Genetics Test 3 Book Problems

26-27 (a) From the information provided, derive the most likely mode of inheritance of this disorder. Using the Hardy-Weinberg law, calculate the frequency of the mutant allele in the population and the frequency of heterozygotes, assuming Hardy- Weinberg conditions. (b) What is the most likely explanation for the high frequency of the disorder in the Pennsylvania Amish community and its absence in other Amish communities?

(a) It is most likely inherited recessively, or else it would be more present. q= 0.045 p= 0.955 2pq= 0.085 (b) The founder effect is the reduction in genetic variation that results when a small subset of a large population is used to establish a new colony. As the Amish are their own colony and breed with each other, the likelihood of a recessive allele frequency is much higher.

24-28 The table in this problem summarizes some of the data that have been collected on mutations in the BRCA1 tumor-suppressor gene in families with a high incidence of both early-onset breast cancer and ovarian cancer. (a) Note the coding effect of the mutation found in kindred group 2082. This results from a single base-pair substitution. Draw the normal double-stranded DNA sequence for this codon (with the 5 ′ and 3 ′ ends labeled), and show the sequence of events that generated this mutation, assuming that it resulted from an uncorrected mismatch event during DNA replication. (b) Examine the types of mutations that are listed in the table, and determine if the BRCA1 gene is likely to be a tumor-suppressor gene or an oncogene. (c) Although the mutations listed in the table are clearly deleterious and cause breast cancer in women at very young ages, each of the kindred groups had at least one woman who carried the mutation but lived until age 80 without developing cancer. Name at least two different mechanisms (or variables) that could underlie variation in the expression of a mutant phenotype, and propose an explanation for the incomplete penetrance of this mutation. How do these mechanisms or variables relate to this explanation?

(a) The coding effect produced by the single base-pair substitution is a nonsense mutation, which is the substitution of a single base that leads to the production of a stop codon, which prematurely ends an mRNA sequence. The two codons that code for Gln are CAG and CAA. The stop codons are UAA, UAG, and UGA. The C-G base pair at the 5' end would need to change to a T-A base pair. This could occur by changing the C to a T in the coding strand (CAG(A)) or by changing G to A on the template strand (GTC(T)). This would result in the coding strand producing TAG(A), which would produce ATC(T) as the template strand and UAG(A) when transcribed. The template strand would also transcribe to UAG(A). This would produce a stop codon. (b) As loss of function, or production of stop codon, causes cancer, it is likely a tumor suppressor gene. (c) One mechanism that could underlie variation in the expression of a mutant phenotype is that a woman could carry a gene whose products provide the same function as the BRCA gene. Therefore, having one mutated gene would not matter as much. Another mechanism could be that the woman could have an immune system that recognizes and destroys the pre-cancerous cells. They may also have mutations in the signal transduction genes so that suppression of cell division occurs in the absence of BRCA. These mechanisms would prevent the progression of cancer.

24-25 Mutations in tumor-suppressor genes are associated with many types of cancers. In addition, epigenetic changes (such as DNA methylation) of tumor-suppressor genes are also associated with tumorigenesis. (a) How might hypermethylation of the TP53 gene promoter influence tumorigenesis? (b) Knowing that tumors release free DNA into certain surrounding body fluids through necrosis and apoptosis, outline an experimental protocol for using human blood as a biomarker for cancer and as a method for monitoring the progression of cancer in an individual.

(a) Tumor suppressor genes work to regulate cell checkpoints and initiate apoptosis. If hypermethylated, or down regulated, the p53 gene would not be produced and cells would not be able to assess checkpoints or undergo apoptosis. (b) Tumors release free DNA into the bloodstream. If one has a baseline blood sample/control, you can compare the progression of cancer by monitoring increases or decreases in the levels of cancerous DNA over time. Blood can act as a biomarker as it will have characteristic properties of cancer if affected.

24-9 Define tumor-suppressor genes. Why is a mutated single copy of a tumor-suppressor gene expected to behave as a recessive gene?

A tumor suppressor gene is any gene that acts to regulate cell-cycle checkpoints and initiate apoptosis. Mutated tumor suppressor genes will exhibit cancerous traits. Tumor suppressor genes are recessive, so they require both alleles to be mutated for them to express cancerous traits.

26-12 Under what circumstances might a lethal dominant allele persist in a population?

A lethal dominant allele could persist if it shows incomplete penetrance or has a late age of onset. Lethal dominant alleles are likely to be selected against, so the probability of it having a significant allele frequency is very low.

24-15 How do translocations such as the Philadelphia chromosome contribute to cancer?

A translocation involving exchange of genetic material between chromosome 9 and 22 is responsible for the generation of the "Philadelphia chromosome." Genetic mapping has told us that certain genes were combined in the translocation to form a hybrid oncogene that encodes a protein that has been implicated in the formation of chronic leukemia.

24-8 What is apoptosis, and under what circumstances do cells undergo this process?

Apoptosis is programmed cell death. Cells will undergo this process when they have reached the end of their development or if they start to exhibit cancerous traits. During apoptosis, DNA becomes fragmented, cellular structures are disrupted, and the cells dissolve. Caspases are the enzyme that allow for apoptosis.

26-29 A number of comparisons of nucleotide sequences among hominids and rodents indicate that inbreeding may have occurred more often in hominid than in rodent ancestry. Scientists suggest that an ancient population bottleneck that left approximately 10,000 humans might have caused early humans to have a greater chance of genetic disease. Why would a population bottleneck influence the frequency of genetic disease?

Bottleneck events involve a drastic reduction in a population that leaves only a few surviving individuals. As those individuals now define the gene pool, they will repopulate and create the new gene pool with new allele frequencies.

24-22 Genetic tests that detect mutations in the BRCA1 and BRCA2 tumor-suppressor genes are widely available. These tests reveal a number of mutations in these genes—mutations that have been linked to familial breast cancer. Assume that a young woman in a suspected breast cancer family takes the BRCA1 and BRCA2 genetic tests and receives negative results. That is, she does not test positive for the mutant alleles of BRCA1 or BRCA2. Can she consider herself free of risk for breast cancer?

Cancer is only hereditary 1-2% of the time. Just because a woman does not test positive for mutant BRCA alleles does not meant that she can not acquire these mutations later and still get breast cancer. She will still have the general population risk of 10%. It is also possible that her genetic test was incorrect or incomplete, as some tests are not FDA approved and most do not test for all types of breast cancer mutations.

26-15 Describe how populations with substantial genetic differences can form. What is the role of natural selection?

During speciation, groups will become genetically distinct. Non-reproductively isolated groups can become separated by a physical barrier, which allows for genetic drift, migration, natural selection, and mutation to change allele frequencies and differentiate the two populations until mating is not possible anymore.

24-20 Epigenetics is a relatively new area of genetics with a focus on phenomena that affect gene expression but do not affect DNA sequence. Epigenetic effects are quasi-stable and may be passed to progeny somatic or germ-line cells. What are known causes of epigenetic effects, and how do they relate to cancer?

Epigenetic effects can be caused by DNA methylation or histone modifications, including acetylation and phosphorylation. These modifications can activate or silence entire genes or chromosomes and can influence the activity of heterochromatin. Patterns of methylation and histone modifications are often different between normal and cancerous cells, which is why epigenetic changes are now one of the indicators of cancer.

26-6 Consider rare disorders in a population caused by an autosomal recessive mutation. From the frequencies of the disorder in the population given, calculate the percentage of heterozygous carriers: (a) 0.0064 (b) 0.000081 (c) 0.09 (d) 0.01 (e) 0.10

Given are the allele frequencies of the disorder population, meaning that all given allele frequencies are homozygous recessive (if inherited in an autosomal recessive manner). (a) q= 0.08 p+0.08=1, so p=0.92 2pq= (0.08)(0.92)= 14.72% (b) q= 0.009 p+0.009=1, so p=0.991 2pq= (0.009)(0.991)= 1.78% (c) q= 0.3 p+0.3=1, so p=0.7 2pq= (0.3)(0.7)= 42% (d) q= 0.1 p+0.1=1, so p=0.9 2pq= (0.1)(0.9)= 18% (e) q= 0.32 p+0.32=1, so p=0.68 2pq= (0.32)(0.68)= 43.2%

24-23 Explain the apparent paradox that both hypermethylation and hypomethylation of DNA are often found in the same cancer cell.

Hypomethylation leads to unrestricted cell growth and is a hallmark of cancer. Hypermethylation can also exist and contribute to cancer as it may occur in CpG islands or other regulatory regions, down regulating their important functions. There can be upregulation/hypomethylation of a oncogene and down regulation/hypermethylation of a tumor suppressor gene.

22-19 Should the FDA regulate direct-to-consumer genetic tests, or should these tests be available as a "buyer beware" product?

I think that FDA approval should be considered as it would decrease the spread of misinformation. But, requiring FDA approval would make the technology less readily available to the general public.

24-6 What is the difference between saying that cancer is inherited and saying that the predisposition to cancer is inherited?

If a trait is inherited, there is an assumption that the phenotype will reflect the cancer. If the predisposition to the trait is inherited, this means that a particular phenotype is expressed in families in a consistent pattern. The phenotype may not be expressed, but individuals could be carriers.

26-5 The use of nucleotide sequence data to measure genetic variability is complicated by the fact that the genes of many eukaryotes are complex in organization and contain 5′ and 3′ flanking regions as well as introns. Researchers have compared the nucleotide sequence of two cloned alleles of the gamma-globin gene from a single individual and found a variation of 1 percent. Those differences include 13 substitutions of one nucleotide for another and three short DNA segments that have been inserted in one allele or deleted in the other. None of the changes takes place in the gene's exons (coding regions). Why do you think this is so, and should it change our concept of genetic variation?

Lots of parts of the eukaryotic genome do not code for proteins, such as introns, and can tolerate nucleotide changes. Changes in exons result in phenotypic changes and are subject to selection.

26-2 Write a short essay describing the roles of mutation, migration, and selection in bringing about speciation.

Mutations are the root source of variation, but mutation alone is too slow to result in allele frequency shifts. Genetic drift, or the drastic shift in a population structure due to a natural event, can decrease variation. Selection can increase or decrease variation. Inbreeding decreases variation. Gene flow can increase variety if populations migrate to the community, and can decrease variety if populations migrate from the community. Inbreeding can decrease variety.

19-18 How are mutations in histone acetylation (HAT) genes linked to cancer?

Mutations in HAT genes would prevent the acetylation of genes and would result in a constant closed chromatin configuration, which prevents gene expression. This can result in the silencing of tumor-suppressor genes and the uncontrolled proliferation of the cell.

24-5 How can mutations in noncoding segments of DNA contribute to the development of cancers?

Mutations in non-coding segments can be part of regulatory sequences that can upregulate or downregulate gene expression. Mutations in proto-oncogenes, tumor suppressor genes, and cell cycle genes can also alter the timing and level of expression that leads to cancer.

24-14 Of the two classes of genes associated with cancer, tumor-suppressor genes and oncogenes, mutations in which group can be considered gain-of-function mutations? In which group are the loss-of-function mutations? Explain.

Mutations that produce oncogenes act in a dominant capacity. Proto-oncogenes normally function to promote or maintain cell division, and in the mutant state, they will maintain uncontrolled cell division. This is considered a gain of function. Loss of function is attributed to tumor suppressor genes, as mutations will cause the TS genes to lose their ability to halt the cell cycle or undergo apoptosis. These mutations are normally recessive.

24-18 How do normal cells protect themselves from accumulating mutations in genes that could lead to cancer? How do cancer cells differ from normal cells in these processes?

Normal cells protect themselves from cancerous cells with checkpoints, DNA repair mechanisms, and induced apoptosis. Through mutations, these mechanisms will promote cancer through increased rates of mutation, chromosomal abnormalities, and genomic instability.

24-13 Distinguish between oncogenes and proto-oncogenes. In what ways can proto-oncogenes be converted to oncogenes?

Proto-oncogenes are genes within our body that promote cell growth and division. They encode for transcription factors that stimulate cell expression, signal transduction molecules, and cell cycle regulators. Oncogenes are mutated proto-oncogenes; they contribute to proliferated cell growth and cancer. Oncogenes are dominant as only one allele needs to be mutated for the oncogene to regulate increased proliferation. Proto-oncogenes are converted to oncogenes through point mutations, gene amplification, translocation, repositioning of regulatory sequences, and other mechanisms.

24-21 Radiotherapy (treatment with ionizing radiation) is one of the most effective current cancer treatments. It works by damaging DNA and other cellular components. In which ways could radiotherapy control or cure cancer, and why does radiotherapy often have significant side effects?

Radiotherapy could control or cure cancer as it can disrupt the damaged DNA processes that are causing uncontrolled cell growth. It can have significant side effects in that it can damage healthy or helpful DNA and continue to encourage mutations or harmful effects.

24-17 DNA sequencing has provided data to indicate that cancer cells may contain tens of thousands of somatic mutations, only some of which confer a growth advantage to a cancer cell. How do scientists describe and categorize these recently discovered populations of mutations in cancer cells?

Scientists differentiate between passenger and driver mutations. Mutations that provide a selective growth advantage, and thus promote cancer development, are driver mutations. Mutations that increase in cancer cells overtime due to genomic instability are passenger mutations. Passenger mutations do not alter fitness. Driver mutations can produce passenger mutations.

26-24 Some critics have warned that the use of gene therapy to correct genetic disorders will affect the course of human evolution. Evaluate this criticism in light of what you know about population genetics and evolution, distinguishing between somatic gene therapy and germ-line gene therapy.

Somatic gene therapy allows individuals to live more normal lives than those not receiving therapy. The ability of those individuals to contribute to the gene pool increases the likelihood that less-fit alleles will enter and be maintained in the gene pool. This consequence of therapy is accepted. Germ-line therapy, could lead to limited, isolated, and infrequent removal of an allele from a gene lineage. It is not likely that this would have too much influence on humankind if done small-scale.

26-23 What genetic changes take place during speciation?

Speciation involves the gradual accumulation of genetic variations until reproductive isolation occurs. This can occur slowly or quickly depending on environmental and geographic conditions. Natural selection, genetic drift, migration, chromosomal mutations, natural causes, and reduction of gene flow are the main causes of speciation.

24-11 Part of the Ras protein is associated with the plasma membrane, and part extends into the cytoplasm. How does the Ras protein transmit a signal from outside the cell into the cytoplasm? What happens in cases where the ras gene is mutated?

The Ras protein encodes for signal transduction molecules that activate the transcription of genes that initiate cell division. Mutant Ras proteins are locked in the "on" position and are continually signaling for cell division. The RAS-GDP complex normally gains an ATP molecule and is then able to regulate the MAP kinase cascade. Once the job is completed, the RAS-GTP complex will then shed the phosphate group. When mutated, the RAS-GTP complex will be unable to shed the phosphate, resulting in an unregulated MAP kinase cascade, which will result in increased cell proliferation.

24-10 Describe the steps by which the TP53 gene responds to DNA damage and/or cellular stress to promote cell-cycle arrest and apoptosis. Given that TP53 is a recessive gene and is not located on the X chromosome, why would people who inherit just one mutant copy of a recessive tumor-suppressor gene be at higher risk of developing cancer than those without the recessive gene?

The TP53 gene is one of the most frequently mutated genes in human cancers. It is called the guardian of the genome due to its importance to genomic integrity. The TP53 gene encodes the p53 transcription factor that either represses or stimulates that transcription of more than 50 different genes. The p53 gene is continuously synthesized and degraded. It is normally bound to MDM2. When present MDM2, will tag p53 for degradation and prevent phosphate and acetyl additions that would activate the protein. Chemical damage to DNA, double stranded breaks due to radiation, and UV light can cause MDM2 to dissociate from p53, unmasking its transcription domain. Increases in the levels of activated p53 will stimulate expression of the MDM2 gene and as MDM2 levels increase, p53 will bind to it and decrease activity in a negative feedback loop. p53 controls the p21 protein, which regulates cell cycle checkpoints. It also regulates the BAX protein, which increases in response to cell damage, and therefore Bcl2 levels, which decrease in response to triggered apoptosis. If those pathways are disrupted by mutation, damaged DNA will not be fixed, apoptosis will not occur, and more mutations will occur. The majority of p53 mutations cause loss of function, but gain of function mutations can occur and will result in increased transcription levels of proteins whose products involve chromatin modifications. This will lead to genome wide changes in histone modifications and therefore altered gene expression. A person with one inherited mutated copy of TP53 will be at higher risk as less mutations will be required for their cells to become cancerous.

24-29 Researchers have identified some tumors that have no recurrent mutations or deletions in known oncogenes or tumor-suppressor genes and no detectable epigenetic alterations. However, these tumors often have large chromosomal deletions. What are some possible explanations that could account for the genetic causes behind these tumors?

The chromosomal deletions could have contained tumor suppressor genes. Without those genes, growth would be unregulated. Another theory is that the regulatory mechanisms for proto-oncogenes could have been deleted, resulting in controlled cell growth.

26-4 The genetic difference between two Drosophila species, D. heteroneura and D. silvestris, as measured by nucleotide diversity, is about 1.8 percent. The difference between chimpanzees (Pan troglodytes) and humans (H. sapiens) is about the same, yet the latter species is classified in a different genera. In your opinion, is this valid? Explain why.

The classification of organisms into different species is based on evidence that the two organisms are reproductively isolated. There must be evidence that gene flow does not occur between the groups for them to be called different species. It is fair that if two organisms cannot produce gene flow, they are to be considered two different species. Categorizations above species do not have genetic basis.

26-14 One of the first Mendelian traits identified in humans was a dominant condition known as brachydactyly. This gene causes an abnormal shortening of the fingers or toes (or both). At the time, some researchers thought that the dominant trait would spread until 75 percent of the population would be affected (because the phenotypic ratio of dominant to recessive is 3 : 1). Show that the reasoning was incorrect.

The distribution of a gene among individuals is determined by mating (population size, inbreeding, etc.) and environmental factors (selection, etc.). Rare alleles are still rare, as their frequency is determined by fitness the allele confers, the mutation rate, and input from migration. They will remain rare, even if they are expressed dominantly.

24-2 Write a short essay describing how epigenetic changes in cancer cells contribute to the development of cancers.

The epigenetic state of normal cells is altered in cancerous cells. They can contain high levels of methylation in CpG islands and other important genes. They can also undergo hypomethylation, which would lead to unrestricted transcription in certain areas. Hypomethylation is a hallmark of cancer cells. Cancer cells may also show disrupted histone modification profiles as mutations in genes encoding HAT and HDAC are linked to the development of cancer. The focus of epigenetic cancer therapy is to reactivate silenced cells.

26-18 What are considered significant factors in maintaining the surprisingly high levels of genetic variation in natural populations?

The presence of selectively neutral alleles (alleles whose fixation are independent of natural selection) and genetic adaptations to varied environments contribute to genetic variation in natural populations.

26-25 List the barriers that prevent interbreeding, and give an example of each.

There are multiple isolating mechanisms that prevent interbreeding. A geographic barrier could involve the two organisms living in different habitats. A seasonal barrier could involve two organisms having different mating cycles. A behavioral barrier could involve two organisms having different/incompatible behavior before mating. A mechanical barrier could involve differences in reproductive structures. A physiological barrier could involve the gametes failing to survive in alien reproductive tracts.

26-26 What are the two groups of reproductive isolating mechanisms? Which of these is regarded as more efficient, and why?

There are prezygotic mechanisms, which prevent fertilization and zygote formation, and postzygotic mechanisms, which involve fertilization taking place but leading to nonviable, weak, or sterile hybrids. Prezygotic mechanisms are more efficient because they occur before resources are expended in the processes of mating. Prezygotic mechanisms include geographic, seasonal, behavioral, mechanical, and physiological barriers. Postzygotic mechanisms include hybrid non-viability or weakness, developmental hybrid sterility (gonads develop abnormally or meiosis cannot be completed), or segregational hybrid sterility (abnormal segregation of chromosomes to gametes).

26-13 Assume that a recessive autosomal disorder occurs in 1 of 10,000 individuals (0.0001) in the general population and that in this population about 2 percent (0.02) of the individuals are carriers for the disorder. Estimate the probability of this disorder occurring in the offspring of a marriage between first cousins. Compare this probability to the population at large.

There is a 1/4 chance that two heterozygous parents will pass the trait to their child. To understand how the parents being first cousins will impact the offspring, calculate the inbreeding coefficient. The inbreeding coefficient for an individual is F = (½)^(n1+n2+1), where n1 and n2 are the numbers of generations separating the individuals in the consanguineous mating from their common ancestor. If a child is inbred through more than one line of descent, the total coefficient of inbreeding is the sum of each of the separate coefficients. If the parents are first cousins, they are separated by two generations. F = (½)^(2+2+1) F= (½)^5 + (½)^5 = 1/16, as first cousins are related through two grandparents and there are two lines of descent The frequency of the recessive allele is 0.01. Multiplying the inbreeding coefficient by the frequency of the recessive allele gives 0.000625%, or 1/1600. This means that the chance of inheriting the trait is almost 6X more likely than in the general population.

24-7 As a genetic counselor, you are asked to assess the risk for a couple with a family history of familial adenomatous polyposis (FAP) who are thinking about having children. Neither the husband nor the wife has colorectal cancer, but the husband has a sister with FAP. What is the probability that this couple will have a child with FAP? Are there any tests that you could recommend to help in this assessment?

To acquire FAP, the individual has to be homozygous recessive for the trait. Assuming both parents are heterozygous, as they do not express FAP, there is a 1/4 chance that the child will have FAP. You could order genetic tests to see if the parents are carriers. You could also order a colonoscopy on the husband to see if he also has polyps and a potential predisposition to the disease.

26-21 In an isolated population of 50 desert bighorn sheep, a mutant recessive allele c when homozygous causes curled coats in both males and females. The normal dominant allele C produces straight coats. A biologist studying these sheep counts four with curled coats. She also takes blood samples from the population for DNA analysis, which reveals that 17 of the sheep are heterozygous carriers of the c allele. What is the inbreeding coefficient F for this population?

To find the expected heterozygosity, calculate 2pq. This is 0.405. Multiply that by the population of sheep to get 20.3. Do the F formula to get 0.162.

26-7 What must be assumed in order to validate the answers in Problem 7?

To validate the answers in problem 9, you must assume Hardy Weinberg equilibrium. This means that you must assume random mating, no genetic drift, infinite population size, no gene flow, no mutation, and no selection.

26-16 Achondroplasia is a dominant trait that causes a characteristic form of dwarfism. In a survey of 50,000 births, five infants with achondroplasia were identified. Three of the affected infants had affected parents, while two had normal parents. Calculate the mutation rate for achondroplasia and express the rate as the number of mutant genes per given number of gametes.

Total allele number is 50,000*2=100,000 The mutation rate is calculated by new alleles/total alleles. As two new mutant alleles were introduced to the population, the mutation rate is 2/100,000.

24-16 Explain why many oncogenic viruses contain genes whose products interact with tumor-suppressor proteins.

Viruses often encode for genes that stimulate growth to encourage infected cells to undergo cell division. Many viruses either inactivate tumor-suppressor genes of the host or bring in other genes that will upregulate growth. By inactivating tumor suppressor genes, oncogenic viruses will work to halt cell checkpoint mechanisms and will not allow the cells to initiate apoptosis.

26-8 In a population where only the total number of individuals with the dominant phenotype is known, how can you calculate the percentage of carriers and homozygous recessives?

You would divide the remaining population by the total population number to get the number of homozygous recessive individuals. This would give you q^2, which you can use to find q and then p.

26-19 A botanist studying water lilies in an isolated pond observed three leaf shapes in the population: round, arrowhead, and scalloped. Marker analysis of DNA from 125 individuals showed the round-leaf plants to be homozygous for allele r1, while the plants with arrowhead leaves were homozygous for a different allele at the same locus, r2. Plants with scalloped leaves showed DNA profiles with both the r1 and r2 alleles. Frequency of the r1 allele was estimated at 0.81. If the botanist counted 20 plants with scalloped leaves in the pond, what is the inbreeding coefficient F for this population?

inbreeding coefficient (F)= (expected heterozygosity - observed heterozygosity)/ expected heterozygosity expected heterozygosity= 2pq= 2 (0.81)(0.19) = 0.2078 0.3078 * 125 = 38.48 F = (38.48 - 20)/38.48 = 0.48

26-22 To increase genetic diversity in the bighorn sheep population described in Problem 21, ten sheep are introduced from a population where the c allele is absent. Assuming that random mating occurs between the original and the introduced sheep, and that the c allele is selectively neutral, what will be the frequency of c in the next generation?

pi' = (1 - m)pi + mpm m = the percentage of the population that are migrants pi = frequency of dominant allele in migrating population pm = frequency of dominant allele in mainland population pi= 1.0, as there are no recessive alleles pm = 0.75, as there are 75% dominant alleles m= 10/60 = 0.17 pi' = (1 - 0.17)(0.75) + (0.17)(1.0) pi' = 0.7925 qi' = 0.2075 So, the frequency of the c allele will have gone from 25% to 20.75%.

26-9 If 4 percent of a population in equilibrium expresses a recessive trait, what is the probability that the offspring of two individuals who do not express the trait will express it?

q= 0.2 p+q=1, so p=0.8 The parents do not have the trait, so they are either AA or Aa. Therefore, the probability that a parent is heterozygous is 0.2*0.8= 0.16. The probability that both parents are heterozygous is 0.16 * 0.16 = 0.0256, or 2.56%.

20-30 A widely used method for calculating the annealing temperature for a primer used in PCR is 5 degrees below the melting temperature, Tm (°C), which is computed by the equation 81.5 + 0.41 * (%GC) - (675/N), where %GC is the percentage of GC nucleotides in the oligonucleotide and N is the length of the oligonucleotide. Notice from the formula that both the GC content and the length of the oligonucleotide are variables. Assuming you have the following oligonucleotide as a primer, 5 ′@TTGAAAATATTTCCCATTGCC@3 ′ compute the annealing temperature for PCR. What is the relationship between Tm (°C) and %GC? Why?

%GC= 7/21= 1/3= 33.33% N=21 Tm= 81.5 + 0.41(1/3) - (675/21) = 63 Increased GC content will increase the annealing temperature as there are three hydrogen bonds within GC bonds compared to the two H-bonds in AT.

21-25 Whole-exome sequencing (WES) is helping physicians diagnose a genetic condition that has defied diagnosis by traditional means. The implication here is that exons in the nuclear genome are sequenced in the hopes that, by comparison with the genomes of nonaffected individuals, a diagnosis might be revealed. (a) What are the strengths and weaknesses of this approach? (b) If you were ordering WES for a patient, would you also include an analysis of the patient's mitochondrial genome?

(a) A strength can be seen in the fact that the information provided by the WGS is heavily correlated to the life cycle of the organism and its environment. The information will also be able to be analyzed for expression levels. Weaknesses can be seen in that incorrect splicing events may be identified as causation for the disease, that there is a lot of information to sift through, and that it is difficult to identify a single issue. (b) WES does not routinely include an analysis of the patients mitochondrial genome as mitochondrial DNA has a high mutation fixation rate and may not contain the data that you are looking for. I don't think it would be harmful to include if if you are trying to be thorough, though.

21-1 (a) How do we know which contigs are part of the same chromosome? (b) How do we know if a genomic DNA sequence contains a protein-coding gene? (c) What evidence supports the concept that humans share substantial sequence similarities and gene functional similarities with model organisms? (d) How can proteomics identify differences between the number of protein-coding genes predicted for a genome and the number of proteins expressed by a genome? (e) How has the concept of a reference genome evolved to encompass a broader understanding of genomic variation in humans? (f) How have microarrays demonstrated that, although all cells of an organism have the same genome, some genes are expressed in almost all cells, whereas other genes show cell- and tissue-specific expression?

(a) Contigs with overlapping end sequences are thought to be a part of the same chromosome. (b) We know that a genomic DNA sequence contains a protein-coding gene if the gene is conserved and if various upstream, downstream, splicing, and punctuation sequences are present and in the proper reading frames. (c) The Human Genome Project found that humans share up to 50% of human genes show similarity to genes in other organisms. Comparative genetics allows for this. Comparative mutation analysis can also indicate similar function between genes of different organisms. (d) Proteomics is the study of the proteome, or entire set of proteins that an organism needs to function. Genome annotation, or the identification of functional elements the in the genome, has allowed for an estimate of the number of protein coding genes. Technologies such as electrophoresis, chromatography, and microarrays allow for the detection of gene and protein expression. ChIP-Seq can then be used to identify where the protein-binding sites are and if they are functional. These techniques allowed researchers to identify the fact that there are many more functional genes produced than there are proteins, which provided the basis for theories on alternative splicing. (e) The reference genome is the most up-to-date template genome that encompasses what we know about human genomics. Comparing it to different human genomes could allow for any abnormalities to arise. (f) Microarrays provide a method for identifying active genes by the hybridization of complementary gene probes to stretches of DNA. Different hybridization patterns indicate that although some genes are expressed in almost all cells, others show cell- and tissue-specific expression. The probes allow for distinction between target DNA and other types of gene expression.

22-24 In 2010, a U.S. District Judge ruled to invalidate Myriad Genetics' patents on the BRCA1 and BRCA2 genes. Judge Sweet noted that since the genes are part of the natural world, they are not patentable. Myriad Genetics also holds patents on the development of a direct-to-consumer test for the BRCA1 and BRCA2 genes. (a) Would you agree with the ruling to invalidate the patenting of the BRCA1 and BRCA2 genes? If you were asked to judge the patenting of the direct-to-consumer test for the BRCA1 and BRCA2 genes, how would you rule? (b) J. Craig Venter has filed a patent application for his "first-ever human-made life form." This patent is designed to cover the genome of M. genitalium. Would your ruling for Venter's "organism" be different from the judge's ruling on patenting of the BRCA1 and BRCA2 genes?

(a) I do agree with this ruling as I do not think it is ethical to make it more difficult to get these genes tested for based on their implications. I would rule that no one should be allowed to hold patents on genes. (b) As Venter's invention does not harm anyone, make information less accessible, or have any equity concerns, I do not think that the same ruling should apply. The creation of a new organism has scientific implications worth exploring.

19-23 Amino acids are classified as positively charged, negatively charged, or electrically neutral. (a) Which category includes lysine? (b) How does this property of lysine allow it to interact with DNA? (c) How does acetylation of lysine affect its interaction with DNA, and how is this related to the activation of gene expression?

(a) Lysine is positively charged. (b) DNA is negatively charged, so its interaction with lysine dramatically alters the charge and shape of the lysine residue by neutralizing its positive charge and increasing its size. (c) Acetylation removes the positive charge on the histones, thereby decreasing the interaction of the N-termini of histones with the negatively charged phosphate groups of DNA. As a consequence, the condensed chromatin is transformed into a more relaxed structure that is associated with greater levels of gene transcription. This relaxation can be reversed by deacetylation catalyzed by HDAC activity.

20-33 (a) Given that STRs are repeats of varying lengths, for example (TCTG)9 - 17 or (TAT)6 - 14, explain how PCR could reveal differences (polymorphisms) among individuals. How could the Department of Justice make use of those differences? (b) Y-STRs from the non-recombining region of the Y chromosome (NRY) have special relevance for forensic purposes. Why? (c) What would be the value of knowing the ethnic population differences for Y-STR polymorphisms? (d) For forensic applications, the probability of a "match" for a crime scene DNA sample and a suspect's DNA often culminates in a guilty or innocent verdict. How is a "match" determined, and what are the uses and limitations of such probabilities?

(a) PCR would allow for the differently polymorphisms among ethnic groups to be amplified. The results could be run through electrophoresis, which would allow for scientists to distinguish between ethnic groups with DNA. (b) This data can be important in forensic evidence as it does not include female DNA, which will make it easier to distinguish the male DNA in a sample. Y-STRs allow for males to be eliminated from the suspect pool in a crime, allow for the identification of paternal lineage of male perpetrators, and can highlight multiple male contributors to a sample. (c) Different ethnicities have different levels of Y-STR polymorphisms. Therefore, knowing the ethnic population differences will allow for a match between a sample and individual to be made easier. (d) A "match" is determined by multiplying the occurrence probabilities of each haplotype to arrive at the overall probability (product) of a genotype occurring in a population. This allows forensic scientists to pinpoint individuals.

22-1 (a) What experimental evidence confirms that we have introduced a useful gene into a transgenic organism and that it performs as we anticipate? (b) How does a positive ASO test for sickle-cell anemia determine that an individual is homozygous recessive for the mutation that causes sickle-cell anemia? (c) From microarray analysis how do we know what genes are being expressed in a specific tissue? (d) How can we correlate the genome with RNA expression data in a tissue or a single cell? (e) From GWAS how do we know which genes are associated with a particular genetic disorder?

(a) Physical evidence that the organism is transgenic could be found with analyses such as PCR, RT-PCR, Southern Blotting, and DNA Sequencing. To determine if the transgenic organism is functional, you could perform a microarray analysis or a direct assay of the gene products. (b) An allele-specific oligonucleotide (ASO) is a short piece of synthetic DNA complementary to the sequence of a variable target DNA. The test involves collecting a single cell from an embryo, amplifying the DNA with PCR, and hybridizing one cell membrane to a mutant probe and one to a wild-type prove. A homozygous recessive individual would hybridize strongly with the mutant probe and not at all with the wild-type probe. (c) Expressed genes will hybridize with the microarray DNA locations and create a readable pattern of red, green, yellow, and black. (d) To correlate the genome with RNA expression data in a tissue or cell, isolate both DNA and RNA in the same cell and compare them. (e) Genome wide association studies compare the genomes of individuals with a disease to those without it, identifying differences between the two that could be conferring risk of developing the disease.

19-19 A developmental disorder in humans called spina bifida is a neural tube defect linked to a maternal diet low in folate during pregnancy. (a) What does this suggest about the cause of spina bifida? (b) Does this exclude genetic mutations as a cause of this condition? (c) Should researchers be looking for mutant alleles of genes that control formation and differentiation of the neural tube?

(a) This indicates that spina bifida is caused by environmental factors, which indicates that it is a result of epigenetic modifications. (b) This does not exclude genetic modifications; a woman could have a higher predisposition to the disease. While taking folic acid may decrease the risk, spina bifida still occurs in children that were provided adequate levels. (c) No, it is unlikely that there will be mutant alleles. They should more likely look for any modifications to the alleles that determine the neural tube.

20-18 To estimate the number of cleavage sites in a particular piece of DNA with a known size, you can apply the formula N/4^n where N is the number of base pairs in the target DNA and n is the number of bases in the recognition sequence of the restriction enzyme. If the recognition sequence for BamHI is GGATCC and the l phage DNA contains approximately 48,500 bp, how many cleavage sites would you expect?

# of cleavage sites = N/4^n N= number of base pairs n= number of bases in sequence # of cleavage sites = 48,500/4^6= 11.8 sites

21-22 Homology can be defined as the presence of common structures because of shared ancestry. Homology can involve genes, proteins, or anatomical structures. As a result of "descent with modification," many homologous structures have adapted different purposes. (a) List three anatomical structures in vertebrates that are homologous but have different functions. (b) Is it likely that homologous proteins from different species have the same or similar functions? Explain. (c) Under what circumstances might one expect proteins of similar function to not share homology? Would you expect such proteins to be homologous at the level of DNA sequences?

(a) The arm of a human, the wing of a bird, and the fin of a whale are all homologous in structure, indicating common ancestry, but have different functions. (b) Homologous proteins may have similar functions but can evolve to have different functions. (c) If proteins evolve by convergent evolution, they will have similar function but will not share homology. Convergent evolution is defined as the process whereby distantly related organisms independently evolve similar traits to adapt to similar necessities. These organisms will have different amino acid sequences and functional groupings.

20-11 In the context of recombinant DNA technology, of what use is a probe?

A probe is used to help identify where a specific gene or protein is located. Southern blot uses nucleic acid probes, northern blot uses RNA or oligodeoxynucleotide probes, and western blot uses primary antibodies as probes. FISH uses fluorescent probes that are injected directly into the chromosome or RNA.

22-3 Why are most recombinant human proteins produced in animal or plant hosts instead of bacterial host cells?

Bacteria do not possess eukaryotic proteins similar to those of plants and animals hosts. Utilizing plant and animal hosts allow for the production of recombinant protein with authentic post-translational modifications. Products commonly made by gene transfer are antibodies, hormones, growth factors, cytokines, and vaccines.

22-10 Does genetic analysis by ASO testing allow for detection of epigenetic changes that may contribute to a genetic disorder? Explain your answer.

No. Epigenetic changes are not reflected in the genotype.

19-12 What is the histone code?

The histone code is the sum of the complex patterns and interactions of histone modifications that change chromatin organization and gene expression.

16-9 What properties demonstrate that the lac repressor is a protein? Describe the evidence that it indeed serves as a repressor within the operon system.

The lac repressor is a protein as it can undergo conformational change with the addition of a substrate. It also only binds to a specific region, is heat liable, and can be isolated. It is noticeably a repressor as once it is bound to the operator region, it represses all transcriptional activity.

19-5 Identical twins each carry the same genome, but over time, can develop different phenotypes. How can you explain this?

They experience different environments that can prompt epigenetic changes.

20-1 In this chapter we focused on how specific DNA sequences can be copied, identified, characterized, and sequenced. At the same time, we found many opportunities to consider the methods and reasoning underlying these techniques. From the explanations given in the chapter, what answers would you propose to the following fundamental questions? (a) In a recombinant DNA cloning experiment, how can we deter- mine whether DNA fragments of interest have been incorporated into plasmids and, once host cells are transformed, which cells contain recombinant DNA? (b) What steps make PCR a chain reaction that can produce mil- lions of copies of a specific DNA molecule in a matter of hours without using host cells? (c) How has DNA-sequencing technology evolved in response to the emerging needs of genome scientists? (d) How can gene knockouts, transgenic animals, and gene editing techniques be used to explore gene function?

(a) We could determine this by determining whether the selectable marker is present. We could also note a potential change in function or any differences between the offspring and parents. (b) There are three main steps to PCR- heating, cooling, and then annealing. The first step involves heating the target DNA to 94 degrees for 0.5-1 minutes in order to separate the two strands. The second step involves cooling the segment at 50-65 degrees for 0.5-1 minute and adding oligonucleotide primer. The third step involves reheating the target DNA to 72 degrees for 0.5-2 minutes and adding Taq polymerase, or another heat-tolerant polymerase. The process requires a DNA template, primers, termo-stable DNA polymerase, and the 4 different dNTPs. (c) It has evolved to not only be quicker, but to be more specific and to fill specific needs. (d) Gene targeting allow for the manipulation of a specific allele, locus, or base sequence in order to learn its function. Gene knockouts are a result of gene targeting and involve an entire gene being taken out of the genome so that the results can be observed. Transgenic animals allow for the ability to see what altering specific genes will do. Different gene editing techniques make it easier for us to a alter genotypes.

19-1 In this chapter, we focused on epigenetic modifications to the genome that regulate gene expression. Several mechanisms are involved, and epigenetic control of gene expression is important in development, cancer, and modulating the genomic response to environmental factors. From the explanations given in the chapter, (a) How do we know how methylation of promoters silences gene expression? (b) What is the evidence that epigenetic changes are involved in cancer? (c) How does an environmental factor like stress generate a response that is transmitted from generation to generation?

(a) We know how methylation of promoters silences gene expression as the methyl groups visibly sit in the major groove and silence genes by blocking the binding of transcription factors. (b) Hypomethylation turns genes on and can lead to high levels of gene expression in oncogenes. Cancer in general is characterized by widespread hypomethylation. Hypermethylation turns genes off and can be seen at CpG islands and in the inactivation of certain genes such as tumor suppressor genes. Epigenetic mechanisms (methylation, acetylation, and non-coding RNA use) has been seen to replace mutations as a way of silencing tumor suppressor genes and in activating oncogenes. (c) Stress is correlated to decreased levels of serotonin, which results from decreases levels of histone acetylation and increases levels of methylation. Glucocorticoid receptor are blocked by methyl groups and serotonin is not produced. This results in a more stressed phenotype. As the epigenome is inheritable, the P generation will pass the stress response onto the F1 generation.

20-19 In a typical PCR reaction, describe what is happening in stages occurring at temperature ranges (a) 92-95°C, (b) 45-65°C, and (c) 65-75°C.

(a) denaturation- the target DNA to be amplified is heated up so that the double strand breaks into two strands and can be read (b) annealing- the primers are added and the segment is cooled (c) extension- a heat-stable polymerase is added so that the DNA can be synthesized

20-23 What is the difference between a knockout animal and a transgenic animal?

A knockout animal is one that was manipulated to have a certain gene in its genome knocked out. Scientists do so to be able to see what happens. To create one, construct a targeting vector that includes the DNA for introduction. The targeting vector will then undergo homologous recombination with the gene of interest to render it nonfunctional. Now that the target vector has a mutated gene of interest, it can be injected into a blastocyst, which is then placed into the uterus of a pseudo-pregnant mouse. The foreign DNA will produce a non-functional protein and will allow scientists to see what happens when the gene of interest is disrupted. A transgenic animal are manipulated in order to express or over express a gene of interest. A vector with the transgene undergoes recombination into the host genome and is then put into embryonic stem cells and injected into the animal. This process allows for the study of effects on appearance and function of a gene in mice.

16-10 Predict the effect on the inducibility of the lac operon of a mutation that disrupts the function of (a) the crp gene, which encodes the CAP protein, and (b) the CAP-binding site within the promoter.

A mutation that disrupts the crp gene would result in CAP not being produced, CAP not binding to cAMP, and the complex not binding to the promoter, which would result in low levels of lactose production. If the CAP binding site on the promoter is disrupted, the cAMP:CAP complex cannot bind to the promoter and there will be low levels of lactose production.

19-11 What are the functions of lncRNAs in epigenetic regulation? Describe each in detail.

All lncRNAs share common properties in that they form RNA-protein complexes with chromatin regulators, deliver those complexes to their designated locations, and participate in chromatin remodeling. lncRNAs will interact with the genome through several mechanisms. They can act as decoys in which lncRNA binding sites will compete with transcription initiation sites to prevent transcription from occurring. They can act as adaptors by serving as a platform for two or more proteins to form active DNA binding protein complexes. They can act as guides in recruiting protein complexes to specific loci. They can also act as enhancers by binding to DNA upstream and forming loops around it in order to regulate gene action and expression.

21-6 Annotation involves identifying genes and gene-regulatory sequences in a genome. List and describe characteristics of a genome that are hallmarks for identifying genes in an unknown sequence. What characteristics would you look for in a bacterial genome? A eukaryotic genome?

Annotating a sequence involves noting differences in a sequence when compared to another. Hallmarks of sequences are regulatory sequences found upstream of genes (promoters), downstream elements (termination sequences), and certain codons. Bacterial genomes are much less complex and do not contain a lot of elements found in eukaryotic genes. Eukaryotic genomes should include enhancers, silencers, and downstream sequences such as the poly-A site.

21-8 BLAST searches and related applications are essential for analyzing gene and protein sequences. Define BLAST, describe basic features of this bioinformatics tool, and give an example of information provided by a BLAST search.

BLAST directs searches through databanks of DNA and protein sequences. A segment of DNA can be compared with any other sequence present in GenBank, which stores all genomic sequences available. BLAST can compute similarity scores between sequences that indicated the level of significance of a match. A limitation of the method is that it can only compare known sequences. One initial approach is to compare a new genomic DNA sequence to the known sequences in the database. An example can be seen in a comparison of a chromosome 12 contig in rats being compared to a portion of chromosome 8 in mice that codes for the insulin receptor gene. A high similarity score indicates that the rat contig sequence contains a gene for the insulin receptor in mice.

21-5 What is bioinformatics, and why is this discipline essential for studying genomes? Provide two examples of bioinformatics applications.

Bioinformatics is the study of the genome using computer-based approaches to organize, share and analyze gate related to gene structure, expression, and protein data. This discipline is essential as there is way too much data involves in genomics to be processed without a computer. One example is a microarray in which thousands of sequences are bound to a slide and fluorescent labeled sampled are added; the analysis of the microarray allows for differing levels of gene expression in the two samples to be made visible. Without a computer to analyze results of thousands of spots on the microarray, the process would not be possible. Another example is ChIP-seq, which combines ChIP and RNA-seq data to provide high resolution sequences of protein binding sites.

16-19 In a theoretical operon, genes A, B, C, and D represent the repressor gene, the promoter sequence, the operator gene, and the structural gene, but not necessarily in the order named. This operon is concerned with the metabolism of a theoretical molecule (tm). From the data provided in the accompanying table, first decide whether the operon is inducible or repressible. Then assign A, B, C, and D to the four parts of the operon. Explain your rationale. (AE = active enzyme; IE = inactive enzyme; NE = no enzyme.)

C= structural gene A= operator B = promoter D= repressor C is the structural gene as when it is mutated, inactive enzymes are produced. Also, when it is introduced to a wild-type F factor, both the inactive and active enzymes are produced. A is the operator as a mutated operator region would not be able to bind the repressor and would therefore always produce the product, even if there is enough present to act as a substrate to the repressor. As the operator is cis-acting, even with the introduction of a wild type operator, the mutant operator will dominate and gene expression will always be active. B is the promoter as a mutated promoter will produce nothing as the RNA polymerase will not be able to bind. If the wild type is introduced by the F factor, there will still be production of the active enzyme in the presence of the metabolite. D is the repressor as a mutated repressor results in constant synthesis without an inducer needed, as the repressor cannot bind to the operator. If the F factor introduces a wild-type genotype to the mutated genotype, the mutated genotype will be dominated over and regular transcription will occur (yes if no lactose, no if lactose present).

21-9 What functional information about a genome can be determined through applications of chromatin immunoprecipitation (ChIP)?

ChIP is a technique that determines whether a protein of interest interacts with a specific DNA sequence. Antibodies are used to isolate a protein of interest once the chromatin has been fragmented. Reverse cross-links will then separate the protein from the DNA and allow computers to identify which DNA sites have protein interactions. ChIP is used to investigate a particular protein-DNA interaction, several protein-DNA interactions, or interactions across the whole genome or a subset of genes.

22-18 The National Institutes of Health created the Genetic Testing Registry (GTR) to increase transparency by publicly sharing information about the utility of their tests, research for the general public, patients, health-care workers, genetic counselors, insurance companies, and others. The Registry is intended to provide better information to patients, but companies involved in genetic testing are not required to participate. Should company participation be mandatory? Why or why not? Explain your answers.

Companies should not be required to participate as that would involve publicly sharing people's private genetic information. It would be helpful if pooled statistical data concerning false positives and negatives as well as population or geographic distribution data was made public. If this occurred, it would be necessary to restrict individual results and name labels.

16-14 Neelaredoxin is a 15-kDa protein that is a gene product common in anaerobic bacteria. It has superoxide-scavenging activity, and it is constitutively expressed. In addition, its expression is not further induced during its exposure to O2 or H2O2. What do the terms constitutively expressed and induced mean in terms of neelaredoxin synthesis?

Constitutively expressed refers to the fact that the gene that codes for the protein is always active and constantly producing, even if inducers are not available. Induced in neelaredoxin synthesis means that the gene expression is not further induced by O2 or H2O2. The gene is not responsive to those molecules.

22-14 Consider ethical issues associated with creating a synthetic human genome. Are there specific applications for a synthetic human genome that you support? Is creating a synthetic genome enhanced with genes for certain kinds of traits one of those applications?

Creating a synthetic human genome could allow for us to "grow" human organs, the production of new vaccines, more effective drugs and novel biofuels. There are real concerns about bioterrorism and inadvertent environmental release, along with the ethical considerations of altering life. I do not think that there is any application that would be ethical as it would not be available to all populations equally.

21-18 What are DNA microarrays? How are they used?

DNA microarrays are sampling techniques that allow researchers to analyze all of a sample's expressed genes simultaneously. Microarrays are prepared by spotting single stranded DNA molecules onto glass slides and using an arrayer to fix the DNA at specific locations, which each contain a unique sequence that can be used as a probe for another gene. The microarray is used by first extracting mRNA from cells or tissues. The mRNA will then be reverse transcribed to synthesize cDNA tagged with fluorescent labels. cDNA from one tissue/cell will be labeled with one color dye (green) and cDNA from another tissue/cell will be labeled with another dye (red). The labeled cDNAs will incubate overnight and hybridize to spots on the microarray that contain complementary sequences. A green spot indicates that the cDNA from only one sample is present. A red spot indicated that the cDNA only from the other sample is present. A yellow spot indicated that both cDNA samples hybridized to the probe. A black spot indicates that nothing hybridizes to the probe in either sample. The intensity of the spot indicates the level to which the gene is expressed. Microarrays not only tell you what genes are expressed, but also by how much they are expressed.

20-34 There are a variety of circumstances under which rapid results using multiple markers in PCR amplifications are highly desired, such as in forensics, pathogen analysis, or detection of genetically modified organisms. In multiplex PCR, multiple sets of primers are used, often with less success than when applied to PCR as individual sets. Numerous studies have been conducted to optimize procedures, but each has described the process as time consuming and often unsuccessful. Considering the information given in Problem 30, why should multiplex PCR be any different than single primer set PCR in terms of dependability and ease of optimization?

Different primers require different annealing temperatures. It is unlikely that multiple primers with different Tm will be effective under a single Tm.

19-2 Write a short essay describing how epigenetic changes in cancer cells contribute to the development and maintenance of cancers.

Epigenetic changes such as DNA hypermethylation, hypomethylation, and gene silencing are present in cancer cells. Hypomethylation turns genes on and will lead to high levels of gene expression in unfavorable oncogenes. Hypomethylation can result in DNA instability and the activation of unnecessary and potentially harmful genes. Hypermethylation of tumor suppressor genes can result in the unfavorable hindered expression of tumor suppressor genes. It can also increase rates of gene mutation and silencing in tumor suppressor genes. Gene silencing is normally triggered by chromatin remodeling and a lack of transcription factors; it can cause important CpG to become methylated and silenced.

19-15 Imprinting disorders do not involve changes in DNA sequence, but only the methylated state of the DNA. Does it seem likely that imprinting disorders could be treated by controlling the maternal environment in some way, perhaps by dietary changes?

Evidence suggests that maternal nutrition does not largely influence global methylation patterns, particularly in nutrient-replete populations; however, an important impact on gene-specific methylation is observed. DNA methylation is a common and significant component of normal gene regulation; it is important to remember that what may be useful for controlling one gene in the genome may have numerous unwanted changes in other parts of the genome.

20-22 How is fluorescent in situ hybridization (FISH) used to produce a spectral karyotype?

FISH involves the hybridization of a fluorescent labeled probe to a complementary stretch of DNA in a chromosome or fragment of RNA. This allows it to locate a specific DNA sequence (gene/gene fragment). This information can be used to create a spectral karyotype, which detects individual chromosomes and can be used to identify abnormalities.

21-3 What is functional genomics? How does it differ from comparative genomics?

Functional genomics is the study of how genes function and how they interact with each other. Gene function is studied in different organisms and comparative genomics allows for the comparison of different organisms. Joint analysis of multiple genomes provides insight into genome function and evolution. The two areas of study allow for the identification of conserved regions, or functional genes that are common between organisms and therefore important.

20-17 Examine the data carefully and choose the best conclusion. (a) None of the offspring are legitimate. (b) Offspring B and C are not the products of these parents and were probably purchased on the illegal market. The data are consistent with offspring A being legitimate. (c) Offspring A and B are products of the parents shown, but C is not and was therefore probably purchased on the illegal market. (d) There are not enough data to draw any conclusions. Additional polymorphic sites should be examined. (e) No conclusion can be drawn because "human" primers were used.

Heavier nucleic acid molecules will not travel as far as the lighter ones. As offspring B and C contain nucleic acid fragments that neither parent could have provided (differing weights), it is likely that these offspring are not legitimate.

19-17 How can the role of epigenetics in cancer be reconciled with the idea that cancer is caused by the accumulation of genetic mutations in tumor-suppressor genes and proto-oncogenes?

Hypomethylation on oncogenes could increase expression of cancer-causing genes. Hypermethylation on tumor suppressor genes could decrease expression of genes that work against cancer. Epigenetic modifications can also change gene expression, which can result in cancer.

22-7 As genetic testing becomes widespread, medical records will contain the results of such testing. Who should have access to this information? Should employers, potential employers, or insurance companies be allowed to have this information? Would you favor or oppose having the government establish and maintain a central database containing the results of individuals' genome scans?

I think that at this point in time, only the individual should have access to their genetic information. I think that, if possible, all genomic data for people could be anonymously placed into a database as to allow for more effective comparative genomics. Making genetic information public could allow for discrimination on the basis of genetic differences that are not even visible; insurance companies could charge more and employees could pass you over.

22-20 Would you have your genome sequenced, if the price was affordable? Why or why not? If you answered yes, would you make your genome sequence publicly available? How might such information be misused?

I would, as I would like to be aware of anything that I would pass onto my kids. It would also benefit me to be aware of any predispositions that I could work towards combatting. I do not think that I would make my information public (tied to my name) as to avoid potential discrimination.

20-24 One complication of making a transgenic animal is that the transgene may integrate at random into the coding region, or the regulatory region, of an endogenous gene. What might be the consequences of such random integrations? How might this complicate genetic analysis of the transgene?

If a transgene integrates itself at random, it will cause a mutation and change the phenotype of the organism in an unexpected way. As two different events are occurring (disruption of one gene and the addition of another), it will be difficult to distinguish which event the phenotype is a result of or if it is a result of both.

20-25 When disrupting a mouse gene by knockout, why is it desirable to breed mice until offspring homozygous (-/-) for the knockout target gene are obtained?

If the mouse is homozygous for the transgene, it will solely express traits that do not include those of the gene that has been knocked out. Heterozygous organisms would still have one copy of the functional gene.

20-7 Restriction sites are palindromic; that is, they read the same in the 5′ to 3′ direction on each strand of DNA. What is the advantage of having restriction sites organized this way?

If they can be read in either direction, this will create single-stranded overhang that allows for the annealing of recombinant molecules. Recognizing a palindromic sequence enables an enzyme to cut both strands of DNA at the "same" site, because the strand will have the same sequence only in different directions at that site. A palindromic sequence also increases the chance that both strands of DNA are cut.

16-4 Contrast the role of the repressor in an inducible system and in a repressible system.

In an inducible system, the binding of the inducer reduces the affinity of the repressor to the operator, which allows for transcription of the operon structural genes to continue. In the presence of the metabolite, the system will be ON. In a repressible system, the pathway is repressed in the presence of its product. The repressor (produced by trpR in tryptophan) will not bind to the promoter if tryptophan is present. Even if the trpR gene is mutated, and the repressor cannot bind to the operator, tryptophan production will still occur as there is an attenuator located after the operator region. In the presence of the metabolite, the system will be OFF.

19-8 Why are changes in nucleosome spacing important in changing gene expression?

In increasing the amount of space in nucleosomes, you are uncoiling the histones. This makes the chromatin available for gene expression. DNA methylation and histone acetylation work together in accomplishing this. When DNA is unmethylated and acetylated, it is open for transcription, When DNA is methylated and de-acetylated, it is closed for transcription.

16-20 A bacterial operon is responsible for the production of the biosynthetic enzymes needed to make the hypothetical amino acid tisophane (tis). The operon is regulated by a separate gene, R. The deletion of R causes the loss of enzyme synthesis. In the wild-type condition, when tis is present, no enzymes are made; in the absence of tis, the enzymes are made. Mutations in the operator gene (O-) result in repression regardless of the presence of tis. Is the operon under positive or negative control? Propose a model for (a) repression of the genes in the presence of tis in wild-type cells and (b) the mutations.

In positive control, an activator binds to the DNA in order for transcription to take place. In negative control, a repressor binds to the operator and blocks transcription until the repressor is removed. This gene product is exerting positive control as the loss of R, which is the activator, causes enzyme synthesis to stop. If R was present and the system was not active, we would know that it was acting as a repressor and that the system was under negative control. (a) When tis is present, no enzymes will be made. When tis is absent, the enzyme will not be made. In order for repression in the presence of tis, the regulatory protein (the activator, or R) must be inactivated and released from the DNA. (b) A mutation of the operon would negate the positive effects of the activator being bound to its DNA binding site. This would result in the less transcription of the operon.

19-22 From the data in Table 19.3, draw up a list of histone H3 modifications associated with gene activation. Then draw up a list of H3 modifications associated with repression. (a) Are there any overlaps on the lists? (b) Are these overlaps explained by different modifications? (c) If not, how can you reconcile these differences?

In the table, the H represents the histone type, the K represents the lysine type, and the S represents the serine type. We can see that lysine 4, 9, 14 , 27, 36, 79, and 20 are associated with activation. We can see that lysine 9, 27, and 79 and serine 10 are associated with repression. (a) Yes, lysine 9, 27, and 79 can cause either activation or repression. (b) Yes, in each case, either activation or repression is triggered by an additional methylation. The only exception is seen in lysine 79, where both activation and repression can occur with the third methylation. (c) This can be reconciled through determining other interactions that lysine 79 may be undergoing. This could be impacting the result of methylation.

16-13 Attenuation of the trp operon was viewed as a relatively inefficient way to achieve genetic regulation when it was first discovered in the 1970s. Since then, however, attenuation has been found to be a relatively common regulatory strategy. Assuming that attenuation is a relatively inefficient way to achieve genetic regulation, what might explain its widespread occurrence?

It has become widespread as it is another means to regulate gene output. In this form, amino acids can control gene expression directly, which has important ramifications in scientific research.

16-21 A marine bacterium is isolated and shown to contain an inducible operon whose genetic products metabolize oil when it is encountered in the environment. Investigation demonstrates that the operon is under positive control and that there is a reg gene whose product interacts with an operator region (o) to regulate the structural genes, designated sg. In an attempt to understand how the operon functions, a constitutive mutant strain and several partial diploid strains were isolated and tested with the results shown in the following table. Draw all possible conclusions about the mutation as well as the nature of regulation of the operon. Is the constitutive mutation in the trans-acting reg element or in the cis-acting o operator element?

It is an inducible system where oil stimulates the gene that metabolizes oil. It is a cis-acting system. grrrrrrrrrrrrrr

21-24 Genomic sequencing has opened the door to numerous studies that help us understand the evolutionary forces shaping the genetic makeup of organisms. Using databases containing the sequences of 25 genomes, scientists examined the relationship between GC content and global amino acid composition. They found that it is possible to identify thermophilic species on the basis of their amino acid composition alone, which suggests that evolution in a hot environment selects for a certain whole organism amino acid composition. In what way might evolution in extreme environments influence genome and amino acid composition? How might evolution in extreme environments influence the interpretation of genome sequence data?

It is expected that there would be increased protein stability in species that live in unfavorable environments. There are multiple factors that could lead to increased protein stability: distribution of ionic interactions on the surface of cells, the density of hydrophobic residues and interactions, and the number of hydrogen and disulfide bonds. A high GC content favors certain amino acids, which likely only allows for certain proteins that are stable in unfavorable environments. Organisms in environments that require extraordinary physical demands will have strong patterns of common decent as individuals were selected for over time.

20-20 We usually think of enzymes as being most active at around 37°C, yet in PCR the DNA polymerase is subjected to multiple exposures of relatively high temperatures and seems to function appropriately at 65-75°C. What is special about the DNA polymerase typically used in PCR?

It is heat resistant, so it will not denature at high temperatures.

16-15 The creation of milk products such as cheeses and yogurts is dependent on the conversion by various anaerobic bacteria, including several Lactobacillus species, of lactose to glucose and galactose, ultimately producing lactic acid. These conversions are dependent on both permease and b-galactosidase as part of the lac operon. After selection for rapid fermentation for the production of yogurt, one Lactobacillus subspecies lost its ability to regulate lac operon expression. Would you consider it likely that in this subspecies the lac operon is on or off? What genetic events would likely contribute to the loss of regulation as described above?

It is likely that the operon is on, as lactose is still being rapidly produced. Loss of gene regulation can result from mutations in several parts of the gene, such as the repressor, operator, or lacI gene.

20-13 In a control experiment, a plasmid containing a HindIII recognition sequence within a kanamycin resistance gene is cut with HindIII, re-ligated, and used to transform E. coli K12 cells. Kanamycin-resistant colonies are selected, and plasmid DNA from these colonies is subjected to electrophoresis. Most of the colonies contain plasmids that produce single bands that migrate at the same rate as the original intact plasmid. A few colonies, however, produce two bands, one of original size and one that migrates much less far down the gel. Diagram the origin of this slow band as a product of ligation.

It is likely that two plasmids were litigated to form a dimer. Dimers are formed in homologous recombination during replication. As dimers are twice the size of the other plasmids, it would not travel as far in electrophoresis.

21-17 Metagenomics studies generate very large amounts of sequence data. Provide examples of genetic insight that can be learned from metagenomics.

Metagenomics uses the WGS to sequence genomes from entire communities of microbes in natural environments in order to teach us more about uncharacterized bacteria and viruses. The study also promises new information about genetic diversity within microbes that is key to understanding interactions between bacteria and their environment. The study also has great potential for identifying new genes. For example, the microbiome of the New York subway was processed and almost half of the DNA sequences did not match any organism on file.

21-16 It can be said that modern biology is experiencing an "omics" revolution. What does this mean? Explain your answer.

Omics is a rapidly evolving, multi-disciplinary, and emerging field that encompasses genomics, epigenomics, transcriptomics, proteomics, and metabolomics. Each of these fields offers the possibility to understand and view biology from a global perspective in a way that was previously unthinkable. Proteomics is the study of proteins in a cell or tissue. Metabolomics is the study of enzymatic pathways. Glycomics is the study of carbohydrates of a cell or tissue. Toxicogenomics is the study of toxic chemicals in the genome. Metagenomics is the study of environmental issues, or the study of a collection of genetic material from different environments. Pharmacogenomics is the study of customized medicine based on the genome. Transcriptomics is the study expressed genes, or the transcriptome specifically.

20-16 To create a cDNA library, cDNA can be inserted into vectors and cloned. In the analysis of cDNA clones, it is often difficult to find clones that are full length—that is, many clones are shorter than the mature mRNA molecules from which they are derived. Why is this so?

One reason may be that the reverse transcriptase may not completely synthesize the DNA from the RNA template. The other reason may be that the 3' end of the copied DNA tends to fold back on itself to make a primer for the DNA polymerase. Additional preparation of the cDNA requires some digestion at the folded region. Since this folded region corresponds to the 5" end of the mRNA, some of the message is often lost.

19-10 How do microRNAs regulate epigenetic mechanisms during development?

Overall, miRNAs allow for rapid changes in multiple parts of the genome. This is a form of post-transcriptional gene expression modification. Once converted from pri-miRNA to miRNA by dicers and RISC, miRNAs will search for miRNA response elements; if there is a perfect match, the mRNA target will be degraded, and if there is a partial match, the mRNA target will be blocked from transcription.

16-3 Contrast positive versus negative control of gene expression.

Positive regulation is the increased expression of genes. It involves transcription factors binding to the promoter region, which is a process enabled by allosteric effector binding. The binding of these transcription factors enables the binding of RNA polymerase to the promoter region and therefore enables gene expression. Negative gene expression involves the decreased expression of genes. A repressor will bind to the operator, preventing the binding of RNA polymerase.

22-17 In 2013 the actress Angelina Jolie elected to have prophylactic double-mastectomy surgery to prevent breast cancer based on a positive test for mutation of the BRCA1 gene. What are some potential positive and negative consequences of this high-profile example of acting on the results of a genetic test?

Potential positive consequences could be that women became more aware of their BRCA status. Negative implications could be that increased numbers of women elected for the surgery not because they needed it, but because someone famous did it.

20-2 Write a short essay or sketch a diagram that provides an overview of how recombinant DNA techniques help geneticists study genes.

Recombinant DNA techniques help geneticists study genes as they provide for different ways to see how adding, getting rid of, or altering a gene can impact function. The use of plasmid technology allows for genes to be inserted into a plasmid and then carried into bacteria via transformation, which normally involves a high intensity shock. Geneticists use DNA libraries, which are collections of genes that have been cloned into vectors so that researchers can identify and isolate the DNA fragments that interest them for further study. PCR is a way to quickly amplify DNA of interest. Southern blot, FISH, western blot, and northern blot techniques allow us to pick out genes or proteins of interest and use to data to understand a variety of chromosomal abnormalities and other genetic mutations.

20-26 What techniques can scientists use to determine if a particular transgene has been integrated into the genome of an organism?

Reporter genes are those introduced into target cells that produce a protein receptor or enzyme that binds a specific injected probe. In placing a reporter gene under the control of the promoter of the gene of interest, it becomes easy to follow how the gene's expression changes in response to different conditions. Surface traits are visible traits. If the gene of interest is expressed or not expressed, it may display differing surface traits. Probes are single stranded sequences that will contain a sequence complementary to the target gene. In qPCR, the probe contains a reporter and quencher dye. Once the two are no longer intact, the reporter dye becomes fluorescent and the gene is effectively labeled. Amplification by PCR will allow for transgenes to be easily distinguished.

20-3 What roles do restriction enzymes, vectors, and host cells play in recombinant DNA studies? What role does DNA ligase perform in a DNA cloning experiment? How does the action of DNA ligase differ from the function of restriction enzymes?

Restriction enzymes make sequence specific cuts to phosphodiester bonds in DNA and produce sticky ends that allow for clean ligation. Vectors are DNA molecules that carry foreign material cells into host bacterial cells. DNA ligase joins sticky ends together, restriction enzymes create sticky ends.

19-7 Describe how reversible chemical changes to DNA and histones are linked to chromatin modification.

Reversible histone modifications influence the structure of chromatin by altering the accessibility of nucleosomes to the transcriptional machinery. Opening and closing genes for transcription allows for variable expression.

19-4 What parts of the genome are reversibly methylated? How does this affect gene expression?

Reversible methylation occurs at CpG rich regions and promoter sequences. Once a gene is imprinted by methylation, it's expression will be more carefully controlled. This can affect gene expression as methylation can be turned off in order to increase expression or on to decrease expression.

22-22 Private companies are offering personal DNA sequencing along with interpretation. What services do they offer? Do you think that these services should be regulated, and if so, in what way? Investigate one such company, 23andMe, at http://www.23andMe .com, before answering these questions.

Services include analysis of your ancestry and health with open access to your data, This data is difficult to interpret and regulation of these companies should be of consumer interest as many consumers do not know the many implications of putting this data out there.

19-6 What are the possible roles of proteins in histone modification?

Some proteins will add chemical groups (methyl, acetyl, phosphate) to histones, some will interpret modifications, and some will remove the added chemical groups. These proteins influence the structure of chromatin by altering the accessibility of the genes and allowing for expression.

22-5 Sequencing the human genome, the development of microarray technology, and personal genomics promise to improve our understanding of normal and abnormal cell behavior. How are these approaches dramatically changing our understanding and treatment of complex diseases such as cancer?

Techniques such as whole-genome and whole-exome sequencing make it possible to identify specific mutations associated with certain conditions, which allows for screening as well as for the more efficient development of treatment plans.

20-9 What are the advantages of using a restriction enzyme whose recognition site is relatively rare? When would you use such enzymes?

The advantage in this is that it will cut less often and that the fragment that is cut out will be larger. You may want to use these enzymes when you are isolating fully intact genes or centromeres.

21-19 Annotation of the human genome sequence reveals a discrepancy between the number of protein-coding genes and the number of predicted proteins actually expressed by the genome. Proteomic analysis indicates that human cells are capable of synthesizing more than 100,000 different proteins and perhaps three times this number. What is the discrepancy, and how can it be reconciled?

The discrepancy is that there are more proteins than there are protein coding genes. This can be explained by alternative splicing, which is a process in which one gene can code for multiple proteins.

19-24 Methylation of H3K9 by itself silences genes, but if H3K4 and H4K20 are also methylated, the combination of modifications stimulates transcription. What conclusions can you draw about this?

The environment of histone modification is as important as the modification itself. There must be an interaction between the three lysine residues that leads to gene activation rather than silencing, as lysine 9 accomplishes by itself.

16-6 For the genotypes and conditions (lactose present or absent) shown in the following table, predict whether functional enzymes, nonfunctional enzymes, or no enzymes are made.

The functional enzyme will be made. The inducer is acting as it should. The operator is constitutive, indicating that the repressor will not bind to the operator and production will be constant. Therefore, it doesn't matter that lactose is present. The Z gene is not mutated, so functional genes will be produced. Nonfunctional enzyme will be made. The I gene is mutated, indicating that it is constitutive and the repressor will not be able to bind to the operator. The repressor will never bind, so it does not matter whether inducer is present and there will be constant production. The Z gene, however, is mutated, so there will not be functional enzyme made. Nonfunctional enzyme will be produced. This example has the same conditions as the last, but lactose is present. As the I gene is still mutated, and the repressor will never bind to the operator, it does not matter if inducer is present. The mutated Z gene results in nonfunctional product. No enzyme would be made. There is a mutated I gene, but as the F factor introduces a wild-type I gene, the wild-type will dominate and the I gene will function normally to produce the repressor. As there is no lactose present, the repressor binds and there will be no transcription. Functional enzyme will be made. The F factor introduces the wild-type lacO gene, but as lacO is cis acting, this will not have an impact on the outcome of the original genotype. The original genotype is Oc, which implies that there is a constitutive mutant and that the operator will not be able to bind the repressor. This means that in the presence of lactose, lactose will bind to the repressor and will not bind to the DNA. Functional enzymes will be produced. Functional and nonfunctional enzyme will be made. The original genotype has a mutant lacZ gene, but as the F factor introduces a wild-type Z gene. The Z gene is cis-acting, so the original genotype will produce nonfunctional enzymes and the F factor genotype will product functional enzymes. No enzyme will be made. The original genotype contains mutant lacI, but as the F factor contains wild-type lacI, the I gene will act as it should as it is trans-acting. As there is no lactose present, the repressor will bind to the DNA and will repress transcription. Neither the normal or mutant lacZ genes will be transcribed. No enzyme will be made. There is a mutation to the lacI gene that results in lactose not being able to bind to the repressor, so it will never change shape and will remain attached to the DNA constantly. This means that no transcription will occur. Functional enzyme will be made. The operator region has a constitutive mutation, so the repressor will never be able to bind. This means that the functional genes will be constantly transcribed, no matter the presence of lactose.

21-11 Describe the human genome in terms of genome size, the percentage of the genome that codes for proteins, how much is composed of repetitive sequences, and how many genes it contains. Describe two other features of the human genome.

The human genome contains about 3.1 billion nucleotides, but only 2% of those code for proteins. At least 50% of the genome is composed of repetitive sequences or transposable elements. It contains at least 20,000 protein-coding genes. As there are many more proteins than protein-coding genes, it is obvious that alternative splicing is common. The human genome also has 50% similarity to genes in other organisms, but up to 40% of that group has no molecular function. Another feature is that the average size of a human gene is 25 kB.

16-7 The locations of numerous lacI- and lacIS mutations have been determined within the DNA sequence of the lacI gene. Among these, lacI- mutations were found to occur in the 5'-upstream region of the gene, while lacIS mutations were found to occur farther downstream in the gene. Are the locations of the two types of mutations within the gene consistent with what is known about the function of the repressor that is the product of the lacI gene?

The lacI- mutation is commonly found in the 5'-upstream region and the lacIS mutation is commonly found further downstream. These locations are consistent with the

20-21 Traditional Sanger sequencing has largely been replaced in recent years by next-generation and third-generation sequencing approaches. Describe advantages of these sequencing methods over first-generation Sanger sequencing.

The main difference is that new generation sequencing allows for higher volumes of DNA fragments to be created. It is also less expensive and less time consuming.

22-12 What is the main purpose of genome-wide association studies (GWAS)? How can information from GWAS be used to inform scientists and physicians about genetic diseases?

The main purpose of GWAS is to identify genes that influence disease. GWAS are used to compare the genomes of healthy individuals to those with the selected disease; the variations presented in these studies can tell scientists and physicians a lot about what genetic variation may be causing the disease.

22-11 Maternal blood tests for three pregnant women revealed they would be having boys, yet subsequent ultrasound images showed all three were pregnant with girls. In each case Y chromosome sequences in each mother's blood originated from transplanted organs they had received from men! This demonstrates one dramatic example of a limitation of genetic analysis of maternal blood samples. What kind of information could have been collected from each mother in advance of these tests to better inform physicians prior to performing each test?

The mothers could have undergone an amniocentesis and further genetic testing, This would have allowed for more insight into not only sex determination, but karyotyping, biochemical analysis, and testing for other genetic disorders.

22-13 Describe how the team from the J. Craig Venter Institute created a synthetic genome. How did the team demonstrate that the genome converted the recipient strain of bacteria into a different strain?

The team compared small genomes to identify 256 genes that represent the minimum genes required for life. They then used transposon-based methods to determine the essential gene number in Mycoplasma. They then synthesized short DNA fragments and assembled them into a synthetic genome that was transformed into a Mycoplasma. They verified the effectiveness of their recipient strain by producing specific proteins unique to the synthetic genome.

22-2 Write a short essay that summarizes the impacts that genomic applications are having on society and discuss which of the ethical issues presented by these applications is the most daunting to society.

The therapeutic applications of genomics involve producing proteins by introducing human genes into bacteria, allowing for the creative of active hormones (insulin) or vaccines (Gardasil). There are also applications in genetic testing of embryos and adults as well as in confirming diagnoses. The agricultural applications of genomics involve insect resistance, herbicide resistance, and genetically modified food and animal products. These applications obviously have a large impact on society and are spurring discussions surrounding guidelines for the technology, intellectual property rights, and even legal implications (requiring genetic screening of infants by law to avoid treatable genetic disorders).

16-16 Both attenuation of the trp operon in E. coli and riboswitches in B. subtilis rely on changes in the secondary structure of the leader regions of mRNA to regulate gene expression. Compare and contrast the specific mechanisms in these two types of regulation with that involving short noncoding RNAs (sRNAs).

The two Riboswitches encompass a variety of mechanisms that can either terminate or allow for transcription. In both cases, there is a formation of intramolecular double-stranded RNA that induces conformational changes. sRNA molecules will bind to the short ribonucleotide sequence (attenuator) in order to induce the conformational change. sRNAs are separate transcripts that are complementary to the mRNA and function through the formation of intermolecular double stranded RNA. The binding of sRNA either prevents of enhances translation.

19-9 What are the similarities and differences in the two types of ncRNAs involved in epigenetic control of gene expression?

The two types of ncRNAs involved are miRNAs and siRNAs. miRNAs have distinct genomic loci and are encoded for by their own genes. miRNAs work to suppress gene expression where necessary. Prior to dicer processing, miRNA is in its primary form and is still single stranded and in hairpin form. Once processed by RISC, it is capable of targeting over 100 mRNA types fully and partially. It represses translation and encourages degradation of mRNA. miRNA is involved in gene expression as it suppresses expression by binding to the mRNA sequence that it is complementary to. siRNAs are encoded by transposons, viruses, and heterochromatin. Prior to dicer processing, it is double stranded. siRNAs are only capable of targeting one mRNA sequence and it must be fully complementary to it. It utilized endonucleolytic cleavage and degrades the mRNA target. siRNAs operate in the RNA interference (RNAi) pathway, effectively controlling gene expression. Both miRNA and siRNA serve as post-transcriptional forms of epigenetic control.

20-28 What do you think are some of the concerns about the use of CRISPR-Cas on humans? Should CRISPR-Cas applications be limited for use on only certain human genes but not others? Explain your answers.

There are concerns in the idea that the genome of babies could be altered to express certain favorable traits later in life. Creating designer babies based on preference is unethical. This technology could be applied to genes that code for diseases and unfavorable traits; however, it is unlikely that every pregnant woman would have equal access to this technology, making it unethical to edit one fetus and not others.

16-2 Write a brief essay that discusses why you think regulatory systems evolved in bacteria (i.e., what advantages do regulatory systems provide to these organisms?), and, in the context of regulation, discuss why genes related to common functions are found together in operons.

There are many evolutionary advantages of gene expression control. Regulatory systems enable transcription of different genes in different environments. They also provide an efficient response to protect bacteria from harmful environmental factors. Genes with common functions are found together in operons as this allows for coordinated responses.

19-3 What are the major mechanisms of epigenetic genome modification?

There are three major mechanisms of epigenetic genome modification. They include histone modification, DNA methylation, and interfering with non-coding RNA molecules. Chromatin is a complex of DNA made up of DNA, proteins, and RNA that make up chromosomes. Histones are positively charges proteins that are associated with DNA. Histones make bonds with the phosphates on chromatin; when tightly coiled together, the genes within chromatin cannot be expressed (heterochromatin) and when loosely coiled, the genes are expressed (euchromatin). In adding acetyl groups to the amino acids in the N-terminal region, the grips of histones are relaxed and the genes are made available for transcription. Adding methyl groups to histones works to silence the genes; the addition impacts when proteins can bind on the DNA and interact with the regulatory region. The addition of phosphate groups by enzyme kinases can also further relax the histones, allowing for more gene expression. DNA methylation is a reaction catalyzed by DNA methyltransferases (DNMTs) that involves the addition of a methyl group to cytosine on DNA. Methylation occurs in CpG islands, which are regions of CpG dinucleotide repeats located around promoter sequences. When methyl groups are added, they occupy major grooves in the DNA and block transcription factors from binding and initiation transcription. This effectively silences genes. There are three non-coding RNA epigenetic mechanisms: siRNAs, miRNAs, and lncRNAs. siRNAs are cut down from double stranded RNA by dicers. Once this happens, they will associate with RISCs, which will cleave and evict one of the siRNA strands of DNA and retain the other as a degradation guide. Once the siRNA:RISC complex finds a piece of mRNA that matches the degradation code, it will be targeted and erased from the transcribable mRNA code. miRNAs are initially pri-miRNAs in hairpin shape which are transported to the cytoplasm. They are then cleaved by RISC and (like siRNAs) further processed to be single stranded. miRNA works to find a miRNA response element (MRE); a perfect match for the MRE will be cleaved and degraded and a partial match for the MRE will be blocked from translation. It is important to remember that siRNAs have ONE target and miRNAs have MULTIPLE targets. lncRNAs are much longer than siRNAs and miRNAs. They regulate gene expression at multiple levels by interacting with DNA, RNA or proteins. Antisense lncRNA overlap protein coding regions and are transcribed in the opposite direction of the opposing gene. Intronic lncRNA are located within introns. Bidirectional lncRNAs have no sequence overlap, but use the promoter of an adjacent sequence to transcribe in the opposite direction. Intergenic lncRNA are independent and use their own promoters. All of these mechanisms will act to compete with transcription initiation sites (decoy), act as a platform for two proteins to form DNA binding complexes (adapter), recruit certain protein complexes (guide), or bind to DNA upstream and guide expression (enhance).

22-15 The family of a sixth-grade boy in Palo Alto, California, was informed by school administrators that he would have to transfer out of his middle school because they believed his mutation of the CFTR gene, which does not produce any symptoms associated with cystic fibrosis, posed a risk to other students at the school who have cystic fibrosis. After missing 11 days of school, a settlement was reached to have the boy return to school. What ethical problems might you associate with this example?

There is an ethical issue in the fact that the school obtained his private genetic information and used it to discriminate against him. They also publicly distributed his private information. While GINA (the genetic information non-discrimination act) only applies to employment and health insurance, the law's widespread potential applications can be seen here.

20-10 In 1975, the Asilomar Conference on Recombinant DNA was organized by Paul Berg, a pioneer of recombinant DNA technology, at a conference center at Asilomar State Beach in California. Physicians, scientists, lawyers, ethicists, and others gathered to draft guidelines for safe applications of recombinant DNA technology. These general guidelines were adopted by the federal government and are still in practice today. Consider the implications of recombinant DNA as a new technology. What concerns might the scientific community have had then about recombinant DNA technology? Might those same concerns exist today?

There were original concerns in creating recombinant organisms and in combining eukaryotes and bacteria. While these concerns have been alleviated with new technology, the most prevalent concern is now the potential danger in genetically modified plants and animals.

22-6 A couple with European ancestry seeks genetic counseling before having children because of a history of cystic fibrosis (CF) in the husband's family. ASO testing for CF reveals that the husband is heterozygous for the ∆508 mutation and that the wife is heterozygous for the R117 mutation. You are the couple's genetic counselor. When consulting with you, they express their conviction that they are not at risk for having an affected child because they each carry different mutations and cannot have a child who is homozygous for either mutation. What would you say to them?

They are heterozygous for different mutations, but both mutations occur on the CF gene. This means that if the child were to inherit both mutations, they would have CF. There is a 25% chance of that happening as both parents are heterozygous.

22-21 Following the tragic shooting of 20 children at a school in Newtown, Connecticut, in 2012, Connecticut's state medical examiner requested a full genetic analysis of the killer's genome. What do you think investigators might be looking for? What might they expect to find? Might this analysis lead to oversimplified analysis of the cause of the tragedy?

They may be looking for certain markers that made him more likely to be a murderer. They may expect to find certain SNPs connected with mental illness or high aggression that would help explain his actions. This data could be used to identify killers, but it could also be used to discriminate against people that have a genetic predisposition to aggression. This could lead to oversimplified analysis of this tragedy, and it is unlikely that any conclusive evidence would be drawn from the study of one individual.

20-6 Using DNA sequencing on a cloned DNA segment, you recover the nucleotide sequence shown below. Does this segment contain a palindromic recognition sequence for a restriction enzyme? If so, what is the double-stranded sequence of the palindrome, and what enzyme would cut at this sequence? (Consult Figure 20.1 for a list of restriction sites.) CAGTATGGATCCCAT

This sequence contains the palindromic recognition sequences of GATCC and GATC, which respectively use the BamHI and Sau3AI restriction enzymes.

16-12 Describe the role of attenuation in the regulation of tryptophan biosynthesis.

Trp attenuators are riboswitches as they bind with small ligands to cause a conformational change, producing the anti-terminator and terminator sequences. The aptamer in a riboswitch binds to the ligand and the expression platform is where conformational change takes place. The trp structural genes are preceded by a leader sequence called an attenuator. The attenuator is another form of regulation in the trp operon. Transcription of the attenuator will occur even if there is trp present (normally, when trp is present, the operon is no longer transcribed). The attenuator sequence will fold into either the anti-terminator (if trp is absent) or terminator hairpin (if trp is present). With low levels of trp, the ribosome will be stalled at the trp codon, which will allow time for a 2-3 stem loop to form, for the RNA polymerase to continue to process, and for trp to be produced. With high levels of trp, the ribosome moves quickly past the codon and into the 1-2 region. This stops the 2-3 stem loop from forming, and transcription cannot proceed. The ribosome falls off.

16-24 Figure 16.13 depicts numerous critical regions of the leader sequence of mRNA that play important roles during the process of attenuation in the trp operon. A closer view of the leader sequence, which begins at about position 30 downstream from the 5'-end, is shown below, running along both columns. Within this molecule are the sequences that cause the formation of the alternative hairpins. It also contains the successive triplets that encode tryptophan, where stalling during translation occurs. Take a large piece of paper (such as manila wrapping paper) and, along with several other students from your genetics class, work through the base sequence to identify the trp codons and the parts of the molecule representing the base-pairing regions that form the terminator and antiterminator hairpins shown in Figure 16.13.

Tryptophan can only be coded for with UGG (base pair ACC), so I know to look for that codon in the code. Base-pairing regions that form the hairpins will match up with the regions in the diagram but I refuse to do that. nasty

21-4 Compare and contrast WGS to a map-based cloning approach.

WGS, or whole genome shotgunning, involves randomly cutting the genome into smaller segments in order to find overlapping sequences and produce the entire genome. There is difficulty in the fact that repetitive regions in the genome makes alignment of contigs difficult. A map-based cloning approach relies on known landmarks in order to align fragments. The map-based approach is much slower and is normally only used when problems arise in WGS.

22-23 Scientists have used single- nucleotide polymorphisms (SNPs) in genome-wide association studies (GWAS) to identify novel risk loci for prostate cancer and Type 2 diabetes, respectively. Each study suggests that disease-risk genes can be identified that significantly contribute to the disease state. Given your understanding of such complex diseases, what would you determine as reasonable factors to consider when interpreting the results of GWAS?

When interpreting GWAS results, I would consider sample size, certain epigenetic/environmental factors, genetic background, population demographics and stratification, etc.

21-14 Explain differences between whole-genome sequencing (WGS) and whole-exome sequencing (WES), and describe advantages and disadvantages of each approach for identifying disease-causing mutations in a genome. Which approach was used for the Human Genome Project?

Whole genome sequencing (WGS) provides results for the entire genome, including noncoding regions, whereas whole-exome sequencing (WES) provides information for only exons, or protein-coding regions. As disease-related variations are more likely to be within exons than introns, WES is more likely to identify these mutations. Only WGS, however, will be able to identify mutations in regulatory regions, as those are located within introns. The WGS was used for the Human Genome Project, as to allow for a wider scope of the human genome.

22-8 Might it make sense someday to sequence every newborn's genome at the time of birth? What are the potential advantages and concerns of this approach?

Widespread screening of newborns would allow for the identification of an infinite number of variables associated with the human genome that may be of both scientific and personal interest. Certain disease states may be identified, which could lead to earlier treatment intervention. The concerns of this are numerous privacy implications and potential discrimination and stigmatization of certain babies.

20-12 If you performed a PCR experiment starting with only one copy of double-stranded DNA, approximately how many DNA molecules would be present in the reaction tube after 15 cycles of amplification?

You would have 16,384 molecules after 15 cycles. The formula to do this is 2^(n-1).

20-4 The human insulin gene contains a number of sequences that are removed in the processing of the mRNA transcript. In spite of the fact that bacterial cells cannot excise these sequences from mRNA transcripts, explain how a gene like this can be cloned into a bacterial cell and produce insulin.

cDNA created from insulin mRNA will not contain introns, so plasmids created from the cDNA of insulin will not face processing issues.

20-14 What advantages do cDNA libraries provide over genomic DNA libraries? Describe cloning applications where the use of a genomic library is necessary to provide information that a cDNA library cannot.

cDNA libraries do not contain introns, or non-coding genes. This means that the cDNA libraries are much less full, which makes it easier to locate specific genes and proteins. You would want to use a DNA library over a cDNA library if you'd like to analyze non-coding gene regions.

16-5 For the lac genotypes shown in the following table, predict whether the structural genes (Z) are constitutive, permanently repressed, or inducible in the presence of lactose.

inducible constitutive constitutive constitutive constitutive repressed repressed