Deck 10: Genome Annotation

Question 1

Choose the phrase from the right column that best fits the term in the left column.

Answer 1

a. DNA Polymorphism - 5. A DNA sequence that occurs in two or more variant forms
b. Phase - 3. Arrangement of alleles of two linked genes in a diploid
c. Informative cross - 8. Allows identification of a gamete as recombinant or non-recombinant
d. ASO - 6. A short oligonucleotide that will hybridize to only one allele at a chosen SNP locus
e. SNP - 2. Two different nucleotides appear at the same position in genomic DNA from different individuals
f. DNA fingerprinting - 7. Detection of genotype at a number of unlinked highly polymorphic loci
g. SSR - 1. DNA element composed of short tandemly repeated sequences
h. Locus - 4. Location on a chromosome
i. Compound heterozygote - 10. Individuals with two different mutations in the same gene
j. Exome - 9. All exons in a genome

Question 2

Would you characterize the pattern of inheritance of anonymous DNA polymorphisms as recessive, dominant, incompletely dominant, or codominant

Answer 2

DNA polymorphism: DNA (Deoxyribose Nucleic Acid) polymorphism is defined as the difference in nucleotides among individuals. Base pair changes, deletions, insertions, or number of copies in DNA sequence are the changes that are seen among the individuals.The locus is said to be polymorphic is two or more alleles exist at a DNA locus. The variations are DNA polymorphisms. Polymorphisms that have no effect on the phenotype are called anonymous DNA polymorphisms.
These polymorphisms can have recessive and dominant inheritance. In some cases the inheritance pattern is codominant. The codominant polymorphisms can either be homozygous or heterozygous.

Question 3

Would you be more likely to find SNPs in the protein-coding or in the non-coding DNA of the human genome

Answer 3

Single nucleotide polymorphism (SNP) is the variation that occurs in the single nucleotide at a specific position in the genome. Every SNP represents the different in the single DNA block known as nucleotide. For example, a SNP may change cytosine to thymine in a DNA sequence. The SNPs occur in every 300 nucleotides, which means about 10 million SNPs in human genome. When the SNPs are present in a gene or near the regulatory region, there are chances of role in disease by altering the function of the gene.The frequency of SNPs is not homogenous but the likelihood of their presence is in the non-coding regions more than the coding regions. The presence of SNPs in non-coding regions results in variation when natural selection is acting. If the SNPs are present in the coding regions they are capable of changing the protein thus translated from that gene. For example, the telomeres are example of a non-coding DNA that sit on the ends of the chromosome and protect the chromosome from errors and degradation. So, most of the SNPs are found in the non-coding regions as the presence of SNPs in a coding may result in change in the expression of gene.

Question 4

A recent estimate of the rate of base substitutions at SNP loci is about 1 × 10 8 per nucleotide pair per gamete.
a. Based on this estimate, about how many de novo mutations (that is, mutations not found in the genomes of your parents) are present in your own genome
b. Where and when did these de novo mutations in your genome most likely occur
c. It has been calculated that each sperm made in a 25-year-old man is the result on average of about 300 rounds of cell division, starting with the first mitotic division of the male zygote. In contrast, each mature oocyte found in a 5-month-old female human fetus is the result of about 25 rounds of division, starting with the first mitotic division of the female zygote. What bearing do these calculations have on the estimate of the rate of base substitutions in humans, and on your answer to part (b)

Question 5

If you examine Fig. 10.5 closely, you will note that there are some regions, such as between nucleotides 116,870K and 116,890K, in which James Watson and Craig Venter share the same SNPs, surrounded by regions in which these two men do not share any SNPs. What does this fact say about the relationship between these two men, and how do you think this pattern of shared and unshared SNPs arose
Figure 10.5 SNP distribution in a 400 kb region. This part of chromosome 7 (from base pairs 116,700,001 to 117,100,000) contains CFTR and two other genes. Vertical marks indicate locations at which a genome is either heterozygous or homozygous for a single nucleotide polymorphism (SNP) different from the human RefSeq. Two rows show SNPs that were read from the personal genomes of Watson and Venter. The third track compiles all SNPs from all human genomes analyzed that were deposited in the central SNP database as of 2009.

Question 6

Approximately 50 million SNPs have thus far been recorded after the characterization of thousands of human genomes.
a. About how many base pairs in the human genome are identical in these thousands of people
b. Do you think that many other SNPs exist among the human population If so, why haven't they been found
c. Almost all of the SNP polymorphisms found to date are bi-allelic; that is, among all the genomes in the population studied to date, only two possible alleles can be found (for example, A and C). Provide a rough estimate for the number of tri-allelic SNP loci that could be found in the same group of humans (that is, the number of loci with three different alleles-for example, A, C, and T). At about how many loci would all four possible nucleotides be found among the human genomes studied to date

Question 7

Mutations at simple sequence repeat (SSR) loci occur at a frequency of 1 × 10 3 per locus per gamete, which is much higher than the rate of base substitutions at SNP loci (whose frequency is about 1 × 10 8 per nucleotide pair per gamete).
a. What is the nature of SSR polymorphisms
b. By what mechanism are these SSR polymorphisms likely generated
c. Copy number variants (CNVs) also mutate at a relatively high frequency. Do these mutations occur by the same or a different mechanism than that generating SSRs
d. The SSR mutation rate is much higher than the mutation rate for new SNPs. Why then have geneticists recorded more than 50 million SNP loci but only about 100,000 SSR loci in human genomes

Question 8

Humans and gorillas last shared a common ancestor about 10 million years ago. Humans and chimps last shared a common ancestor about 6 million years ago. The table below shows the corresponding genomic region from two gorilla gametes, three chimpanzee gametes, and three human gametes.
a. Draw a cladogram similar to that in Fig. 10.6b to show the evolutionary relationships among these three species.
b. The data reveal six polymorphisms among these eight genomes, at positions 2 (A or G), 3 (A or T), 4 (G or T), 7 (C or T), 8 (C or T), and 9 (G or T). On your cladogram, indicate approximately when the mutations that produced each of these polymorphisms occurred. For each allele, state whether it is ancestral, derived, or that you can't tell.
c. Infer the sequences in (i) the last common ancestor of humans and chimpanzees, and (ii) the last common ancestor of all three species. Use a question mark ( ) to represent any uncertainty.
Figure 10.6 Inferring the evolutionary history of SNPs.
(b) Cladogram (diagram of evolutionary lineages). Locus 1 ( light blue ) differs between chimps and humans, but all humans have the same allele (G).The mutation causing the locus 1 difference must have occurred since the species diverged, either in the lineage leading to chimps or that leading to humans; the allele at this position in the most recent common ancestor of the two species cannot be determined. At locus 2 ( orange ) , the C allele shared between chimps and some humans must be ancestral, while the T allele in other humans must be derived (that is, caused by a recent mutation specifically in the lineage of some humans).

Question 9

Using PCR, you want to amplify a ~1 kb exon of the human autosomal gene encoding the enzyme phenylalanine hydroxylase from the genomic DNA of a patient suffering from the autosomal recessive condi­tion phenylketonuria (PKU).
a. Why might you wish to perform this PCR amplifi­cation in the first place, given that the sequence of the human genome has already been determined
b. Calculate the number of template molecules that are present if you set up a PCR reaction using 1 nanogram (1 × 10 grams) of chromosomal DNA as the template. Assume that each haploid genome contains only a single gene for phenylalanine hydroxylase and that the molecular weight of a base pair is 660 grams per mole. The human genome contains 3 × 10
c. Calculate the number of PCR product molecules you would obtain if you perform 25 PCR cycles and the yield from each cycle is exactly twice that of the previous cycle. What would be the mass of these PCR products taken together

Question 10

Which of the following set(s) of primers could you use to amplify the target DNA sequence below, which is part of the last protein-coding exon of the CFTR gene
a.
b.
c.
d.

Question 11

Problem 10 above raises several interesting questions about the design of PCR primers.
a. How can you be sure that the two 18-nucleotide-long primers you chose as your answer to Problem 10 will amplify only an exon of the CFTR gene, but no other region, from a sample of human genomic DNA
b. Primers used in PCR are generally at least 16 nucleotides long. Why do you think the lower limit would be approximately 16
c. Suppose one of the primers in your answer to Problem 10 had a mismatch with a single base in the genomic DNA of a particular individual. Would you be more likely to obtain a PCR product from this genomic DNA if the mismatch was at the 5 end or at the 3 end of the primer Why
Problem 10
Which of the following set(s) of primers a-d could you use to amplify the target DNA sequence below, which is part of the last protein-coding exon of the CFTR gene
a. 5 GGAAAATTCAGATCTTAG 3 ;
5 TGGGCAATAATGTAGCGC 3
b. 5 GCTAAGATCTGAATTTTC 3 ;
3 ACCCGTTATTACATCGCG 5
c. 3 GATTCTAGACTTAAAGGC 5 ;
3 ACCCGTTATTACATCGCG 5
d. 5 GCTAAGATCTGAATTTTC 3 ;
5 TGGGCAATAATGTAGCGC 3

Question 12

You sequence a PCR product amplified from a person's genome, and you see a double peak such as that seen in Fig. 10.11b. Most of the time, this result indicates that the person is a heterozygote for a SNP at that position. But it is also possible that the result is due to a mistake in DNA replication during the PCR amplification, with DNA polymerase misincorporating the wrong nucleotide.
a. If you saw a double peak in the sequence trace, did the mistake happen in the first few rounds of PCR amplification or in the last few rounds
b. Whether or not you see a double peak, is it more likely that a mistake would happen in the first few rounds of PCR amplification or in the last few rounds
c. Given that mistakes can happen during PCR amplification, what could you do to ever be sure of a person's genotype Why would this degree of certainty be difficult to achieve if you were doing preimplantation genotyping of embryos
d. PCR relies on heat-stable DNA polymerases from thermophilic bacteria that grow in hot springs. The DNA polymerase originally used for PCR, from the bacterium Thermus aquaticus, lacks the 3 -to-5 exonuclease found in other DNA polymerases such as that from E. coli (review Fig. 7.9). Why do scientists now most often use DNA polymerase from a different thermophilic bacterium ( Pyrococcus furiosa ) that does contain this exonuclease function
Figure 10.11 Detection of the sickle-cell mutation by sequencing of PCR products. (b) Sequencing of PCR products made using genomic DNA templates from individuals with normal and mutant alleles of this SNP locus. Note that the sequence of the PCR product from a heterozygous carrier shows both the normal ( Hb A ) and mutant ( Hb s ) nucleotides at the position of the substitution.
(b) Genotyping for sickle-cell anemia
Figure 7.9 DNA polymerase's proofreading function. If DNA polymerase mistakenly adds an incorrect nucleotide at the 3 end of the strand it is synthesizing, the enzyme's 3 -to-5 exonuclease activity removes this nucleotide, giving the enzyme a second chance to add the correct nucleotide.

Question 13

Problem 8 in this chapter showed three different sequences of the same autosome in human populations. These sequences are each from a single chromatid. You know this to be true because the PCR amplifications were from individual haploid gametes. If you wanted to obtain the same information by PCR amplification of genomic DNA from somatic cells, the problem would be somewhat more complicated because the starting cells are diploid. Each PCR product to be sequenced would thus actually be amplified from two homologous chromosomes. You could still verify the existence of the three different haploid sequences shown in Problem 8 by analyzing the somatic genomic DNA from as few as two people (if they happened to be the right people).
a. Indicate the diploid genotypes of two people from whom you could identify these three different haploid sequences. Account for all 10 nucleotides in the sequences. There are three possible correct answers; you only need to show one.
b. If you PCR amplified DNA from somatic genomic DNA from a person with one particular genotype, you would not be able to conclude that their genome contains any of these three sequences. What is the genotype of this person Explain why you could not reach this conclusion.
Problem 8
Humans and gorillas last shared a common ancestor about 10 million years ago. Humans and chimps last shared a common ancestor about 6 million years ago. The table below shows the corresponding genomic region from two gorilla gametes, three chimpanzee gametes, and three human gametes.
a. Draw a cladogram similar to that in Fig. 10.6b to show the evolutionary relationships among these three species.
b. The data reveal six polymorphisms among these eight genomes, at positions 2 (A or G), 3 (A or T), 4 (G or T), 7 (C or T), 8 (C or T), and 9 (G or T). On your cladogram, indicate approximately when the mutations that produced each of these polymorphisms occurred. For each allele, state whether it is ancestral, derived, or that you can't tell.
c. Infer the sequences in (i) the last common ancestor of humans and chimpanzees, and (ii) the last common ancestor of all three species. Use a question mark ( ) to represent any uncertainty.
Figure 10.6 Inferring the evolutionary history of SNPs.
(b) Cladogram (diagram of evolutionary lineages). Locus 1 ( light blue ) differs between chimps and humans, but all humans have the same allele (G).The mutation causing the locus 1 difference must have occurred since the species diverged, either in the lineage leading to chimps or that leading to humans; the allele at this position in the most recent common ancestor of the two species cannot be determined. At locus 2 ( orange ) , the C allele shared between chimps and some humans must be ancestral, while the T allele in other humans must be derived (that is, caused by a recent mutation specifically in the lineage of some humans).

Question 14

The trinucleotide repeat region of the Huntington disease ( HD ) locus in six individuals is amplified by PCR and analyzed by gel electrophoresis as shown in the following figure; the numbers to the right indicate the sizes of the PCR products in bp. Each person whose DNA was analyzed has one affected parent.
a. Which individuals are most likely to be affected by Huntington disease, and in which of these people is the onset of the disease likely to be earliest
b. Which individuals are least likely to be affected by the disease
c. Consider the two PCR primers used to amplify the trinucleotide repeat region. If the 5'-end of one of these primers is located 70 nucleotides upstream of the first CAG repeat, what is the maximum distance downstream of the last CAG repeat at which the 5'-end of the other primer could be found

Question 15

Sperm samples were taken from two men just beginning to show the effects of Huntington disease. Individual sperm from these samples were analyzed by PCR for the length of the trinucleotide repeat region in the HD gene. In the graphs that follow, the horizontal axes represent the number of CAG repeats in each sperm, and the vertical axes represent the fraction of total sperm of a particular size. The first graph shows the results for a man whose mutant HD allele (as measured in somatic cells) contained 62 CAG repeats; the man whose sperm were analyzed in the second graph had a mutant HD allele with 48 repeats.
a. What is the approximate CAG repeat number in the HD + alleles from both patients
b. Assuming that these results indicate a trend, what can you conclude about the processes that give rise to mutant HD alleles In what kinds of cells do these processes take place
c. How do these results explain why approximately 5-10% of Huntington disease patients have no family history of this condition
d. Predict the results if you performed this same PCR analysis on single blood cells from each of these patients instead of single sperm.

Question 16

In 1993, the courts for the first time accepted plant DNA as evidence in a murder trial. The accused defendant owned a pickup truck in which the police found a few seed pods from a Palo Verde, the state tree of Arizona. The murdered woman was found abandoned in the Arizona desert. How could a prosecutor use DNA from the seed pods to build a strong case against the defendant

Question 17

a. It is possible to perform DNA fingerprinting with SNPs instead of SSRs as DNA markers, but in general you would need to examine more SNP markers than the 13 SSRs used in the CODIS database to be sure of a match. Explain why.
b. DNA fingerprinting has been used to verify pedigrees of valuable animals such as show dogs, racing greyhounds, and thoroughbred horses. However, the technology is much harder to apply in these cases than it is in forensic applications for humans. In particular, many more DNA markers must be examined in domesticated animals to establish the identity or close familial relationship of two DNA samples. Why would you need to look at more polymorphic loci in these animals than you would in humans

Question 18

On July 17, 1918, Tsar Nicholas II; his wife the Tsarina Alix; their daughters Olga, Tatiana, Maria, and Anastasia; their son, the Tsarevitch (Crown Prince) Alexei; and four loyal retainers were murdered by Bolshevik revolutionaries. The bodies were not recovered for many years, fueling legends that Grand Duchess Anastasia had escaped, and allowing a woman named Anna Anderson to claim that she was Anastasia. In 1991 and in 2007, two mass graves with a total of nine skeletal remains were unearthed at Ekaterinburg in Russia's Ural Mountains. The table on the next page presents partial DNA fingerprint analysis (using only five SSR loci and the sex chromosome marker Amel) of these skeletons. Entries separated by commas indicate alleles (number of repeating units).
a. What is the most likely identification for each skeleton [ Note: You cannot differentiate among any of the daughters based on this information alone.]
b. Three PCR reactions failed to yield PCR products. If the reactions had worked properly, what alleles would you expect to see in each case
c. Are any of the daughters identical twins
d. What kind of evidence could you obtain from the skeletons to differentiate among the daughters
e. How do these DNA fingerprints repudiate the claims of Anna Anderson
The DNA fingerprint data in the table is certainly consistent with the idea that some of these skeletons were members of the Tsars family, but they do not prove the hypothesis. To investigate further, forensic scientists obtained blood samples from Prince Philip (the consort of Queen Elizabeth II of Great Britain) and compared his DNA fingerprint to those obtained from the skeletal remains in Russia. A family genealogy is provided below. The results validated that the Tsar's family was indeed interred in these graves.
f. For autosomal DNA markers, what percentage of alleles in the Tsarina's skeleton should match with alleles in Prince Philips genome
g. For autosomal DNA markers, what percentage of alleles in the Tsarevitch's skeleton should match with alleles in Prince Philips genome
h. A question for genealogy aficionados : What is Prince Philips relationship with the Tsarina

Question 19

The figure below shows DNA fingerprint analysis of the genomic DNA from semen associated with a rape (***) and from mouth swabs (somatic cells) of individuals 1-4. This analysis involves the PCR amplification of six SSR loci, each from a different (nonhomologous) chromosome. All PCR primers used are 20 nucleotides long; the primers for each locus have fluorescent tags in a locus-specific color. In the gel, some bands are thicker because relatively more of the corresponding PCR product was obtained. The figure has dots aligned on both sides that you can use to find the critical bands, using the edge of a piece of paper as a guide.
a. Sperm are haploid, but the semen sample shows two different-sized PCR products for certain loci. How is this possible
b. Is any locus on the X chromosome If so, identify it.
c. Is any locus on the Y chromosome If so, which one
d. Explain why these results demonstrate that none of the four individuals is the rapist. What pattern would you expect by analyzing mouth swab DNA from the rapist
e. Do these results nonetheless provide any information that could help catch the rapist If so, be as specific as possible.
f. The two orange bands amplified by PCR from the semen are 200 and 212 bp long. How many tandem repeats of the microsatellite repeating unit are found in the two alleles of this locus in the rapists genomic DNA (Assume that the PCR products are the shortest possible and that the repeating unit for this locus is TCCG.)

Question 20

Microarrays were used to determine the genotypes of seven embryos (made by in vitro fertilization) with regard to sickle-cell anemia. Each pair of squares in the figure below represents two ASOs, one specific for the Hb A allele (A) and the other for the HB S allele (T), attached to a chip of silicon and hybridized with fluorescently labeled PCR product from a single cell from one of the embryos. The hybridizations were performed at three different temperatures (80°C, 60°C, and 40°C) as shown.
a. Why do you think the PCR step is needed for this microarray analysis
b. Make a sketch of the location in genomic DNA of the PCR primers relative to the sickle-cell mutation. Indicate the 5 -to-3 polarities of all DNA molecules involved.
c. Why is no hybridization seen at 80°C
d. Why do you see strong hybridization of all genomic DNA probes to both ASOs at 40°C
e. What are the genotypes of the seven embryos Which of these embryos would you choose to implant into the mother s uterus so as to avoid the possibility that the child would have sickle-cell anemia

Question 21

A partial sequence of the wild-type Hb A allele is shown here (the top strand is the RNA-like coding strand, and the location of the disease-causing mutation is underlined):
and the sickle-cell allele HB s sequence is:
Design two 21-nucleotide-long ASOs that could be attached to a silicon chip for the microarray analysis performed in Problem 20. Two possibilities exist for each ASO; you only need to show one possibility for each.
Problem 20
Microarrays were used to determine the genotypes of seven embryos (made by in vitro fertilization) with regard to sickle-cell anemia. Each pair of squares in the figure below represents two ASOs, one specific for the Hb A allele (A) and the other for the HB S allele (T), attached to a chip of silicon and hybridized with fluorescently labeled PCR product from a single cell from one of the embryos. The hybridizations were performed at three different temperatures (80°C, 60°C, and 40°C) as shown.
a. Why do you think the PCR step is needed for this microarray analysis
b.Make a sketch of the location in genomic DNA of the PCR primers relative to the sickle-cell mutation. Indicate the 5 -to-3 polarities of all DNA molecules involved.
c. Why is no hybridization seen at 80°C
d. Why do you see strong hybridization of all genomic DNA probes to both ASOs at 40°C
e. What are the genotypes of the seven embryos Which of these embryos would you choose to implant into the mother s uterus so as to avoid the possibility that the child would have sickle-cell anemia

Question 22

a. In Fig. 10.17b, PCR is performed to amplify genomic DNA for genotyping on microarrays. This PCR reaction requires only a single primer, but normally, PCR requires two primers. Why does only a single primer suffice in this case
b. Again in Fig. 10.17b, genomic DNA was cut with a restriction enzyme before its PCR amplification. What kind of restriction enzyme would be most effective for this purpose: Would it create sticky ends or blunt ends Would it recognize a site made of 4 bp, 6 bp, or 8 bp
Figure 10.17 DNA microarrays. (b) Method to amplify genomic DNA for microarray analysis. Genomic DNA is cut with a restriction enzyme (RE), and the ends produced are ligated to a double-stranded oligonucleotide adapter. PCR then amplifies all genomic fragments using a single primer that hybridizes to part of the adapter. The resultant DNA fragments are denatured and fluorescently tagged ( red ).

Question 23

The figure at the top of the next page shows a partial microarray analysis for members of a nuclear family. The eight SNP loci examined are evenly spaced at about 10 Mb intervals on chromosome 4, and they are shown on the microarray in their actual order on this chromosome. For the time being, focus your attention only on the two parents and ignore whether they are affected or unaffected.
a. Write out the complete genotype for all the DNA markers in both parents.
b. The microarray data indicate that one SNP locus has three alleles in this family. Which one
c. How would you know that these loci are in fact on chromosome 4 and are about 10 Mb apart
d. About what percentage of the total length of chromosome 4 is present in the region between DNA markers 1 and 8 (Chromosome 4 is 191 Mb long; it is the fourth largest in the human genome.)

Question 24

The microarray shown at the top of this page analyzes genomic DNA from a nuclear family in which the father, one son, and one daughter have rare, late onset polycystic kidney disease; while the mother, the second son, and the second daughter are unaffected. As stated in the previous problem, the eight SNP loci examined are evenly spaced at about 10 Mb intervals on chromosome 4, and they are shown on the microarray in their actual order on this chromosome.
a. Is the allele responsible for the disease dominant or recessive with respect to wild type Is the disease gene autosomal or X-linked
b. For each of the four siblings, indicate the genotype of the sperm from which they were created. For each of these sperm, write the alleles for each of the eight loci on chromosome 4 in order.
c. Identify the two SNP loci that are uninformative in this family (that is, you cannot determine whether or not either of these loci is linked to the disease gene).
d. Assuming for the sake of simplicity that the four children shown would be completely representative even if the parents had 100 children, the data in the figure indicate that one locus is unlinked to the disease gene. Which one
e. The microarray results indicate that during meiosis in the father, two different recombination events occurred in the region including the disease gene and the SNP loci that are genetically linked to it. Draw a map of chromosome 4 showing the locations of the disease gene, the linked SNP loci, and the two recombination events. Your map should indicate any uncertainties in these positions.
f. Diagram the location and arrangement (phase) of all alleles of all genes that are on the two chromosomes in the father s diploid genome.

Question 25

The figure below shows the pedigree of a family in which a completely penetrant, autosomal dominant disease is transmitted through three generations, together with microarray analysis of each individual for a biallelic SNP locus (the alleles are C and T).
a. Do the data suggest the existence of genetic linkage between the SNP locus and the disease locus If so, what is the estimated genetic distance between the two loci
b. Calculate the maximum Lod score for linkage between the SNP and the disease locus for this pedigree. What does this value of the Lod score signify

Question 26

One of the difficulties faced by human geneticists is that matings are not performed with a scientific goal in mind, so pedigrees may not always provide desired information. As an example, consider the following matings (W, X, Y, and Z):
a. Which of these matings are informative and which noninformative for testing the linkage between anonymous loci A and B (A1 and A2 are different alleles of locus A, B1 and B2 are different alleles of locus B, etc.) Explain your answer for each mating.
b. Is locus A more likely to be a SNP or an SSR What about locus B Explain.

Question 27

Now consider a mating between consanguineous people involving a recessive genetic disease. The figures below show the genotype of these two people at four SNP loci (1-4).
a. For which of these loci is it possible to obtain information about the linkage of the SNP to the disease gene Explain your answer for each locus and describe any special conditions that may apply.
b. For any of the loci for which the mating is potentially informative, how would you tell whether the child is the product of a recombinant or nonrecombinant gamete (That is, how could you solve the phase problem ) Be as specific as possible.

Question 28

The pedigree shown in Fig. 10.22 was critical to the identification of the Huntington disease gene HD, which is located on chromosome 4.
a. The data show that the DNA marker G8 is clearly linked to HD. For the large majority of the people in the pedigree with Huntington disease, which allele of G8 (A, B, C, or D) did they inherit, along with the dominant disease-causing allele of HD, on the copy of chromosome 4 from their affected parent
b. How many people in the pedigree can you categorize absolutely as the product of parental or recombinant gametes from their affected parent, without making any assumptions at all (including the assumption of linkage)
c. If you now make the assumption that G8 and HD are linked, how many of the people in this pedigree must be the product of a recombinant gamete from their affected parent
d. Based solely on the data from this pedigree, what would be the best estimate for the map distance between G8 and HD
e. Considering your answers to parts (b) through (d), calculate the maximum Lod score. The pedigree contains 46 people resulting from informative matings. (Note that 0° = 1.) What does this Lod score signify
Figure 10.22 A marker closely linked to the Huntington disease locus. Detection of linkage between the DNA marker G8 and a locus responsible for Huntington disease (HD) was the first step in the cloning of the HD gene. The pedigree shows an extended Venezuelan family affected by HD. Alleles at the G8 marker locus are indicated (A, B, C, and D). Cotransmission of marker alleles with the mutant and wild-type alleles at the HD locus is obvious.

Question 29

You have identified a SNP marker that in one large family shows no recombination with the locus causing a rare hereditary autosomal dominant disease. Furthermore, you discover that all afflicted individuals in the family have a G base at this SNP on their mutant chromosomes, while all wild-type chromosomes have a T base at this SNP. You would like to think that you have discovered the disease locus and the causative mutation but realize you need to consider other possibilities.
a. What is another possible interpretation of the results
b. How would you go about obtaining additional genetic information that could support or eliminate your hypothesis that the base-pair difference is responsible for the disease

Question 30

Problem show that you can make predictions about a child's genotype by genotyping linked markers even if you don't examine directly the disease-causing mutation. This method can be valuable for diseases showing high allelic heterogeneity if the linkage is extremely tight.
The pedigrees indicated here were obtained with three unrelated families whose members express the same completely penetrant disease caused by a dominant mutation that is linked at a distance of 10 cM from an SSR marker locus with three alleles numbered 1, 2, and 3. The SSR alleles present within each live genotype are indicated below the pedigree symbol. The phenotypes of the newly born labeled individuals-A, B, C, and D-are unknown.
a. What is the probability of disease expression in each of these newborn babies
b. Why would a human geneticist be unlikely to use this SSR marker for diagnosis of the genetic disease

Question 31

Problem show that you can make predictions about a child's genotype by genotyping linked markers even if you don't examine directly the disease-causing mutation. This method can be valuable for diseases showing high allelic heterogeneity if the linkage is extremely tight.
Approximately 3% of the population carries a mutant allele at the CFTR gene responsible for the recessive disease cystic fibrosis. A genetic counselor is examining a family in which both parents are known to be carriers for a CFTR mutation. Their first child was born with the disease, and the parents have come to the counselor to assess whether the new fetus inside the mother's womb is also diseased, is a carrier, or is homozygous wild type at the CF locus. DNA samples from each family member and the fetus are tested by PCR and gel electrophoresis for an SSR marker within one of the CFTR gene's introns. The following results are obtained:
a. What is the probability that the child who will develop from this fetus will exhibit the disease
b. When this child grows up and mates, what is the probability that any one of her children will be afflicted with the disease
c. The cystic fibrosis gene displays extensive allelic heterogeneity: More than 1500 different mutations of the CFTR gene have been shown to be associated with cystic fibrosis worldwide. With this fact in mind, why might human geneticists choose to test the fetus in the indirect manner described in this problem rather than focusing directly on the mutations that actually caused the disease in the first child

Question 32

The drug ivacaftor has recently been developed to treat cystic fibrosis in children with the rare G551D mutant allele of CFTR.
a. Do you think that ivacaftor would be effective only in patients homozygous for the G551D mutation, or might it work as well in compound heterozygotes in which one copy of chromosome 7 had G551D and the other copy a different allele of CFTR, such as the more prevalent allele F508 (The protein encoded by G551D folds up properly and inserts into the cell membrane, but is inefficient in chloride ion transport. Ivacaftor increases the efficiency of G551D's ion transport. The F508 protein does not fold up properly and therefore does not get inserted into the cell membrane.)
b. Why do you think ivacaftor would be more effective in children than in older cystic fibrosis patients
c. The scientists who developed ivacaftor had a model for cystic fibrosis: a line of cells that grow in culture and that are homozygous for G551D. These cells accumulate mucus at their surfaces that prevent cilia (tiny hairs on the outside of cells) from beating. Explain how the scientists could use this disease model to screen for drugs that would be effective against G55ID-associated cystic fibrosis.

Question 33

In the high-throughput DNA sequencing protocol shown in Fig. 10.24:
a. What is the purpose of adding poly-A to fragments of single-stranded genomic DNA, and why is the poly-A added to the 3 end of these fragments
b. Why at the end of every synthesis cycle do you need to remove the fluorescent tag on the incorporated nucleotide
c. Why do the incorporated nucleotides have a blocking group, and why does this blocking group need to be removed each cycle
Figure 10.24 One method for high-throughput, single molecule DNA sequencing. (a) Millions of fragments of single-stranded genomic DNA to which poly-A has been enzymatically added at the 3 end are hybridized to oligo-dT molecules attached to the surface of a special microarray called a flowcell. (b) Using the genomic fragment as template and the oligo-dT as primer, DNA polymerase synthesizes new DNA containing nucleotides with colored, base-specific fluorescent tags. These nucleotides are also blocked at their 3 ends so that only one nucleotide can be added at a time. This chemical block is reversible. (c) After a high-resolution camera photographs the fluorescence, chemicals applied to the flowcell remove the tag and blocking group from the just-added nucleotide. (d) Each subsequent cycle begins by infusing the flowcell with a new dose of tagged nucleotides and polymerase, and is followed by an iteration of step (c).The sequencing machine takes about 100 pictures that record a sequence of colored flashes at each of millions of spots on a flowcell where a single DNA molecule is being synthesized. The machine's computer rearranges the data into millions of short sequence reads of about 100 nucleotides, and then assembles the genome sequence.

Question 34

A researcher sequences the whole exome of a patient suffering from Usher syndrome, a rare autosomal recessive condition that is nonetheless the leading cause for simultaneous deafness and blindness. The exome sequence does not show homozygosity for any polymorphisms different from the human RefSeq.
a. Flow could the researcher examine the data already gathered to try to find the disease gene, assuming the sequence is accurate
b. If the attempt described in part (a) was unsuccessful, the researcher might contemplate sequencing the patient's whole genome. What are the potential pitfalls of this strategy

Question 35

As explained in the text, the cause of many genetic diseases cannot yet be discerned by analyzing whole-exome/genome sequences. But in some of these seemingly intractable cases, important clues can be obtained by looking at mRNAs or proteins, rather than at the DNA.
a. As you will see in more detail in Chapters 15 and 16, it is possible to use single-molecule methods to sequence cheaply cDNA copies of millions of mRNA molecules from any particular tissue. How could you sometimes use such information to find a disease gene When would this information be noninformative
b. A technique called Western blotting allows you to examine any protein for which you have an antibody; it is possible to see differences in size or amount of that protein. How could you sometimes use such information to find a disease gene When would this information be noninformative

Question 36

Figure 10.26 portrayed the analysis of Miller syndrome through the sequencing of four complete genomes: those of a brother and sister both affected by the disease, and of both their parents.
a. Researchers made the assumption that Miller syndrome is a recessive trait. Could Miller syndrome instead be due to a dominant mutation If so, what scenarios would make this possible
b. Why is it highly unlikely that Miller syndrome in this family is due to de novo mutations that occurred in the germ line of the mother, or of the father, or of both parents Describe a scenario based on your understanding of cell divisions in human ovaries or testes (see Figs. 4.18 and 4.19 ) that make the de novo mutation hypothesis at least theoretically possible even if very unlikely.
c. On Fig. 10.26b, indicate the location on chromosome 16 closest to the DHOD gene at which recombination took place during meiosis in one of the parents of the Miller syndrome patients. In which parent did this recombination occur
d. Do the number of crossovers you see in Fig. 10.26b fit previous estimates that in the human genome, 1 centiMorgan corresponds to about 1 Mb Chromosome 16 is about 90 Mb long; chromosome 17 is about 81 Mb long.
e. How could researchers use all the sequence data from this family to estimate the per-nucleotide rate of mutation in humans
Figure 10.26 The first family with completely sequenced genomes. (a) Pedigree for the inheritance of Miller syndrome. (b) Map showing the inheritance of parental alleles from chromosomes 16 and 17 in the affected children. Identical regions are those in which the affected brother and sister shared the same maternally and paternally derived alleles. Nonidentical regions are those in which the siblings share no alleles. In haploidentical maternal regions, the siblings have the same allele from the mother but different alleles from the father. In haploidentical paternal regions, the brother and sister share a common allele from the father but have different alleles from the mother. If Miller syndrome is recessive, the responsible gene should lie in an identical region. This prediction was upheld when mutations in the DHOD gene on chromosome 16 were found to be causative for the disease.
Figure 4.18 In humans, egg formation begins in the fetal ovaries and arrests during the prophase of meiosis I. Fetal ovaries contain about 500,000 primary oocytes arrested in the diplotene substage of meiosis I. If the egg released during a menstrual cycle is fertilized, meiosis is completed. Only one of the three cells produced by meiosis serves as the functional gamete, or ovum.
Figure 4.19 Human sperm form continuously in the testes after puberty. Spermatogonia are located near the exterior of seminiferous tubules in a human testis. Once they divide to produce the primary spermatocytes, the subsequent stages of spermatogenesis-meiotic divisions in the spermatocytes and maturation of spermatids into sperm-occur successively closer to the middle of the tubule. Mature sperm are released into the central lumen of the tubule for ejaculation.

Question 37

A research paper published in the summer of 2012 presented a method to obtain the whole-genome sequence of a fetus without any invasive procedure such as amniocentesis that could very rarely cause miscarriage. This new technique is based on the fact that some fetal cells leak into the mother's bloodstream and then break down, releasing their DNA. Assume that exactly 10% of the DNA fragments in the mother's blood serum come from the fetus, while the remaining 90% of the DNA fragments in the serum come from the mother's genome.
The investigators collected cell-free DNA from a pregnant woman's bloodstream and subjected it to an advanced high-throughput sequencing method. The table at the end of this problem looks at seven unlinked loci; the number of "reads" of particular alleles (identified by Greek letters) are shown. You should assume for the sake of simplicity that all numerical differences are statistically significant (even though actual data is never this clean).
a. Determine whether each locus is autosomal, X-linked, or Y-linked.
b. Describe the diploid genomes of the mother and fetus by using Greek letters for the alleles, or a dash (-) if no Greek letter is appropriate.
c. Is the fetus male or female
d. At an eighth locus, 1500 reads of a single type of sequence were found. Provide a possible explanation for this result, being as specific as possible.

Question 38

Table 10.2 and Fig. 10.27 together portray the search for the mutation causing Nic Volker's severe inflammatory bowel disease. Neither of Nic's parents had the condition, so geneticists narrowed their investigation by focusing on rare variants that showed a recessive pattern and those on the X chromosome.
a. For candidate variants on an autosome, would the researchers have looked only for variants for which Nic is homozygous Explain.
b. Apart from the recessive and X-linked hypotheses, do any other possible explanations exist for Nic's condition
c. The causative mutation was pinpointed by analyzing only Nic's exome, because at the time of these investigations, whole-genome or whole-exome sequencing was too expensive to perform on his parents. How could you determine inexpensively whether or not this mutation occurred de novo in the germ line of one of his parents (that is, during the formation of the particular egg or sperm that produced Nic) Your answer should not involve whole-genome or whole-exome sequencing.
Figure 10.27 A mutation in a conserved amino acid in the XIAP protein. Amino acids 195-211 (numbering from the N terminus) of the XIAP protein, written in the one-letter code. Compared to the XIAP protein encoded by the human RefSeq (second row), Nic Volker's XIAP (first row) has an amino acid substitution at position 203, from cysteine (C) to tyrosine (Y). In all other species examined, cysteine is found at this position; this evolutionary conservation suggests that the mutation in Nic's genome might alter XIAP function.

Question 39

The human RefSeq of the entire first exon of a gene involved in Brugada syndrome (a cardiac disorder characterized by an abnormal electrocardiogram and an increased risk of sudden heart failure) is:
The genomic DNA of four people (1-4), three of whom have the disorder, was subjected to single-molecule sequencing. The following sequences represent all those obtained from each person. Nucleotides different from the RefSeq are underlined.
a. The first exon of the RefSeq copy of this gene includes the start codon. Write as much of the amino acid sequence of the encoded protein as possible, indicating the N-to-C polarity.
b. Are any of these individuals homozygotes If so, which person and what allele
c. Is the inheritance of Brugada syndrome among these individuals dominant or recessive
d. Is Brugada syndrome associated with allelic heterogeneity
e. Are any of these individuals compound heterozygotes
f. Do the data show any evidence for locus heterogeneity
g. Which person has normal heart function
h. For each variant from the RefSeq, describe: (i) what the mutation does to the coding sequence; and (ii) whether the variation is a loss-of-function allele, a gain-of-function allele, or a wild-type allele.
i. For each variant, indicate which of the following terms apply: null, hypomorphic, hypermorphic, nonsense, frameshift, missense, silent, SNP, DIP, SSR, anonymous.
j. Is the function of this gene haploin sufficient Explain.

Question 40

Estimate how many DNA polymorphisms differentiate the genomes of any two people.

Question 41

Explain why most of these DNA polymorphisms are not responsible for the phenotypic differences between people.

Question 42

Differentiate among different classes of DNA variants in terms of their structures, mechanisms of formation, and frequency in genomes.

Question 43

Outline the steps by which the polymerase chain reaction (PCR) amplifies a specific region of a genome.

Question 44

Describe how the sequencing or sizing of PCR products can elucidate genotypes.

Question 45

Explain how PCR can be used to genotype fetuses in utero or embryos prior to implantation.

Question 46

Explain why a relatively small number of SSR loci are sufficient to provide a DNA fingerprint of an individual.

Question 47

Describe how a DNA microarray is constructed and how to genotype millions of loci on this microarray.

Question 48

Describe the process of positional cloning and how it allows mapping of disease-causing mutations.

Question 49

Examine the limitations of pedigree analysis in providing the information needed for positional cloning.

Question 50

Explain how a Lod score is obtained and what information it provides.

Question 51

Discuss the consequences of allelic heterogeneity, compound heterozygosity, and locus heterogeneity.

Question 52

Describe a high-throughput, automated method by which millions of DNA templates may be sequenced simultaneously.

Question 53

Summarize a sequence of investigative steps that can narrow the candidates for a disease-causing variant.

Question 54

Explain how databases that catalog sequence variation in many people can facilitate the diagnosis of genetic diseases.