Genetics Exam 2

The genome of an organism is 7,500 base pairs long and is 15% cytosine. A restriction enzyme has a known recognition sequence of 5'-CATCG-3'. How many times would the enzyme be expected to cut the genome?

(0.15)(0.35)(0.35)(0.15)(0.15)=0.0004134 1/0.0004134= 2,419 bp 7500/2419= 3.1 Enzyme would cut 3 times

What are two major limitations of PCR?

(1) Requires the use of specific primers, meaning there must be sequence info available for the DNA to be amplified in order for primers to be designated. (2) The length of DNA that can be amplified by PCR is limited by the enzyme and conditions to about 40 kb.

Describe the three "types" of cloning:

1. Molecular cloning is where copies of specific gene fragment are produced. (breaking part of gene) 2. Cellular cloning- single celled organisms with exact genetic content of the original cell. Think of binary fission. 3. Organism cloning- reproductive cloning where multicellular clone is created

What are the stages of Mitosis and what occurs during each stage?

1. Prophase - this is when the chromosomes appear and the nucleoli disappear. The mitotic spindle forms. 2. Prometaphase - The nuclear envelope breaks into fragments with the microtubules attaching to the kinectochore. 3. Metaphase - The centrosomes move to opposite poles while the chromosomes line up at the metaphase plate. 4. Anaphase - The chromatid pairs split when the cohesins are cleaved. The individual chromatids now move to opposite sides of the cell. 5. Telophase - The daughter cell nuclei forms and the nucleoli reappear. The chromosomes are now at individual poles.

Define the different types of mutations:

1. Spontaneous mutations- happen naturally and are random. 2. Induced mutation- induced by exposure to chemical and physical mutagens. 3. Somatic mutations- occur in any cell except germ cells. 4. Germ-line mutations- occur in gametes and are inherited.

*10.35 The enzyme Tsp45I recognizes the 5-bp site -G-T-(either C or G)-A-C- . This site appears in exon 4 of the human gene for -synuclein, where, in a rare form of Parkinson disease, it is altered by a single G-to-A mutation. (Note: Not all forms of Parkinson disease are caused by genetic mutations.) a. Suppose you have primers that can be used in PCR to amplify a 200-bp segment of exon 4 containing the Tsp45I site, and that the Tsp45I site is 80 bp from the right primer. Describe the steps you would take to determine if a patient with Parkinson disease has this -synuclein mutation. b. What different results would you see in homozygotes for the normal allele, homozygotes for the mutant allele, and in heterozygotes? c. How would you determine, in heterozygotes, if the mutant allele is transcribed in a particular tissue?

10.35 Answer: a. Use the PCR-RFLP method: Isolate genomic DNA from the individual with Parkinson disease, and use PCR to amplify the 200-bp segment of exon 4; purify the PCR product, digest it with Tsp45I, and resolve the digestion products by size using gel electrophoresis. The normal allele will contain the Tsp45I site, and so produce 120- and 80-bp fragments. The mutant allele will not contain the Tsp45I site, and so produces only a 200-bp fragment. b. Homozygotes for the normal allele will have 120- and 80-bp fragments; homozygotes for the mutant allele will have a 200-bp fragment; heterozygotes will have 200-, 120-, and 80-bp fragments. c. Use RT-PCR to amplify a DNA copy of the mRNA, and digest the RT-PCR product with Tsp45I. First, isolate RNA from the tissue. Then make a single-stranded cDNA copy using reverse transcriptase and an oligo(dT) primer. Then amplify exon 4 of the cDNA using PCR, digest the product with Tsp45I, and separate the digestion products by size using gel electrophoresis. If a 200-bp fragment is identified in a heterozygote, then the mutant allele is transcribed. If only 120- and 80-bp fragments are identified, then the mutant allele is not transcribed. Note that to assess expression of either allele accurately, it is essential that the RT-PCR reaction be performed on a purified RNA template without contaminating genomic DNA.

*10.38 Abbreviations used in genomics typically facilitate the quick and easy representation of longer tonguetwisting terms. Explore the nuances associated with some abbreviations by stating whether an RFLP, VNTR, or STR could be identified as an SNP? Explain your answers.

10.38 Answer: A SNP is a single nucleotide polymorphism. Since a single base change can alter the site recognized by a restriction endonuclease, a SNP can also be a RFLP, or restriction fragment length polymorphism. Since simple tandem repeats (STRs) and variable number tandem repeats (VNTRs) are based on tandemly repeated sequences (repeats that are 2 to 6 bp long for STRs, and from seven to tens of bp long for VNTRs), they will not usually be SNPs.

10.41 What is DNA fingerprinting and what different types of DNA markers are used in DNA fingerprinting? How could this method be used to establish parentage? How is it used in forensic science laboratories?

10.41 Answer: DNA fingerprinting is the characterization of an individual in terms of the set of DNA markers the person has. For DNA fingerprinting, DNA markers are usually chosen so that they are highly polymorphic in a population, and they may include RFLPs, simple tandem repeat polymorphisms that are 2 to 6 bp long (STRs, microsatellites), and variable numbers of tandem repeat polymorphisms that are from seven to tens of bp long (VNTRs, minisatellites). VNTRs can be derived either from just one locus or from more than one locus (monolocus or multilocus probes). Since all individuals except for identical (monozygotic) twins have different genomes, individuals of a species differ in terms of their DNA fingerprints. Each parent donates one allele at a DNA marker to his or her offspring. Therefore, the set of DNA markers a child has come from his or her parents. If two parents and a child are tested, all of the DNA markers present in the child must also be present in the set of markers in the two parents. However, the parents can have additional markers, since they donate only one of their two alleles at each marker. If a child has no alleles in common with a suspected parent, the suspected individual can be excluded from consideration as the child's parent. If alleles are shared, the likelihood that the set of alleles came from the suspected parent must be calculated and compared with the likelihood that the set of alleles came from another person. If the alleles present in the child and one known parent are known, then one can infer at least some of the alleles that were contributed by the other parent. Usually, multiple DNA markers are used to minimize possible inaccuracy and provide for a better statistical evaluation of the results, showing that the results obtained are not based on sampling variation. The calculation of the relative odds that a set of alleles derives from a suspected parent or from another person depends on knowing the frequencies of the marker alleles in the child's ethnic population. These types of data should be used together with other non-DNA-based evidence to support a claim of parentage. In forensic science laboratories, DNA fingerprinting is used to match or exclude the DNA fingerprint of a suspected individual with the DNA fingerprint provided by physical evidence (blood, skin, hair, semen, saliva, etc.) gathered at the scene of a crime. As in the analysis for parentage, DNA fingerprinting is very useful for excluding suspects who do not share alleles with that found in the physical evidence. Inclusion results, results where an individual is positively identified as being responsible for a crime, are more difficult to obtain. As for parentage, a calculation of the relative odds that a set of alleles derive from a suspect or from another person must be made, and this requires good population statistics. In addition, there must be evidence that there were no errors in collecting or processing samples. Thus, DNA evidence is often more useful for excluding suspects than for proving their guilt, and when used to support guilt, must usually be supported by additional, non-DNA-based evidence.

12.14 Does mitosis or meiosis have greater significance in the study of heredity? Explain your answer.

12.14 Answer: Meiosis has greater significance. While mitosis generates progeny cells that are genetically identical to a parent cell, meiosis generates gametes that are genetically diverse. Gamete diversity is obtained when nonparental combinations of genes are obtained in two ways: the random assortment of maternal and paternal chromosomes at anaphase I of meiosis I and crossing-over events in prophase I of meiosis I. When gametes from different parents fuse at fertilization, more diversity is obtained. Hence, meiosis provides a means for genetic variation.

1a. A linear double stranded DNA sequence is cut with a restriction enzyme. There are four restriction sites for this enzyme. How many bands are expected to be viewed on an agarose gel? 1b. How many bands would be expected to be viewed on an agarose gel if the DNA strand was circular?

1a. Five bands 1b. Four bands

You are studying a new organism named Undergraduentus studentia. This organism has a genome size of 2,000,000 base pairs and has 40% GC-content. The restriction enzyme EcmI cuts DNA at the 6-base recognition sequence 5'-CCAGTG-3. Approximately how many times would you expect this enzyme to cut the genome? YOU CAN JUST SHOW YOUR WORK, YOU DON'T NEED TO CALCULATE THE NUMBER (4)

20% C, 20% G, 30% A, 30% T Would expect a frequency of (0.2 x 0.2 x 0.3 x 0.2 x 0.3 x 0.2) = 0.000144 (or every 6,944 base pairs). If the genome is 2,000,000 bp long, it would cut: 2,000,000 / 6,944 = approximately 288 times (or you can calculate it as 0.000144 x 2,000,000 = approximately 288 times) (This was very similar to HW question 8.5a)

7.19 DNA damage by mutagens has serious consequences for DNA replication. Without specific base pairing, the replication enzymes cannot specify a complementary strand, and gaps are left after the passing of a replication fork. a. What response has E. coli developed to large amounts of DNA damage by mutagens? How is this response coordinately controlled? b. Why is the response itself a mutagenic system? c. What effects would loss-of-function mutations in recA or lexA have on E. coli's response?

7.19 a. Large amounts of DNA damage trigger the SOS response in which the RecA protein becomes activated and stimulates the LexA protein to cleave itself. Since the LexA protein functions as a repressor for about 17 genes whose products are involved in DNA damage repair, this results in the coordinate transcription of those genes. Following the repair of DNA damage and inactivation of RecA, newly synthesized LexA coordinately represses their transcription. b. The response is mutagenic because a DNA polymerase for translesion DNA synthesis is produced during the SOS response. When this polymerase encounters a lesion, it incorporates one or more nucleotides not specified by the template strand into the new DNA across from the lesion. These nucleotides may not match the wild-type template sequence, and so this polymerase introduces mutations. c. In mutants having loss-of-function mutations in both recA and lexA, or only in lexA, there would be no functional LexA protein to repress transcription of the 17 genes whose protein products are involved in the SOS response; this would result in constitutive activation of the SOS response. If the lossof-function mutation is only in recA, however, heavy DNA damage would not trigger RecA protein activation, so RecA could not stimulate the LexA protein to cleave itself to induce the SOS response. Instead, the LexA protein would continue to repress the DNA repair genes in the SOS system. Such a mutant would be highly sensitive to mutagens such as UV light and X-rays.

What is the difference between a dideoxy and a deoxynucleotide?

A dideoxy has a 3'-H, which leads to termination of the growing chain. Deoxy nucleotides contain a 3'-OH, which the 5'-phosphate of the next nucleotide in the growing chain forms a phosphodiester bond with.

What are the two ways mistakes in DNA replication fixed?

A. 3'-5' exonuclease activity in which DNA polymerase reverses its direction, excises the incorrect base and inserts the correct one B. 2nd line of defense is Mismatch repair- which correct errors that remain after proofreading. DNA is recognized based on DNA methylation of the parental strand in prokaryotes, it is currently unknown how the parental strand is identified in eukaryotes

7.17 DNA polymerases from different organisms differ in the fidelity of their nucleotide insertion; however, even the best DNA polymerases make mistakes, usually mismatches. If such mismatches are not corrected, they can become fixed as mutations after the next round of replication. a. How does DNA polymerase attempt to correct mismatches during DNA replication? b.What mechanism is used to repair such mismatches if they escape detection by DNA polymerase? c. How is the mismatched base in the newly synthesized strand distinguished from the correct base in the template strand?

A. Dna polymerase recognizes the mismatched base because of the incorrect watson-crick pairing. and cuts out a strand of the dna using 3'-5' exonuclease excision then dna polymerase puts in the right pair. b. Mismatch repair proteins recognize the mismatch and cut out the incorrect nucleotide. c. If the dna polymerase keeps going and doesn't see it, both strands are being replicated so the mismatch activity takes place after it has been replicated. In bacteria the old strand is methylated, and the new strand does not, so when the mismatch machinery comes in, it looks for the strand that is methylated to determine which strand is wrong and fixes it.

Q: Would a nonsense mutation in a somatic cell affect the offspring? Define a nonsense mutation and explain if it would be passed on.

A: A nonsense mutation is a point mutation that changes a nucleotide to create a premature stop codon, thus the protein made is almost always nonfictional. This mutation will not be passed on to the offspring because only mutations in germ-line cells are passed on to offspring.

*7.8 The mutant lacZ-1 was induced by treating E. coli cells with acridine, whereas lacZ-2 was induced with 5BU. What kinds of mutants are these likely to be? Explain. How could you confirm your predictions by studying the structure of the -galactosidase in these cells?

Acridine is an intercalating agent that induces frameshift mutations. lacZ-1 probably is a frameshift mutation that results in a completely altered amino acid sequence after some point, although it might be truncated due to the introduction of an out-of-frame nonsense codon. In either case, the protein produced by lacZ-1 would most likely have a different molecular weight and charge. During gel electrophoresis (see Figure 4.8, p. 70), it would migrate differently than the wild-type protein. 5BU is incorporated into DNA in place of T. During DNA replication, it can be read as C by DNA polymerase because of a keto-to-enol shift. This results in point mutations, usually TAto- CG transitions. lacZ-2 is likely to contain a single amino-acid difference, due to a missense mutation; although it, too, could contain a nonsense codon. A missense mutation might lead to the protein having a different charge, while a nonsense codon would lead to a truncated protein that would have a lower molecular weight. Both would migrate differently during gel electrophoresis.

*8.21 In a sequencing reaction using dideoxynucleotides that are labeled with different fluorescent dyes, the DNA chains produced by the reaction are separated by size using capillary gel electrophoresis and then detected by a laser eye as they exit the capillary. A computer then converts the differently colored fluorescent peaks into a pseudocolored trace. Suppose green is used for A, black for G, red for T, and blue for C. What pattern of peaks do you expect to see on a sequencing trace if you carry out a dideoxy sequencing reaction after the primer 5¿-CTAGG-3¿ is annealed to the following single stranded DNA fragment? 3¿-GATCCAAGTCTACGTATAGGCC-5¿

After the primer anneals to the fragment, DNA polymerase will extend the primer at its 3′ end by adding nucleotides that are complementary to the template. As the primer is extended, some chains will be prematurely terminated when a labeled dideoxynucleotide is incorporated. Since the four dideoxynucleotides are labeled with different fluorescent dyes, the extension products that terminate with the same base will have the same fluorescent label and be detected as distinct, labeled bands after separation using capillary gel electrophoresis. Therefore, the order of fluorescent peaks can be used to determine the sequence.

What is aneuploidy?

An abnormal of chromosomes in a genome, it can be either too many or too few.

Compare the error rate, Throughput, and common uses for Dideoxy Sequencing, Pyrosequencing and next-gene sequencing

Answer: Dideoxy sequencing - low error rate, low throughput, Sequencing a gene Pyrosequencing - medium error rate, medium throughput, Sequencing a few genes Next-Gen sequencing - high error rate, high throughput, Sequencing entire genomes

List the four major steps needed to complete a southern blot:

Answer: The first step is extracting the DNA, the second is Electrophoresis to separate the different size strands of DNA. The third step is to transfer the DNA to membrane and then finally the sequence is determined using labeled probes that hybridize certain sequences in the DNA.

*9.11 If you assume that each step of the PCR process is 100% efficient, how many copies of a template would be amplified after 30 cycles of a PCR reaction if the number of starting template molecules were

As shown in Figure 9.3, two unit-length, double-stranded DNA molecules (called amplimers) are produced after the third cycle of PCR from one double-stranded DNA template molecule. If each step of the PCR process is 100% efficient, the number of amplimers geometrically increases in each subsequent cycle: In the 4th cycle there will be 4 amplimers, in the 5th cycle there will be 8 amplimers, and more generally, in the nth cycle there will be 2^(n-2) amplimers. In the 30th cycle, there will be 2^28=2.68 x 10^8 molecules. A larger number of initial template molecules will lead to a proportional increase in amplimer production. a. 10!2^28=2.68!10^9 molecules b.1,000!2^28=2.68!10^11 molecules c.10,000!2^28=2.68!10^12 molecules Consider these answers with respect to the experimental observation that about 5 ng of DNA (about copies of a 200-bp DNA fragment) is detected readily on an ethidium bromide stained agarose gel.

7.1 Mutations are (choose the correct answer) a. caused by genetic recombination. b. heritable changes in genetic information. c. caused by faulty transcription of the genetic code. d. usually, but not always, beneficial to the development of the individuals in which they occur.

B

8.22 How does pyrosequencing differ from dideoxy chain-termination sequencing? What advantages does it have for large-scale sequencing projects?

Both dideoxy chain-termination sequencing and pryosequencing start with a DNA template, DNA polymerase, and a sequencing primer. Dideoxy chain-termination sequencing detects the sequence of a nucleotide chain using a chain-termination mechanism, while pyrosequencing does not. In dideoxy chain-termination sequencing, four different fluorescently labeled dideoxynucleotides are present in a DNA synthesis reaction where the sequencing primer is extended by DNA polymerase. When a radioactively or fluorescently labeled dideoxynucleotide is incorporated into a growing polynucleotide chain, chain termination occurs. The extension products are separated by size using either: 1) gel electrophoresis (for radiolabeled ddTNPs), and the sequence is determined by analyzing the pattern of the bands in the four lanes of the sequencing gel, or by 2) capillary gel electrophoresis (for fluorescently labeled dNTPs), where the order of fluorescent peaks can be used to determine the sequence. In contrast, pyrosequencing determines a sequence by enzymatic detection of the pyrophosphate (diphosphate) released when a nucleotide is incorporated by DNA polymerase as it extends a new DNA strand. The pyrosequencer detects and quantifies the amount of light released and correlates it to which dNTP was present in the reaction. In this way, the base added to the growing DNA strand is determined. The process is continued cyclically. Reverse-terminator sequencing (sometimes called DNA-Seq, RNA-Seq or Next Generation Sequencing) utilizes non-specific primers that bind to all DNA fragments. This results in the ability to sequence entire genomes in short amounts of time. Dideoxy (Sanger) sequencing is most advantageous for obtaining sequence information for shorter, specific pieces of DNA, as this method uses specific primers, has lower throughput but higher accuracy. Pyrosequencing is advantageous for obtaining sequence information for longer, specific pieces of DNA, as it has higher throughput, but slightly lower accuracy. Reverse-terminator sequencing is most advantageous for sequencing many different pieces of DNA, including entire genomes, as it is high-throughput, but has the lowest accuracy of the three.

In meiosis 1, ______ line up at the metaphase plate and in meiosis 2, _____ line up at the metaphase plate.

Chromosomes, sister chromatids

Polymerase chain reaction (PCR) is a method to make large amounts of specific DNA sequences from small amounts sequences. Explain when cDNA would need to be made and with what process.

Complementary DNA is in the process of reverse-transcription PCR (RT-PCR) when you want to measure the levels of mRNA between different samples. cDNA is made from mRNA using reverse transcriptase. This is made possible because mRNA contains a Poly-A tail. Reverse transcriptase generates cDNA with an oligo (dt) primer will make cDNA from every mRNA containing a poly A tail. This primer will help make a copy of all the RNA because they all contain the same Poly A tail sequence. This double stranded cDNA that is made can then be used as a template for PCR.

Explain the role of dNTPs and ddNTPs in dideoxy sequencing and how this is set up of a gel.

Dideoxy sequencing is based on the principles of DNA replication in vitro, but only using one strand 3' to 5' as a template and dNTPs and ddNTPs. However ddNTPs lack an additional OH group on the sugar, so when a ddNTP is incorporated, it is unable to continue to make a strand. Four reactions (ddATP, ddGTP, ddTTP, and ddCTP) are run on separate lanes of the gel, and based on this the complementary DNA sequence can be determined because these ddNTPs are added randomly to the sequence. By combining all four reactions together the complementary strand of DNA can be determined.

Compare the error rate, throughput, and common uses for Dideoxy Sequencing, Pyrosequencing and next-gen sequencing

Dideoxy- low error rate, low throughput, Sequencing a gene Pyro- medium error rate, medium throughput, Sequencing a few genes Next-Gen- high error rate, high throughput, Sequencing entire genomes

. Match the types of mutation and the mechanism by which they are repaired. (6) DNA Polymerase Error Thymine Dimer (via UV light) Deamination events

Dna polymerase Error - mismatch repair Thymine Dimer- Nucleotide excision repair Deamination events- Base excision repair

What is the difference between spermatogenesis and oogenesis?

During spermatogenesis four spermatozoa are formed while in oogenesis only one ovum and three polar bodies are formed.

7.3 Which of the following is not a class of mutation? a. frameshift b. missense c. transition d. transversion e. none of the above; all are classes of mutation

E- Frameshift- Missense- Transition- not on exam point mutation A-->G Transversion- not on exam both point mutations C-->T

True or false: In dideoxy sequencing, special DNA polymerase is used

False - in dideoxy sequencing, different nucleotides are used

True or False. If false, make it correct. The number of chromosomes double during the S phase.

False, the number of chromatids double during the S phase.

*12.18 Explain whether the following statement is true or false: "All the sperm from one human male are genetically identical."

False. Genetic diversity in the male's sperm is achieved during meiosis. There is crossing-over between nonsister chromatids as well as independent assortment of the male's maternal and paternal chromosomes during spermatogenesis.

*7.2 Answer true or false: Mutations occur more frequently if there is a need for them.

False. Mutations occur spontaneously at more or less a constant frequency, regardless of selective pressure. Once they occur, however, they can be selected for or against, depending on the advantage or disadvantage they confer.

12.17 Explain whether the following statement is true or false: "Meiotic chromosomes can be seen after appropriate staining in nuclei from rapidly dividing skin cells

False. Skin cells consist entirely of mitotic, not meiotic, cells.

What are the phases of the Cell Cycle and what happens during those phases?

G1 phase. Metabolic changes prepare the cell for division. At a certain point - the restriction point - the cell is committed to division and moves into the S phase. S phase. DNA synthesis replicates the genetic material. Each chromosome now consists of two sister chromatids. G2 phase. Metabolic changes assemble the cytoplasmic materials necessary for mitosis and cytokinesis. M phase. A nuclear division (mitosis) followed by a cell division (cytokinesis). The period between mitotic divisions - that is, G1, S and G2 - is known as interphase.

How will the length of the DNA strand be affected after two replication cycles if (a) an intercalating agent inserts itself in the template strand of the DNA and (b) an intercalating agent inserts itself in the newly synthesized DNA?

If the intercalating agent inserts itself in the template strand it will be paired with a random base pair in the first replication cycle. After another replication the intercalating agent will be lost and have resulted in a one base pair insertion. If it inserted in place of a base in the newly synthesized DNA the agent is lost after one cycle, and after two cycles will result in a one base pair deletion.

What is the difference in the amount of cells you begin with and end with between Mitosis and Meiosis?

In Mitosis you start with one cell and end up with two daughter cells. In Meiosis you start with one cell and end up with four daughter cells. Meiosis results in 4 daughter cells that have varying genetic material to be passed onto offspring (results in genetic variation) and mitosis results in 2 daughter cells that are genetically identical to the parent.

What occurs during base excision repair?

Involves use of specific DNA glycosylases to remove abnormal bases used in cases of deamination of adenine, cytosine, and guanine.

quiz 4C When you perform the RT-PCR, you are unable to see any band on your gel (suggesting no mRNA for this gene was present in the cell). Where in the gene do you think the mutation occurred and why? (3)

Likely in the promoter, since there was no expression of the mRNA. Mutations in the promoter region can inhibit the initiation of transcription. If the mutation occurred in the coding region, it would have no effect on transcription (you would still see a band). This is very similar to the figure in the Grape Transposon paper.

List the complete steps of Meiosis. (Meiosis I and meiosis II).

Meiosis I: Prophase I - DNA condenses into chromosomes, homologous chromosomes pair together, crossing over occurs between homologous pairs. Prometaphase I - Nuclear envelope dissolves. Metaphase I - Homologous chromosomes align at midline. Anaphase I - Homologous chromosomes disconnect, homologous chromosomes move to opposite sides. Telophase I - Nuclear envelope reforms, chromosomes decondense, cell pinches apart (cytokinesis). Meiosis II: Prophase II - DNA condenses into chromosomes. Prometaphase II - Nuclear envelope dissolves. Metaphase II - Chromosomes align at midline. Anaphase II - Sister chromatids disconnect, sister chromatids move to opposite sides. Telophase II - Nuclear envelope reforms, sister chromatids decondense, cell pinches apart (cytokinesis).

Compare the differences between Mitosis and Meiosis

Mitosis • One division which includes Prophase, Metaphase, Anaphase and Telophase • DNA replication occurs during interphase before mitosis begins • Two daughter cells, each diploid (2n) and genetically identical to the parent cell Meiosis • The role is to enable multicellular adult to arise from zygote; produces cells for growth, repair, and, in some species, asexual reproduction • Two divisions including Prophase, Metaphase, Anaphase and Telophase • DNA replication occurs during interphase before meiosis I begins • Four daughter cells, each haploid (n), containing half as many chromosomes as the parent cell; genetically different from the parent cell and from each other • Role is to produce gametes; reduces number of chromosomes by half and introduces genetic variability among the gametes

In gel electrophoresis, DNA runs from negative to positive due to what charge of the phosphodiester backbone?

Negative

Define northern blot vs southern blot tests in terms of what they are measuring.

Northern blot is running RNA on an agarose Gel, then you take a membrane and transfer the rna from the gel onto the membrane. The membrane is then exposed to a single RNA probe that has radioactivity on it, so the presence of bars will be seen. meaning the sequence was present in the cells. Measures if a particular gene was being expressed. Another way to test this is RT-PCR (rna needs to have poly a tails to work)

8.23 Do all SNPs lead to an alteration in phenotype? Explain why or why not.

Not all SNPs lead to an alteration in phenotype. Some are silent. For example, if a SNP does not lie in a DNA sequence that is transcribed, or does lie in a transcribed sequence but after mRNA processing does not alter the amino acid inserted into a polypeptide chain, it will not cause a missense or nonsense mutation and could be silent. If a SNP does also not lie in a gene regulatory region, it will not affect a gene's function and could also be silent. regulate these.

7.32 As genes have been cloned for a number of human diseases caused by defects in DNA repair and replication, striking evolutionary parallels have been found between human and bacterial DNA repair systems. Discuss the features of DNA repair systems that appear to be shared in these two types of organism.

One of the most striking parallels between human and bacterial DNA repair systems has come from the analysis of mutations that affect mismatch repair (see Question 7.17 and its answer above). Mutations in any of the four human genes hMSH2, hMLH1, hPMS1, and hPMS2 result in an increased accumulation of mutations in the genome and a hereditary predisposition to hereditary nonpolyposis colon cancer (HNPCC). Mutations in the E. coli genes mutS and mutL also result in an increased accumulation of mutations, and the normal alleles of these genes are essential for the initial stages of mismatch repair. The human gene hMSH2 is homologous to E. coli mutS, and the other three human genes are homologous to E. coli mutL, indicating that aspects of mismatch repair are shared in these two organisms. Examples of additional human genetic defects that lack functions required for DNA repair include xeroderma pigmentosum, ataxia telangiectasia, Fanconi anemia, Bloom syndrome, and Cockayne syndrome (see text Table 7.1, p. 150). While the specific DNA repair process affected by these mutations may not yet be sufficiently characterized for a direct comparison with a homologous process in E. coli, it has become clear that there are parallels in the types of repair processes that exist in the two organisms. For example, xeroderma pigmentosum and Fanconi anemia both affect repair of DNA damage induced by UV irradiation, whose mechanism of repair is well understood in E. coli.

Describe the process of Polymerase Chain Reaction and what is necessary for this process:

PCR is done many times in vitro to replicate DNA. For this process, these items will be necessary: i. DNA template ii. Two primers iii. Nucleotides iv. Magnesium v. DNA Polymerase The first stage of PCR is denaturation, where the strands of double stranded DNA denature due to high heat, typically at 95 degrees Celsius. After denaturation, the reaction is cooled to allow annealing of the two primers at their target sequences. The reaction is then heated up to 72 degrees, where elongation occurs, adding nucleotides one at a time. These stages are repeated for 30-40 times, until billions of copies of the target sequence is made. This is made possible by using a thermostable DNA polymerase, such as Taq, which is able to withstand high temperatures (95 degrees).

7.18 Two mechanisms in E. coli were described for the repair of thymine dimer formation after exposure to ultraviolet light: photoreactivation and excision (dark) repair. Compare these mechanisms, indicating how each achieves repair.

Photoreactivation requires the enzyme photolyase, which, when activated by a photon of light with a wavelength between 320 and 370 nm, splits the dimers apart. Dark repair does not require light but requires several different enzymes. First, the uvrABC endonuclease makes two single-stranded nicks, on the 5 side and the 3 side of the dimer. Then an exonuclease excises the 12-nucleotide-long segment of one strand between the nicks, including the dimer. Next, DNA polymerase I fills in the single-stranded region in the 5-to-3 direction. Finally, the gap is sealed by DNA ligase.

Write the complementary strand strand of DNA created by dideoxy sequencing. Primer: 5' GTCA3' Template: 3' AAGTCTACGGCATATG...5'

Primer: 5'GTCA'3 Template: 3' AAGTCTACGGCATATG...5' 5' TTCAGATGCCGTATAC...3'

What are the 5 general stages in either mitosis or meiosis?

Prophase, Prometaphase, Metaphase, Anaphase, Telophase

Which of these isn't considered a major class of DNA polymorphism?

RFLP

8.3 Restriction endonucleases are naturally found in bacteria. What purposes do they serve?

Restriction enzymes serve to protect their hosts from infection by invading viruses and degrade any potentially infectious foreign DNA taken up by the cell (for example, by transformation). Since restriction enzymes digest DNA (restrict it) at specific sites, any foreign DNA will be cut up. To protect its own DNA from digestion by its restriction enzyme(s), a bacterium modifies (methylates) the sites recognized by its own restriction enzymes. This prevents cleavage at these sites.

What is the role of restriction enzymes? What are the 2 types?

Restrictions enzymes recognize and cut DNA at specific sites which result in the production of either sticky ends or blunt ends. Sticky ends are staggered cuts which bind to complementary fragments that were made by similar restriction enzymes. Blunt ends are non-specific, because of this, it allows cuts made by different restriction enzymes to be combined easily.

What two enzymes are used by class 1 retrotransposons?

Reverse transcriptase and integrase

Why is it important for DNA to get packaged as chromosomes?

So it can be split easily and divide the genetic material in the process of mitosis and meiosis

*8.4 A new restriction endonuclease is isolated from a bacterium. This enzyme cuts DNA into fragments that average 4,096 base pairs long. Like many other known restriction enzymes, the new one recognizes a sequence in DNA that has twofold rotational symmetry. From the information given, how many base pairs of DNA constitute the recognition sequence for the new enzyme?

The average length of the fragments produced indicates how often, on average, the restriction site appears. If the DNA is composed of equal amounts of A, T, C, and G, the chance of finding one specific base pair (A-T, T-A, G-C, or C-G) at a particular site is 1⁄4. The chance of finding two specific base pairs at a site is (1⁄4)2. In general, the chance of finding n specific base pairs at a site is (1⁄4)n. Here, 1⁄4,096 = (1⁄4)6, so the enzyme recognizes a six-base-pair site.

In 1964, scientists isolated a mutant form of E. coli that had high levels of mutation in the dark after UV exposure. What type of repair mechanism do these cells use? How is this different from the repair mechanism used by non-mutated bacteria?

The mutant E. coli isolated in 1964 showed mutations occurring in the dark after exposure to UV light. This is different from mutations in normal E. coli exposed to UV light because normal mutations occur only in light through photoreactivation. The mutated E. coli had a mutation in the photoreactivation system which made them rely on a different repair mechanism called nucleotide excision repair. Nucleotide excision repair can occur in the absence of light as opposed to photoreactivity where as the light is needed to activate the enzyme photolyase, which splits apart the dimers caused by UV exposure. Photoreactivation is present only in bacteria, eukaryotes only utilize nucleotide excision repair.

Quiz 4B . What is present in the mRNA molecule that allows you to perform the above, even if you don't know the sequence of the gene? (3)

The presence of a polyA tail in the mRNA

How does RT-PCR and Real-time PCR differ from regular PCR? How are they similar?

They are different because both RT-PCR and Real-Time PCR begin with RNA as a template for amplification of cDNA and regular PCR begins with DNA as a template for amplification. They are similar because they can only amplify short segments.

*9.8 What information and materials are needed to amplify a segment of DNA using PCR?

To amplify a specific region, one needs to know the sequences flanking the target region so that primers able to amplify the target region can be designed. Once primers are synthesized, the polymerase chain reaction can be assembled. It contains a DNA template (genomic DNA, cDNA, or cloned DNA), the pair of primers that flank the DNA segment targeted for amplification, a heat-resistant DNA polymerase (such as Taq), the four dNTPs (dATP, dTTP, dGTP, and dCTP), and an appropriate buffer (see Figure 9.3, p. 222)

How does UV light damage DNA and what two processes can fix this damage?

UV light induces the formation of pyrimidine dimers, which are abnormal bonds between adjacent pyrimidines. The problem with these mutations is that polymerase normally cannot replicate past these abnormal bonds (and when it does, it often inserts an incorrect nucleotide, which is why exposure to UV light can induce cancer). The two ways that dimers can be fixed is by photoreactivation and nucleotide excision repair.

Explain how UV light can induce DNA damage, and how this type of damage is repaired.

UV light is capable of inducing damage in DNA because it can increase the chemical energy of the pyrimidine molecules. This change in energy often occurs between two thymine molecules, forming an abnormal bond between them and producing what is known as a thymine dimer. Thymine dimmers are fixed by the process of nucleotide excision repair. As the name suggests, nucleotide excision repair removes the thymine dimer and a small portion of the surrounding region of DNA. DNA polymerase then repairs the gap by adding the correct nucleotides back in place. DNA ligase finishes nucleotide excision repair by joining the nucleotides together, completing the strand.

12.3 Chromatids joined together by a centromere are called a. sister chromatids. b. homologs. c. alleles. d. bivalents (tetrads).

a

A human egg cell contains 40ng of DNA in its chromosomes during the G2 phase. How much DNA is contained in a cell during the following parts of the cell cycle? a) G1 Phase. b) Prophase I. c) Metaphase II. d) After Telophase II. e) After fertilization by sperm cell.

a) G1 Phase. 20ng b) Prophase I. 40ng c) Metaphase II. 20ng d) After Telophase II. 10ng e) After fertilization by sperm cell. 20ng

Name two major differences between mitosis and meiosis.

a)Mitosis completes cell division in one cycle, meiosis needs two cycles in order to complete cell division. b)Mitosis results in 2 daughter cells with their DNA being identical to their parent, meiosis results in 4 daughter cells that are genetically different from their parent.

*8.34 One powerful approach to annotating genes is to compare the structures of cDNA copies of mRNAs to the genomic sequences that encode them. Indeed, a large collaboration involving 68 research teams analyzed 41,118 full-length cDNAs to annotate the structure of 21,037 human genes (see http://www.h-invitational.jp/). a. What types of information can be obtained by comparing the structures of cDNAs with genomic DNA? b. During the synthesis of cDNA (see Figure 8.15), reverse transcriptase may not always copy the entire length of the mRNA and so a cDNA that is not fulllength can be generated. Why is it desirable, when possible, to use full-length cDNAs in these analyses? c. The research teams characterized the number of loci per Mb of DNA for each chromosome. Among the autosomes, chromosome 19 had the highest ratio of 19 loci per Mb while chromosome 13 had the lowest ratio of 3.5 loci per Mb. Among the sex chromosomes, the X had 4.2 loci per Mb while the Y had only 0.6 loci per Mb. What does this tell you about the distribution of genes within the human genome? How can these data be reconciled with the idea that chromosomes have gene-rich regions as well as gene deserts? d. When the research teams completed their initial analysis, they were able to map 40,140 cDNAs to the available human genome sequence. Another 978 cDNAs could not be mapped. Of these 978 cDNAs, 907 cDNAs could be roughly mapped to the mouse genome. Why might some (human) cDNAs be unable to be mapped to the human genome sequence that was available at the time although they could be mapped to the mouse genome sequence? (Hint: Consider where errors and limited information might exist.)

a. Comparison of cDNA and genomic DNA sequences can define the structure of transcription units by elucidating the location of the intron-exon boundaries, poly(A) sites, and the approximate locations of promoter regions. Comparison of different full-length cDNAs representing the same gene can identify the use of alternative splice sites, alternative poly(A) sites, and alternative promoters. b. The analysis of full-length cDNAs provides information about an entire open reading frame, information about the site at which transcription starts and where the promoter lies, and the location of the poly(A) site. Partial-length cDNAs might provide some but not all of this information. While partiallength cDNAs could be compared and assembled to obtain more information, their assembly as challenging because alternative splice sites, alternative promoters, and/or alternative poly(A) sites can be used. c. Genes are not uniformly distributed among different chromosomes, and some chromosomes have more genes than others. While consistent with the finding that chromosomes have gene-rich regions and gene deserts, more data is needed to infer the relationship between the density of genes on a chromosome and how gene-rich it is. For example, a chromosome with many small genes could still have regions classified as gene deserts. d. Two possible explanations are that: (1) some regions of the genome sequence were not yet correctly assembled (e.g., due to the large numbers of repetitive sequences they contain), so the cDNAs are unable to be mapped to just one region; and (2) some of the genes are in regions that have not yet been assembled (e.g., because they are difficult to clone or sequence). As the genome sequence is revised, these issues should be resolved.

*8.7 The average size of fragments (in base pairs) observed after genomic DNA from eight different species was individually cleaved with each of six different restriction enzymes is shown in Table 8.B. a. Assuming that each genome has equal amounts of A, T, G, and C, and that on average these bases are uniformly distributed, what average fragment size is expected following digestion with each enzyme? b. How might you explain each of the following? i. There is a large variation in the average fragment sizes when different genomes are cut with the same enzyme. ii. There is a large variation in the average fragment sizes when the same genome is cut with different enzymes that recognize sites having the same length (e.g., ApaI, HindIII, SacI, and SspI). iii. Both SrfI and NotI, which each recognize an 8-bp site, cut the Mycobacterium genome more frequently than SspI and HindIII, which each recognize a 6-bp site.

a. In a sequence that has a uniform distribution of A, G, C, and T, the chance of finding a 6-bp site is and (1/4)6=1/4,096,) the chance of finding an 8-bp site is In such a (1/4)8=1/65,536. sequence, ApaI, HindIII, SacI, and SspI should produce fragments that average 4,096 bp in size, and SrfI and NotI should produce fragments that average 65,536 bp in size. b. i. The large variation in average fragment sizes when one restriction enzyme is used to cleave different genomes could reflect: (1) the nonrandom arrangements of base pairs in the different genomes (e.g., there is variation in the frequencies of certain sequences that are part of the restriction site in the different genomes); and/or (2) the different base compositions of the genomes (e.g., genomes that are rich in A-T base pairs should have fewer sites for enzymes recognizing sites containing only G-C base pairs).

*7.20 After a culture of E. coli cells was treated with the chemical 5-bromouracil, it was noted that the frequency of mutants was much higher than normal. Mutant colonies were then isolated, grown, and treated with nitrous acid; some of the mutant strains reverted to wild type. a. In terms of the Watson-Crick model, diagram a series of steps by which 5BU may have produced the mutants. b. Assuming that the revertants were not caused by suppressor mutations, indicate the steps by which nitrous acid may have produced the back mutations.

a. In its normal state, 5-bromouracil is a T analog that can base pair with A. In its rare state, it resembles C and can basepair with G. It will induce an AT-to-GC transition. b. Nitrous acid can deaminate C to U, resulting in a CG-to-TA transition.

*12.11 Give the name of each stage of mitosis and meiosis at which each of the following events occurs: a. Chromosomes are located in a plane at the center of the spindle. b. The chromosomes move away from the spindle equator to the poles.

a. Metaphase: Metaphase in mitosis, metaphase I and metaphase II in meiosis b. Anaphase: Anaphase in mitosis, anaphase I and anaphase II in meiosis

*7.9 a. The sequence of nucleotides in an mRNA is Assuming that ribosomes could translate this mRNA, how many amino acids long would you expect the resulting polypeptide chain to be? b. Hydroxylamine is a mutagen that results in the replacement of an A-T base pair for a G-C base pair in the DNA; that is, it induces a transition mutation. When hydroxylamine was applied to the organism that made the mRNA molecule shown in part (a), a strain was isolated in which a mutation occurred at the 11th position of the DNA that coded for the mRNA. How many amino acids long would you expect the polypeptide made by this mutant to be? Why?

a. Six b. Three, since the UGG codon would be replaced by UAG, a nonsense (chain termination) codon, to give 5 -AUG ACC CAU UAG -3 .

What is a somatic mutation? What is a germ-line mutation? How are these two mutations different?

a. Somatic Mutation: it's the change in genetic material of a body (somatic) cell. The mutation could affect the phenotype of the person it occurs in. b. Germ-Line Mutation: a change in the genetic material of a germ-line cell, which make up gametes. This mutation can be passed down giving rise to offspring with mutant genotype in both germ-line and somatic cells. c. The two mutations are different in that a germ-line mutation is heritable and a somatic mutation is not. That means that a germ-line mutation can be passed on from parent to child and a somatic mutation is just specific to that one person.

*12.15 Consider a diploid organism that has three pairs of chromosomes. Assume that the organism receives chromosomes A, B, and C from the female parent and A¿, B¿, and C¿; from the male parent. Answer the following questions, assuming that crossing-over does not occur: a. What proportion of the gametes of this organism would be expected to contain all the chromosomes of maternal origin? b. What proportion of the gametes would be expected to contain some chromosomes of both maternal and paternal origin

a. The chance that a gamete would have a particular maternal chromosome is 1⁄2. Apply the product rule to determine the chance of obtaining a gamete with all three maternal chromosomes: P(A and B and C) = P(A) x P(B) x P(C) x (1⁄2)3 = 1⁄8. b. There are two approaches to this question. 1. List all of the possible options to obtain gametes that satisfy the specified condition—that some maternal and paternal chromosomes are present. Use the product rule to determine the probability of each and then, using the sum rule, determine the proportion of the gametes having maternal and paternal chromosomes. Gametes that satisfy the condition that both maternal and paternal chromosomes are present are ABC', AB'C, A'BC, AB'C', A'BC', and A'B'C. For each one of these six gamete types, the chance of obtaining it is 1⁄8. This is based on applying the product rule and logic similar to that used in part (a) of this question. Apply the sum rule to determine the chance of obtaining one of these types of gametes: P(ABC' or AB'C or A'BC or AB'C' or A'BC' or A'B'C) = 1⁄8 + 1⁄8 +1⁄8 +1⁄8 +1⁄8 +1⁄8 = 6⁄8 = 3⁄4. 2. Realize that the set of gametes with some maternal and paternal chromosomes is composed of all gametes except those that have only maternal or only paternal chromosomes. That is, P(gamete with both maternal and paternal chromosomes) 5 1 2 P(gamete with only maternal or only paternal chromosomes). From part (a), the chance of a gamete having chromosomes from only one parent is 1⁄8. Using the sum rule, P(gamete with both maternal and paternal chromosomes) = 1 - (1⁄8 + 1⁄8) = 3⁄4.

*8.5 An endonuclease called AvrII ("a-v-r-two") cuts DNA whenever it finds the sequence . a. About how many cuts would AvrII make in the human genome, which contains about base pairs of DNA and in which 40% of the base pairs are G-C? b. On average, how far apart (in base pairs) will two AvrII sites be in the human genome? c. In the cellular slime mold Dictyostelium discoidium, about 80% of the base pairs in regions between genes are A-T. On average, how far apart (in base

a. The enzyme recognizes a sequence that has two G-C base pairs, two C-G base pairs, one A-T base pair, and one T-A base pair in a particular order. Since 40% of the genome is composed of G-C base pairs, the chance of finding a G-C or C-G base pair is 0.20, and the chance of finding an A-T or a T-A base pair is 0.30. The chance of finding these six base pairs with this sequence is (0.20)4 × (0.3)2 = 0.000144. A genome with 3 × 109 base pairs will have about 3 × 109 different groups of 6-bp sequences. Thus, the number of sites in the human genome is (0.000144) × (3 × 109) = 432,000. b. 3 × 109 bp/432,000 sites = 1/0.000144 = 6,944 bp between sites. c. The chance of finding these six base pairs in a sequence having 80% A-T base pairs is (0.10)4 × (0.4)2 = 0.000016, so two AvrII sites will be 1/0.000016 = 62,500 bp apart.

Which of the following best describes a "mutator" gene? (3) a. A gene where mutations occur at a much higher rate than other genes. b. A gene that codes for a protein involved in DNA damage repair. c. A gene that is a class I transposon. d. A gene that can move around in the genome, mutating other genes.

b. A gene that codes for a protein involved in DNA damage repair.

*12.1 Interphase is a period corresponding to the cell cycle phases of a. mitosis. b. S. c. G1 ; S ; G2. d. G1 ; S ; G2 ; M.

c

*12.4 Mitosis and meiosis always differ in regard to the presence of a. chromatids. b. homologs. c. bivalents. d. centromeres. e. spindles.

c

*7.4 Ultraviolet light usually causes mutations by a mechanism involving (choose the correct answer) a. one-strand breakage in DNA. b. light-induced change of thymine to alkylated guanine. c. induction of thymine dimers and their persistence or imperfect repair. d. inversion of DNA segments. e. deletion of DNA segments. f. all of the above.

c. The key to this answer is the word "usually." The other choices might apply rarely, but not usually.

12.10 Which of the following does not occur in prophase I of meiosis? a. chromosome condensation b. pairing of homologs c. chiasma formation d. formation of a telomere bouquet e. segregation

e

Quiz 4a. What is the next step after you isolate mRNA from the cells? (3)

mRNA is converted into cDNA by reverse transcriptase