of the results in (e), the frequency of the colorblind allele will be 0.2 in males (the same as in the females of the previous generation) and 1/2( )+0.12=0.4 in females 0.8X 0.32X/X 0.48X/Xcb 0.8X/Y 0.2Xcb 0.08X/Xcb 0.12Xcb/Xcb 0.2Xcb/Y

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Problem 5 A woman (II2 in the pedigree) wishes to know the probability that she is a carrier of Duchenne muscular dystrophy. a. What is the probability if she has another affected male child b. What is the probability if she has another unaffected male child c. What is the probability if she has another unaffected female child d. If she has another unaffected male child what is the probability she will have another affected child I II III

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Answer Problem 5 A woman (II2 in the pedigree) wishes to know the probability that she is a carrier of Duchenne muscular dystrophy. This problem is best solved using Bayes’ Theorem It is easiest to make a table I II III Ancestral information Hypothesis 1 II-2 Is a carrier Hypothesis 2 II-4 Is Not a carrier Prior probability 1/2 Conditional probability (½)3=1/8 1 Joint probability 1/8X1/2=1/16 1/2x1 Posterior probability (1/16)/(1/16+1/2)=1/9 1/2/(1/16+1/2) =8/9 Considers children So the probability she is a carrier is 1/9

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**Answer Problem 5 If she has an affected child then she is a carrier =1**

a. What is the probability if she has another affected male child If she has an affected child then she is a carrier =1 I II III

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Answer Problem 5 b. What is the probability if she has another unaffected male child This problem is best solved using Bayes’ Theorem It is easiest to make a table I II III Hypothesis 1 II-2 Is a carrier Hypothesis 2 II-4 Is Not a carrier Prior probability 1/2 Conditional probability (½)4=1/16 1 Joint probability 1/8X1/2=1/32 1/2x1 Posterior probability (1/32)/(1/32+1/2)=1/17 1/2/(1/32+1/2) =16/17 If she has another unaffected male child then the conditional probability will change So the probability she is a carrier is 1/17

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Answer Problem 5 c. What is the probability if she has another unaffected female child This problem is best solved using Bayes’ Theorem It is easiest to make a table I If she has a female child no additional information is gained because the disease is X-linked recessive Hypothesis 1 II-2 Is a carrier Hypothesis 2 II-4 Is Not a carrier Prior probability 1/2 Conditional probability (½)3=1/8 1 Joint probability 1/8X1/2=1/16 1/2x1 Posterior probability (1/16)/(1/16+1/2)=1/9 1/2/(1/16+1/2) =8/9 II III So the probability she is a carrier is 1/9

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Answer Problem 5 d. If she has another unaffected male child what is the probability she will have another affected child This problem is best solved using Bayes’ Theorem It is easiest to make a table I Hypothesis 1 II-2 Is a carrier Hypothesis 2 II-4 Is Not a carrier Prior probability 1/2 Conditional probability (½)4=1/16 1 Joint probability 1/8X1/2=1/32 1/2x1 Posterior probability (1/32)/(1/32+1/2)=1/17 1/2/(1/32+1/2) =16/17 If she has another unaffected male child then the conditional probability will change II III So the probability she is a carrier is 1/17 if she had four unaffected boys. To pass on the recessive allele is ½ and the chance the child is male is 1/2. So the chance she has an affected child is 1/17X1/2X1/2=1/68

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