Genetics Midterm

A2Q9 - Now that you have come up with an equation that describes the relationship between amounts of different nucleotide bases in DNA, can you use it to predict the amounts of all four nucleotide bases when you only know the amount of one type of base? Approximately 21% of the human genome is comprised of nucleotides containing C. Given this information, calculate the percentage of the human genome that is comprised of nucleotides containing G, T, and A. (a) _____% of the human genome is comprised of G. (b) _____% of the human genome is comprised of T. (c) _____% of the human genome is comprised of A.

(a) 21% of the human genome is comprised of G. (b) 29% of the human genome is comprised of T. (c) 29% of the human genome is comprised of A. Nucleotide bases on one strand of DNA are complementary to the corresponding nucleotide bases on the other strand of DNA. Cytosine (C) binds with guanine (G), and adenine (A) binds with thymine (T). Because of this complementarity, you can calculate the amount of any nucleotide in a genome when the amount of only a single nucleotide is known.

A2Q2 - Avery, MacLeod, and McCarty performed an experiment to narrow down what type of molecule the transforming factor is. They added enzymes to the heat-killed S strain to target different types of molecules in each test. If they had then injected the different mixtures into mice, what would the results have been? Drag the labels to indicate the appropriate explanation and conclusion from each test result.

1. Control: No components destroyed Mice die 2. Protease added: Protein is destroyed Mice die 3. RNase added: RNA is destroyed Mice die 4. DNase added: DNA is destroyed Mice live DNA must be the transforming factor When DNA is destroyed with DNase, the heat-killed S strain of bacteria is no longer able to transform the R strain into a virulent strain. Destroying RNA (with RNase) or protein (with protease) does not prevent the S strain from transforming the R strain into a virulent strain. This experiment showed that DNA--not RNA or protein--must be the transforming factor.

A5Q2 - Mitosis and cell cycle terminology As in most areas of biology, the study of mitosis and the cell cycle involves a lot of new terminology. Knowing what the different terms mean is essential to understanding and describing the processes occurring in the cell. 1. DNA replication produces two identical DNA molecules, called ____ which separate during mitosis. 2. After chromosomes condense, the ____ is the region where the identical DNA molecules are most tightly attached to each other. 3. During mitosis, microtubules attach to chromosomes at the ____. 4. In dividing cells, most of the cell's growth occurs during ____. 5. The ____ is a cell structure consisting of microtubules, which forms during early mitosis and plays a role in cell division. 6. During interphase, most of the nucleus is filled with a complex of DNA and protein in a dispersed form called ____. 7. In most eukaryotes, division of the nucleus is followed by ____, when the rest of the cell divides. 8. The ____ are the organizing centers for microtubules involved in separating chromosomes during mitosis.

1. DNA replication produces two identical DNA molecules, called sister chromatid(s), which separate during mitosis. 2. After chromosomes condense, the centromere(s) is the region where the identical DNA molecules are most tightly attached to each other. 3. During mitosis, microtubules attach to chromosomes at the kinetochore(s). 4. In dividing cells, most of the cell's growth occurs during interphase. 5. The mitotic spindle(s) is a cell structure consisting of microtubules, which forms during early mitosis and plays a role in cell division. 6. During interphase, most of the nucleus is filled with a complex of DNA and protein in a dispersed form called chromatin. 7. In most eukaryotes, division of the nucleus is followed by cytokinesis, when the rest of the cell divides. 8. The centrosome(s) are the organizing centers for microtubules involved in separating chromosomes during mitosis. The key structures involved in mitosis are labeled in this diagram of an animal cell that shows the two sister chromatids of each duplicated chromosome beginning to attach to the mitotic spindle by means of their kinetochores. The centrosomes anchor the mitotic spindle at opposite ends of the cell.

A2Q2 - The virulent form of the bacteria S. pneumoniae is called the S strain because it is surrounded by a polysaccharide coat that makes it appear smooth under a microscope. Sometimes the S strain mutates into a non-virulent form (called the R strain), which lacks the polysaccharide coat and appears rough. Frederick Griffith performed an experiment in which he injected mice with different combinations of these bacterial strains. Drag the labels to indicate whether the mice live or die after each injection and the explanation for each test result.

1. Living S strain Mice die because the S strain is virulent. 2. Heat-killed S strain Mice live because heat-killing the S strain makes it non-virulent. 3. Living R strain Mice live because the R strain is not virulent. 4. Heat-killed S strain + Living R strain Mice die because a transforming factor from the heat-killed S strain made the R strain virulent The S strain of the pneumococcal bacteria is virulent and kills mice. When the S strain is heat-killed, it is no longer virulent and doesn't kill mice. The R strain is non-virulent. However, when the heat-killed S strain is mixed with the live R strain, a transformed virulent R strain is created.

A3Q7 - When cells divide or multiply, the DNA must be copied, or replicated. All known organisms replicate their DNA by a method known as semi-conservative replication, in which each of the two strands of the parent DNA becomes part of the daughter DNA. Complete the following vocabulary exercise related to DNA replication. 1. ____ are the short sections of DNA that are synthesized on the lagging strand of the replicating DNA. 2. After replication is complete, the strand of new DNA, called ____, is complementary to the parental strand. 3. The enzyme that can replicate DNA is called ____. 4. During DNA replication, an open section of DNA, in which a DNA polymerase can replicate DNA, is called a ____. 5. The new DNA strand that grows continuously in the 5' to 3' direction is called the ____.

1. Okazaki fragments are the short sections of DNA that are synthesized on the lagging strand of the replicating DNA. 2. After replication is complete, the strand of new DNA, called daughter DNA, is complementary to the parental strand. 3. The enzyme that can replicate DNA is called DNA polymerase. 4. During DNA replication, an open section of DNA, in which a DNA polymerase can replicate DNA, is called a replication fork. 5. The new DNA strand that grows continuously in the 5' to 3' direction is called the leading strand. DNA replication is a central process in biochemistry and must be done with the highest precision. If a mistake is made in the replication of DNA, the error will propagate into future generations. An error that occurs during replication is called a mutation.

A3Q9 - Synthesis of the lagging strand In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B. Drag the labels to their appropriate locations in the flowchart below, indicating the sequence of events in the production of fragment B. (Note that pol I stands for DNA polymerase I, and pol III stands for DNA polymerase III.)

1. pol III binds to 3' end of primer B --> 2. pol III moves 5' to 3', adding DNA nucleotides to primer B --> 3. pol I binds to 5' end of primer A --> 4. pol I replaces primer A with DNA --> 5. DNA ligase links fragments A and B

A2Q14 - A double stranded DNA molecule is 100,000 bp (100 kb) long. How many complete turns are there in the molecule? (Remember that there are 10 bp per turn in the B-DNA form typically seen within cells.)

10,000 complete turns

A2Q15 - What was the percent identity for the top three matches?

100.00 %

A2Q14 - A double stranded DNA molecule is 100,000 bp (100 kb) long. How many nucleotides does it contain?

200,000 nucleotides

A2Q14 - A double stranded DNA molecule is 100,000 bp (100 kb) long. Within a DNA molecule, the centers of adjacent base pairs are 0.34 nm apart. Indicate the length (in nm) of the DNA molecule described above.

34,000 nm

A5Q1 - How many haploid sets of chromosomes are present in an individual cell that is tetraploid (4n)?

4

A5Q3 - You come across a media story about an animal alleged to be the hybrid of a rabbit and a cat. Given that the cat (Felis domesticus) has a diploid chromosome number of 38 and a rabbit (Oryctolagus cuniculus) has a diploid chromosome number of 44, what would be the expected chromosome number in the somatic tissues of this alleged hybrid?

41 38 / 2 = 19 44 / 2 = 22 19 + 22 = 41

A3Q13 - Dideoxynucleotide DNA sequencing The diagram below shows an autoradiograph of a DNA sequencing gel. Type the 5' to 3' sequence of the template strand ("inferred strand") based on the pattern in this gel. Use only capital letters for the sequence.

5' - CAACTGGTCCAT - 3' The shortest fragment generated by the dideoxynucleotide DNA sequencing reaction migrates to the bottom of the gel. That fragment represents the 5' nucleotide of the sequenced strand. Therefore, the 5' to 3' sequence of nucleotides in the sequenced strand can be determined by reading the bands in all lanes from the bottom of the gel to the top: 5'-ATGGACCAGTTG-3', in this example. The template strand is complementary and antiparallel to the sequenced strand: 5'-CAACTGGTCCAT-3', in this example.

A2Q7 - What is the complementary DNA sequence for the DNA strand shown in the figure below?

5'- TAACGTCATGG -3' This sequence is complementary and in the correct orientation.

A2Q7 - Drag and drop each base to match with its partner in order to maximize hydrogen bonding.

Adenine and thymine are complementary and join via two hydrogen bonds. Guanine and cytosine are complementary and join via three hydrogen bonds.

A2Q8 - The nucleic acids DNA and RNA are made from chains of nucleotides. Nucleotides consist of three components: a five-carbon sugar (either ribose or deoxyribose), a nitrogenous base attached to the sugar's 1'-carbon, and a phosphate group attached to the sugar's 5'-carbon. Sort these nucleotide building blocks by their name or classification.

All of the genetic material in all living organisms is made from these basic building blocks of nucleotides.

A2Q15 - A BLAST search can also be done by entering the unique accession number for a sequence. When you search with the accession number NM_007305, what is the name of the gene that you find?

BRCA1 gene

A2Q15 - DNA sequence information can be stored in databases (e.g., GenBank), shared, compared, and analyzed. An NCBI application called BLAST (Basic Local Alignment Search Tool) is a program that allows scientists to search through such biological databases to compare an entered query sequence and to sequences in databases. Your textbook (pp. 178-179) offers an exercise to acquaint you with the use of the BLAST tool (http://blast.ncbi.nlm.nih.gov/Blast.cgi). Please follow the instructions for this exercise (pp. 178-179) and then answer the following questions: Notice that the top three matches to your query sequence (CCAGAGTCCAGCTGCTGCTCATACTACTGATACTGCTGGG) all refer to a single gene. What is the name of this gene?

BRCA1 gene

A3Q2 - After publication by Watson and Crick of the double helical structure of DNA, researchers proposed three possible models to describe how the two strands of the helix could serve as templates for synthesis of a new double helix in DNA replication. Matthew Meselson and Franklin Stahl used a technique called density-gradient centrifugation to identify the correct model.

Background info

A5Q6 - Meiosis guarantees that in a sexual life cycle, offspring will inherit one complete set of chromosomes (and their associated genes and traits) from each parent. The transmission of traits from parents to offspring is called heredity. Another important aspect of meiosis and the sexual life cycle is the role these processes play in contributing to genetic variation. Although offspring always resemble their parents, they are genetically different from both of their parents and from one another. The degree of variation may be tremendous. The following processes are associated with meiosis and the sexual life cycle: -DNA replication before meiosis -crossing over -chromosome alignment in metaphase I and separation in anaphase I -chromosome alignment in metaphase II and separation in anaphase II -fertilization

Both heredity and genetic variation: -DNA replication before meiosis -crossing over -chromosome alignment in metaphase I and separation in anaphase I -chromosome alignment in metaphase II and separation in anaphase II -fertilization In organisms that reproduce sexually, the processes of DNA replication, the precise pairing of homologs during crossing over, chromosome alignment and separation in meiosis I and II, and fertilization ensure that traits pass from one generation to the next. Unlike with asexual reproduction, offspring of sexual reproduction are genetically different from each other and from both of their parents. Mechanisms that contribute to genetic variation include: -errors (mutations) that occur during DNA replication -the production of recombinant chromosomes due to crossing over -the independent assortment of homologous chromosomes in meiosis I -the separation of sister chromatids (which are no longer identical due to crossing over) in meiosis II -the random fusion of male and female gametes during fertilization

A5Q9 - Write a short essay that contrasts mitosis and meiosis, including their respective roles in organisms, the mechanisms by which they achieve their respective outcomes, and the consequences should either process fail to be executed with absolute fidelity. Compared with ____, which maintains chromosomal constancy, ____ provides for a reduction in chromosome number and an opportunity for exchange of genetic material between homologous chromosomes. In ____ there is no change in chromosome number or kind in the two daughter cells, whereas in ____ numerous potentially different haploid (n) cells are produced. During oogenesis, only one of the four products of ____ is functional; however, four of the four products of ____ of spermatogenesis are potentially functional.

Compared with mitosis, which maintains chromosomal constancy, meiosis provides for a reduction in chromosome number and an opportunity for exchange of genetic material between homologous chromosomes. In mitosis there is no change in chromosome number or kind in the two daughter cells, whereas in meiosis numerous potentially different haploid (n) cells are produced. During oogenesis, only one of the four products of meiosis is functional; however, four of the four products of meiosis of spermatogenesis are potentially functional.

A2Q11 - Define base complementarity. Complementarity is based on ____ bonding between a partial ____ charge of the hydrogen atom and a partial ____ charge of the bonded oxygen and nitrogen atoms. Complementarity is responsible for the chemical attraction between ____ and ____ (uracil in RNA) and guanine and ____. This results in DNA and, to a limited extent, RNA, assuming a ____ conformation.

Complementarity is based on hydrogen bonding between a partial positive charge of the hydrogen atom and a partial negative charge of the bonded oxygen and nitrogen atoms. Complementarity is responsible for the chemical attraction between adenine and thymine (uracil in RNA) and guanine and cytosine. This results in DNA and, to a limited extent, RNA, assuming a double-stranded conformation.

A2Q8 - Sort the parts of a nucleic acid according to whether each occurs exclusively in DNA, exclusively in RNA, or in both types of nucleic acid.

DNA is used for storage of genetic information. The presence of deoxyribose as the sugar in DNA makes the molecule more stable and less susceptible to hydrolysis. The 2'-oxygen on the ribose found in RNA makes RNA much more susceptible to breakdown. It is important that mRNA be easily broken down, to ensure that the correct levels of protein are maintained in the cell.

A3Q8 - In this activity, you will demonstrate your understanding of antiparallel elongation at the replication forks. Keep in mind that the two strands in a double helix are oriented in opposite directions, that is, they are antiparallel. Drag the arrows onto the diagram below to indicate the direction that DNA polymerase III moves along the parental (template) DNA strands at each of the two replication forks. Arrows can be used once, more than once, or not at all.

DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand. Because the two parental DNA strands of a double helix are antiparallel (go from 3' to 5' in opposite directions), the direction that DNA pol III moves on each strand emerging from a single replication fork must also be opposite. For example, in the replication fork on the left, the new strand on top is being synthesized from 5' to 3', and therefore DNA pol III moves away from the replication fork. Similarly, the new strand on the bottom of that same replication fork is being synthesized from 5' to 3'. But because the bottom parental strand is running in the opposite direction of the top parental strand, DNA pol III moves toward the replication fork. In summary, at a single replication fork, one strand is synthesized away from the replication fork, and one strand is synthesized toward the replication fork. When you look at both replication forks, note that a single new strand is built in the same direction on both sides of the replication bubble.

The replication bubble and antiparallel elongation

DNA replication always begins at an origin of replication. In bacteria, there is a single origin of replication on the circular chromosome, as shown in the image here. Beginning at the origin of replication, the two parental strands (dark blue) separate, forming a replication bubble. At each end of the replication bubble is a replication fork where the parental strands are unwound and new daughter strands (light blue) are synthesized. Movement of the replication forks away from the origin expands the replication bubble until two identical chromosomes are ultimately produced.

A3Q13 - The mechanism of DNA replication The diagram below shows a double-stranded DNA molecule (parental duplex). Drag the correct labels to the appropriate locations in the diagram to show the composition of the daughter duplexes after one and two cycles of DNA replication. In the labels, the original parental DNA is blue and the DNA synthesized during replication is red.

During DNA replication, each strand in the parental duplex serves as the template for the production of a daughter strand by complementary base pairing. Therefore, one cycle of replication will produce two daughter duplexes, each with one parental strand and one newly synthesized strand. During a second cycle of replication, all four strands in the two duplexes will serve as templates, resulting in four duplexes (eight strands of DNA).

A3Q9 - RNA primers on the leading and lagging strands The diagram below shows a replication bubble with synthesis of the leading and lagging strands on both sides of the bubble. The parental DNA is shown in dark blue, the newly synthesized DNA is light blue, and the RNA primers associated with each strand are red. The origin of replication is indicated by the black dots on the parental strands. Rank the primers in the order they were produced. If two primers were produced at the same time, overlap them.

Earliest a , h b , g c , f d, e Latest As soon as the replication bubble opens and the replication machinery is assembled at the two replication forks, the two primers for the leading strands (primers a and h) are produced. The production of the first primers on the lagging strands (those closest to the origin of replication, b and g) is delayed slightly because the replication forks must open up further to expose the template DNA for the lagging strands. After completion of the first segments of the lagging strands, additional template DNA must be exposed before the second primers (c and f) can be produced. And after completion of the second segments, additional template DNA must be exposed before the third primers (d and e) can be produced. In summary, because of the way the replication bubble expands, the lagging strand primers near the origin of replication were produced before the primers near the replication forks.

A4Q6 - The second order of chromatin packing occurs when nucleosomes coil together to form a fiber that is 300 nm in diameter. True False

False The second order of chromatin packing occurs when nucleosomes coil together to form a solenoid fiber that is 30 nm in diameter.

A2Q9 - Can you identify the bases that will be added to this parent strand during DNA replication? Drag the labels to the appropriate targets to identify the sequence and orientation of the daughter strand. Blue labels can be used once, more than once, or not at all.

For complementary base-pairing in DNA, purines bind with pyrimidines. A (a purine) binds with T (a pyrimidine) through the formation of two hydrogen bonds. G (a purine) binds with C (a pyrimidine) through the formation of three hydrogen bonds. The two strands are considered to be antiparallel because they are orientated in different directions, from 5' to 3'. Thus, the 5' end of one strand aligns with the 3' end of the opposite strand.

A2Q9 - Now that you have identified the sequence of the daughter strand based on the sequence on the parent strand, can you come up with an equation that explains the relative proportions of each base in a DNA macromolecule? (Note that in the equation below, "G" represents the percentage of bases that are guanine, "T" represents the percentage of bases that are thymine, "A" represents the percentage of bases that are adenine, and "C" represents the percentage of bases that are cytosine.) Drag the labels to the appropriate targets to complete the equation.

G + A = C + T The key to solving this problem is to recall that there will be the same number of A's as T's, and there will be the same number of G's as C's. This is because of complementary base pairing: G binds to C, and T binds to A.

A4Q9 - Complete a comprehensive definition of heterochromatin. Heterochromatin is chromosomal material that ____ and remains ____ when other parts of chromosomes, such as euchromatin, are otherwise ____. Heterochromatic regions replicate ____ in ____ and are relatively ____ in a genetic sense either because there are ____ genes present or because the genes present are ____.

Heterochromatin is chromosomal material that stains deeply and remains condensed when other parts of chromosomes, such as euchromatin, are otherwise deconsended. Heterochromatic regions replicate late in S phase and are relatively inactive in a genetic sense either because there are few genes present or because the genes present are repressed.

A3Q8 - The role of DNA polymerase III In DNA replication in bacteria, the enzyme DNA polymerase III (abbreviated DNA pol III) adds nucleotides to a template strand of DNA. But DNA pol III cannot start a new strand from scratch. Instead, a primer must pair with the template strand, and DNA pol III then adds nucleotides to the primer, complementary to the template strand. Each of the four images below shows a strand of template DNA (dark blue) with an RNA primer (red) to which DNA pol III will add nucleotides. In which image will adenine (A) be the next nucleotide to be added to the primer?

In the example above, DNA pol III would add an adenine nucleotide to the 3' end of the primer, where the template strand has thymine as the next available base. You can tell which end is the 3' end by the presence of a hydroxyl (-OH) group. The structure of DNA polymerase III is such that it can only add new nucleotides to the 3' end of a primer or growing DNA strand (as shown here). This is because the phosphate group at the 5' end of the new strand and the 3' -OH group on the nucleoside triphosphate will not both fit in the active site of the polymerase.

A5Q8 - Meiosis ensures the transmission of traits from one generation to the next. At the same time, it is a key process that introduces genetic variation into the traits that offspring inherit from their parents. In this tutorial, you will explore the genetic context of meiosis.

Knowing the terms and relationships shown in this concept map will help you understand the role that meiosis plays in heredity, sexual reproduction, and genetic variability.

A5Q2 - Phases of the cell cycle The cell cycle represents the coordinated sequence of events in the life of a cell from its formation to its division into two daughter cells. Most of the key events of the cell cycle are restricted to a specific time within the cycle. In this exercise, you will identify when various events occur during the cell cycle. Recall that interphase consists of the G1, S, and G2 subphases, and that the M phase consists of mitosis and cytokinesis. Drag each label to the appropriate target.

Many organisms contain cells that do not normally divide. These cells exit the cell cycle before the G1 checkpoint. Once a cell passes the G1 checkpoint, it usually completes the cell cycle--that is, it divides. -The first step in preparing for division is to replicate the cell's DNA in the S phase. -In the G2 phase, the centrosome replicates. -In early M phase, the centrosomes move away from each other toward the poles of the cell, in the process organizing the formation of the mitotic spindle. -At the end of the M phase when mitosis is complete, the cell divides (cytokinesis), forming two genetically identical daughter cells.

A5Q8 - Animal life cycles In the life cycle of an organism, meiosis is paired with the process of fertilization. Understanding the life cycle of an organism is the key to understanding how sexual reproduction ensures the inheritance of traits from both parents and also introduces genetic variation. Complete the diagram to show the life cycle of a typical animal. Follow these steps: 1. First, drag blue labels onto blue targets only to identify each stage of the life cycle. 2. Next, drag pink labels onto pink targets only to identify the process by which each stage occurs. 3. Then, drag white labels onto white targets only to identify the ploidy level at each stage.

Meiosis creates gametes (eggs and sperm) with only a single chromosome set (haploid or n) from parental cells with two chromosome sets (diploid or 2n). During fertilization, the haploid sperm (n) and egg (n) fuse, producing a diploid zygote (2n). The cells of the zygote then divide by mitosis (which does not change the ploidy level) to produce an adult organism (still 2n) of the next generation. In sexual life cycles, meiosis and fertilization keep the number of chromosomes constant from generation to generation.

A5Q5 - Meiosis is mechanistically similar to mitosis in many ways, although it involves two sequential nuclear and cellular divisions rather than one. The two stages of meiosis are Meiosis I, which consists of prophase I, metaphase I, anaphase I, and telophase I (followed by cytokinesis) Meiosis II, which consists of prophase II, metaphase II, anaphase II, and telophase II (followed by cytokinesis) Can you recognize the eight stages of meiosis based on the location and behavior of the chromosomes? Drag the diagrams of the stages of meiosis onto the targets so that the four stages of meiosis I and the four stages of meiosis II are in the proper sequence from left to right. (Note that only one of the two daughter cells is shown for meiosis II.)

Meiosis involves two sequential cellular divisions. In meiosis I, homologous chromosomes pair and then separate. Thus, although the parent cell is diploid (containing two chromosome sets, one maternal and one paternal), each of the two daughter cells is haploid (containing only a single chromosome set). In meiosis II, the sister chromatids separate. The four daughter cells that result are haploid.

A5Q2 - Mitosis and the Cell Cycle

Mitosis is the part of eukaryotic cell division in which the nucleus divides and distributes the chromosomes to the two daughter nuclei, ensuring that each daughter cell receives chromosomes identical to those of the parent cell. Mitosis is just one part of the cell cycle, the overall sequence of events from the formation of a cell by cell division to the cell's own division into two new daughter cells.In this tutorial, you will review key terms associated with mitosis and the cell cycle, cellular processes that occur during the main phases of the cell cycle, and changes in DNA structure during the cell cycle. Before beginning this tutorial, watch the Mitosis animation. Pay particular attention to the structural changes that occur in the DNA before and during cell division.

A3Q2 - Experimental results: Banding pattern predictions for each model of replication Meselson and Stahl designed an experiment that would allow them to discern whether DNA replication occurs in a dispersive, semiconservative, or conservative manner. -They started with E. coli that had been growing for many generations in medium containing 15N. -They then transferred the bacteria into medium containing only 14N, and allowed the bacteria to undergo two rounds of DNA replication. -After each round of replication, the scientists performed density-gradient centrifugation of the DNA. The scientists reasoned that each of the three models would predict different DNA banding patterns after the two rounds of replication. Can you identify the banding patterns predicted by each model after the first round of replication? (Then, in Part C, you will identify the banding patterns predicted after the second round of replication.) Drag the test tubes to the appropriate locations in the table to show the banding patterns that each model predicts. Test tubes may be used once, more than once, or not at all.

Notice that after one round of replication, the dispersive and semiconservative models predict identical results, whereas the conservative model predicts different results. By continuing the experiment through a second round of replication, would all three models lead to different predictions?

A5Q6 - Consider a diploid cell where 2n = 6. During metaphase I of meiosis, as the pairs of homologous chromosomes line up on the metaphase plate, each pair may orient with its maternal or paternal homolog closer to a given pole. There are four equally probable arrangements of the homologous pairs at metaphase I. (Note that this problem assumes that no crossing over has occurred.)

One aspect of meiosis that generates genetic variation is the random orientation of homologous pairs of chromosomes at metaphase I. Each pair can orient with either its maternal or paternal homolog closer to a given pole; as a result, each pair sorts into daughter cells independently of every other pair.Due to independent assortment alone, a diploid cell with 2n chromosomes can produce 2 n possible combinations of maternal and paternal chromosomes in its daughter cells. For the cell in this problem (n = 3), there are 23, or 8, possible combinations; for humans (n = 23), there are 223, or 8.4 million, possible combinations. Note that when crossing over occurs, the number of possible combinations is even greater.

A4Q11 - The base composition of an RNA virus, a DNA virus, and dolphin DNA were analyzed by a very disorganized laboratory technician. He performed the analyses correctly, but lost the identification tags to the samples. Identify the source of the nucleic acid for each sample. Is the DNA virus single or double-stranded? Is the RNA virus single or double-stranded? ____ contains the dolphin DNA. ____ contains the DNA virus. The DNA in this virus is ____. ____ contains the RNA virus. The RNA in this virus is ____.

Sample 3 contains the dolphin DNA. Sample 1 contains the DNA virus. The DNA in this virus is single-stranded. Sample 2 contains the RNA virus. The RNA in this virus is single-stranded.

A4Q10 - Define satellite DNA. Describe where it is found in the genome of eukaryotes and its role as part of chromosomes. Satellite DNA is identified by ____ as additional peaks that represent DNA of a slightly different density. Satellite DNA is highly repetitive and consists of a relatively ____ number of ____ sequences. Such sequences are clustered in ____ areas devoid of genes, typically flanking the ____.

Satellite DNA is identified by sedimentation equilibrium centrifugation as additional peaks that represent DNA of a slightly different density. Satellite DNA is highly repetitive and consists of a relatively large number of short sequences. Such sequences are clustered in heterochromatic areas devoid of genes, typically flanking the centromere.

Synthesis of the lagging strand is accomplished through the repetition of the following steps.

Step 1: A new fragment begins with DNA polymerase III binding to the 3' end of the most recently produced RNA primer, primer B in this case, which is closest to the replication fork. DNA pol III then adds DNA nucleotides in the 5' to 3' direction until it encounters the previous RNA primer, primer A. Step 2: DNA pol III falls off and is replaced by DNA pol I. Starting at the 5' end of primer A, DNA pol I removes each RNA nucleotide and replaces it with the corresponding DNA nucleotide. (DNA pol I adds the nucleotides to the 3' end of fragment B.) When it encounters the 5' end of fragment A, DNA pol I falls off, leaving a gap in the sugar-phosphate backbone between fragments A and B. Step 3: DNA ligase closes the gap between fragments A and B. These steps will be repeated as the replication fork opens up. Try to visualize primer C being produced to the right (closest to the replication fork). Fragment C would be synthesized and joined to fragment B following the steps described here.

A3Q8 - The chemical structure of DNA and its nucleotides The DNA double helix is composed of two strands of DNA; each strand is a polymer of DNA nucleotides. Each nucleotide consists of a sugar, a phosphate group, and one of four nitrogenous bases. The structure and orientation of the two strands are important to understanding DNA replication. Drag the labels to their appropriate locations on the diagram below.

The DNA double helix is constructed from two strands of DNA, each with a sugar-phosphate backbone and nitrogenous bases that form hydrogen bonds, holding the two strands together. Each DNA strand has two unique ends. The 3' end has a hydroxyl (-OH) group on the deoxyribose sugar, whereas the 5' end has a phosphate group. In the double helix, the two strands are antiparallel, that is, they run in opposite directions such that the 3' end of one strand is adjacent to the 5' end of the other strand.

A3Q2 - When a solution of cesium chloride (CsCl) is subjected to high-speed centrifugation, a stable density gradient is formed. Meselson and Stahl found that when cell contents were subjected to centrifugation with a CsCl solution, a band of DNA formed at the CsCl density that matched the density of the DNA. This technique is called density-gradient centrifugation. The test tubes below show the results of density-gradient centrifugation of five different DNA samples. Drag the description of each DNA sample to the appropriate location to identify the expected appearance of the DNA band(s) after density-gradient centrifugation.

The densities of 14N/14N, 14N/15N, and 15N/15N double helices differ from each other and thus form bands in different positions. 14N/14N forms a band toward the top, 14N/15N in the middle, and 15N/15N toward the bottom. a. DNA from cells grown in 15N contains only 15N/15N double helices. b. DNA from cells grown in 14N contains only 14N/14N double helices. c. A 1:1 mixture of DNA from cells grown in 14N and cells grown in 15N contains both 14N/14N and 15N/15N double helices. d. A 1:1 mixture of DNA from cells grown in 14N and cells grown in 15N, heated and then cooled, contains 14N/14N, 14N/15N, and 15N/15N double helices. e. DNA containing one strand of 15N-DNA and one strand of 14N-DNA contains only 14N/15N double helices.

A3Q1 - In general, DNA replicates semiconservatively and bidirectionally. True False

True

A5Q4 - The end result of meiosis is four haploid daughter cells. True False

True

A2Q7 - Guanine and adenine are purines found in DNA. True False

True Guanine and adenine are indeed purines found in DNA; thymine and cytosine are the pyrimidines found in DNA.

A3Q2 - Experimental results: Banding pattern predictions after the second round of replication Can you identify the banding patterns predicted by each model after the second round of replication? Drag the test tubes to the appropriate locations in the table to show the banding patterns that each model predicts. Test tubes may be used once or not at all.

When Meselson and Stahl performed this experiment, their results were consistent with the pattern predicted by semiconservative replication, confirming that as the correct model. When DNA replicates, the two parent strands separate, and each strand serves as a template for the synthesis of a new DNA strand. Note that Meselson and Stahl were able to rule out the conservative model based on the results after one round of replication. However, they could not rule out the dispersive model at that point because that model predicts the same pattern as the semiconservative model. But, the results after two rounds did enable them to rule out the dispersive model.

A2Q2 - What conclusion(s) could Griffith draw from his experiment? Select all that apply. a. A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent. b. The transforming factor is DNA. c. The transforming factor is RNA. d. The transforming factor is protein.

a. A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent. From this experiment, Griffith concluded that a transforming factor from the heat-killed S strain transformed the R strain, making the R strain virulent. The experiment showed that there was a transforming factor, but it didn't demonstrate the nature of the transforming factor.

A4Q9 - Which of the following are examples of heterochromatin in mammals? Select the three correct examples. a. Barr body b. repetitive DNA c. X chromosome d. centromeric DNA e. telomeric DNA f. nucleosome

a. Barr body d. centromeric DNA e. telomeric DNA

A2Q11 - How do covalent bonds differ from hydrogen bonds? Drag the appropriate items to their respective bins: a. Covalent bonds b. Hydrogen bonds

a. Covalent bonds Strong bonds (difficult to break) Sharing of electrons between 2 or more atoms b. Hydrogen bonds Weak bonds (easy to break) Electrostatic charge attraction Responsible for base complementarity

A3Q5 - Is this a DNA or an RNA molecule? a. DNA b. RNA

a. DNA

A5Q10 - Which of the following statements regarding diploidy are true? Select the two correct statements. a. Diploid eukaryotic cells contain their genetic information in pairs of homologous chromosomes, with one member of each pair being derived from the maternal parent and one from the paternal parent. b. Diploidy is a term that applies to chromosome that has an even number of genes. c. Diploidy means that both members of a homologous pair of chromosomes are present. d. Diploidy is a term often used in conjunction with the symbol 2n, where n = 23 for all organisms.

a. Diploid eukaryotic cells contain their genetic information in pairs of homologous chromosomes, with one member of each pair being derived from the maternal parent and one from the paternal parent. c. Diploidy means that both members of a homologous pair of chromosomes are present.

A4Q8 - These all relate to how DNA is organized in viral, bacterial, and eukaryote chromosomes. Contrast the major differences between the organization of DNA in viruses and bacteria versus eukaryotes. Drag the appropriate characteristics to their respective bins. a. Eukaryotic genomes b. Viral or bacterial genomes

a. Eukaryotic genomes Contain large amounts of DNA Contain a mixture of both unique and repetitive DNA sequences Contain short or long, linear DNA molecules Contain DNA associated with histones that form nucleosomes Consist mostly of noncoding DNA sequences b. Viral or bacterial genomes Relatively little amount of DNA Consist of mostly unique DNA sequences Contain short, circular, or linear DNA molecule Contain DNA associated with Hu and H-NS proteins that form coils Consists mostly of coding DNA sequences

A3Q8 - Unwinding the DNA As DNA replication continues and the replication bubble expands, the parental double helix is unwound and separated into its two component strands. This unwinding and separating of the DNA requires three different types of proteins: helicase, topoisomerase, and single-strand binding proteins. Sort the phrases into the appropriate bins depending on which protein they describe. a. Helicase b. Topoisomerase c. Single-strand binding protein

a. Helicase Binds at the replication fork Breaks H-bonds between bases b. Topoisomerase Binds ahead of replication fork Breaks covalent bonds in DNA backbone c. Single-strand binding protein Binds after the replication fork Prevents H-bonds between bases At each replication fork, helicase moves along the parental DNA, separating the two strands by breaking the hydrogen bonds between the base pairs. (This makes the two parental DNA strands available to the DNA polymerases for replication.) As soon as the base pairs separate at the replication fork, single-strand binding proteins attach to the separated strands and prevent the parental strands from rejoining. As helicase separates the two parental strands, the parental DNA ahead of the replication fork becomes more tightly coiled. To relieve strain ahead of the replication fork, topoisomerase breaks a covalent bond in the sugar-phosphate backbone of one of the two parental strands. Breaking this bond allows the DNA to swivel around the corresponding bond in the other strand and relieves the strain caused by the unwinding of the DNA at the helicase.

A4Q3 - Why is DNA synthesis expected to be more complex in eukaryotes than in bacteria? Check the THREE that apply. a. In eukaryotic cells, DNA is complexed with histones. b. Eukaryotic cells contain much more DNA. c. Eukaryotic chromosomes are linear rather than circular. d. DNA replication in eukaryotes is known to proceed with much more fidelity than in bacteria.

a. In eukaryotic cells, DNA is complexed with histones. b. Eukaryotic cells contain much more DNA. c. Eukaryotic chromosomes are linear rather than circular.

A3Q9 - Drag each phrase to the appropriate bin depending on whether it describes the synthesis of the leading strand, the synthesis of the lagging strand, or the synthesis of both strands.

a. Leading strand Made continuously Only 1 primer needed Daughter strand elongates TOWARD replication fork b. Lagging strand Made in segments Multiple primers needed Daughter strand elongates AWAY from replication fork c. Both strands Synthesize 5' to 3' Because DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand and because the two parental DNA strands are antiparallel, synthesis of the leading strand differs from synthesis of the lagging strand. -The leading strand is made continuously from a single RNA primer located at the origin of replication. DNA pol III adds nucleotides to the 3' end of the leading strand so that it elongates toward the replication fork. -In contrast, the lagging strand is made in segments, each with its own RNA primer. DNA pol III adds nucleotides to the 3' end of the lagging strand so that it elongates away from the replication fork. In the image below, you can see that on one side of the origin of replication, a new strand is synthesized as the leading strand, and on the other side of the origin of replication, that same new strand is synthesized as the lagging strand. The leading and lagging strands built on the same template strand will eventually be joined, forming a continuous daughter strand.

A2Q12 - Assume that you are interested in separating two size of DNA molecules (8kb from 13kb) using polyacrylamide gels and the technique of electrophoresis. What is the relationship between DNA molecule size and the rate of migration through a gel? a. Shorter DNA molecules migrate at a faster rate than longer DNA molecules. b. Longer DNA molecules migrate at a faster rate than shorter DNA molecules.

a. Shorter DNA molecules migrate at a faster rate than longer DNA molecules.

A5Q7 - Compare and contrast properties of sister chromatids and homologous chromosomes. a. Sister Chromatids b. Homologous Chromosomes

a. Sister Chromatids Separate during anaphase of mitosis Separate during anaphase II of meiosis Contain identical nucleotide sequences prior to crossing over b. Homologous Chromosomes In a pair, one is of maternal origin, the other of paternal origin Separate during anaphase I of meiosis Crossover between them contributes to genetic diversity Homologous chromosomes are a pair of chromosomes, one maternal and one paternal, that come together during fertilization. They have the same centromere position, and the same genetic loci (genes); however the DNA sequence is not identical. Crossing over between homologous chromosomes during prophase I of meiosis generates genetic diversity in the offspring. Homologous chromosomes pair up along the midline during metaphase I of meiosis, and move apart during anaphase I. Sister chromatids are the result of the replication of a single chromosome. They are identical in DNA sequence (apart from mutation or crossing over with a chromatid from a homologous chromosome). During metaphase of mitosis, chromosomes line up down the midline and sister chromatids separate during anaphase. In meiosis, sister chromatids separate during meiosis II.

A4Q3 - How is DNA synthesis similar in the two types of organisms? Check the THREE that apply. a. Synthesis is continuous on one strand and discontinuous on the other. b. Synthesis is accompanied with disruption and reassembly of nucleosomes. c. Synthesis proceeds from multiple replication origins. d. Synthesis involves the action of telomerase enzyme. e. Synthesis requires the template, primer and the same set of raw molecular species. f. Synthesis is bidirectional.

a. Synthesis is continuous on one strand and discontinuous on the other. e. Synthesis requires the template, primer and the same set of raw molecular species. f. Synthesis is bidirectional.

A4Q5 - Eukaryotic chromosomes contain two general domains that relate to the degree of condensation. These two regions are ________. a. called heterochromatin and euchromatin b. each void of typical protein-coding sequences of DNA c. uniform in the genetic information they contain d. separated by large stretches of repetitive DNA e. void of introns

a. called heterochromatin and euchromatin

A3Q11 - DNA sequencing by the Sanger method employs which of the following for chain termination? a. dideoxynucleotides b. dinucleotides c. ribonucleotides d. deoxynucleotides

a. dideoxynucleotides dNTPs are nucleotide analogs that cause chain termination due to lack of a free 3'-hydroxyl group.

A3Q5 - In the orientation shown in the diagram, which end of the tetranucleotide is the 3' end? a. the bottom b. the top

a. the bottom

A2Q1 - All EXCEPT which of the following are characteristics of the genetic material? a. It must be replicated accurately. b. It is composed of protein. c. It must be capable of change. d. It contains all the information needed for growth, development, and reproduction of the organism.

b. It is composed of protein. Although early observations favored protein as the genetic material, subsequent experiments demonstrated that the genetic material was DNA.

A3Q10 - If the analysis of DNA from two different microorganisms demonstrated very similar base compositions, are the DNA sequences of the two organisms also nearly identical? a. Yes. The DNA sequences of the two organisms will be nearly identical. b. No. Even though the base composition of two species may be similar, sequences can vary considerably. c. No. The DNA sequences of the two organisms will not be nearly identical. They will be the same. d. No. In order for the sequences to be nearly identical, they should exhibit not only similar base compositions but also the rate of mutations.

b. No. Even though the base composition of two species may be similar, sequences can vary considerably.

A2Q4 - The basic structure of a nucleotide includes ________. a. phosphorus and sulfur b. base, sugar, and phosphate c. phosphorus and zinc d. mRNA, rRNA, and tRNA e. amino acids

b. base, sugar, and phosphate

A2Q10 - Considering the structure of double-stranded DNA, which kind(s) of bonds hold one complementary strand to the other? a. hydrophobic and hydrophilic b. hydrogen c. covalent d. ionic e. van der Waals

b. hydrogen

A3Q4 - DNA polymerase I is thought to add nucleotides ________. a. to the 5' end of the primer b. in the place of the primer after it is removed c. to the 3' end of the primer d. in a 5' to 5' direction e. on single-stranded templates without need for an primer

b. in the place of the primer after it is removed

A3Q5 - Given that the DNA strand that served as a TEMPLATE for the synthesis of this tetranucleotide was composed of the bases 5' - A C A G - 3', what are the bases in this tetranucleotide (from 5' to 3')? a. 5' - A C A G - 3' b. 5' - G A C A - 3' c. 5' - C T G T - 3' d. 5' - U G U C - 3' e. 5' - T G T C - 3'

c. 5' - C T G T - 3'

A2Q3 - Which of the following statements best represents the central conclusion of the Hershey-Chase experiments? a. When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates radioactive label. b. Phage T2 is capable of replicating within a bacterial host. c. DNA is the identity of the hereditary material in phage T2. d. Some viruses can infect bacteria.

c. DNA is the identity of the hereditary material in phage T2. Because phage DNA and not protein was associated with bacteria at the end of the experiment, it could be concluded that DNA - not protein - must be the genetic material.

A4Q4 - Viral chromosomes exist in a variety of conformations and can be made up of ________. a. RNA only b. protein- or lipid-coding sequences c. DNA or RNA d. DNA, RNA, or protein e. DNA only

c. DNA or RNA

A2Q6 - Is the accompanying figure DNA or RNA? Is the circle closer to the 5' or 3' end? a. DNA; 3' b. DNA; 5' c. RNA; 3' d. RNA; 5'

c. RNA; 3'

A4Q12 - The plot shown here represents a Cot analysis of the genome of a generic eukaryote. Thus, we can see that eukaryotes have three main classes of DNA: highly-repeated DNA, moderately-repeated DNA, and single-copy (or unique) DNA. Although snap-back DNA does not represent a bimolecular reaction, we can also include it on this plot. Snap-back DNA folds back instantly after denaturation, because it is an inverted repeat present within a single strand. No time is needed to find a complementary second strand. Snap-back DNA represents only a small portion of the genome and is often found in regions where the secondary structure of DNA is needed for a particular function, such as an interaction with a regulatory protein. Where would snap back DNA would be found on a Cot plot? a. Within moderately-repeated DNA b. To the right of single-copy DNA c. To the left of highly-repeated DNA d. Impossible to predict

c. To the left of highly-repeated DNA

A2Q8 - Identify three possible components of a DNA nucleotide. a. deoxyribose, phosphate group, uracil b. cytosine, phosphate group, ribose c. deoxyribose, phosphate group, thymine d. cytidine, phosphate group, ribose e. adenine, phosphate group, ribose f. guanine, phosphate group, ribose

c. deoxyribose, phosphate group, thymine DNA and RNA have similar structures: a pentose sugar with a nitrogenous base and a phosphate group. DNA and RNA differ in the type of pentose sugar each possesses (DNA has deoxyribose; RNA has ribose) and in one base (DNA has thymine; RNA has uracil).

A4Q2 - All EXCEPT which of the following are related to telomeres? a. the telomerase enzyme b. links to the aging process c. found in eukaryotes and prokaryotes d. short tandem repeats located at the ends of telomeres

c. found in eukaryotes and prokaryotes Telomeres are found in eukaryotes but not prokaryotes because prokaryote chromosomes are usually circular.

A2Q5 - Regarding the structure of DNA, the covalently arranged combination of a deoxyribose and a nitrogenous base would be called a(n) ________. a. ribonucleotide b. nucleotide c. nucleoside d. oligonucleotide e. monophospate nucleoside

c. nucleoside

A4Q13 - Which of the following is NOT a component of chromatin (i.e., would never be found within chromatin)? a. RNA b. histone proteins c. ribosomes d. DNA

c. ribosomes

A2Q7 - What is the complementary DNA sequence to 5′ ATGCTTGACTG 3′? a. 5′ ATGCTTGACTG 3′ b. 5′ TACGAACTGAC 3′ c. 5′ ACTCTACGTAG 3′ d. 5′ CAGTCAAGCAT 3′

d. 5′ CAGTCAAGCAT 3′ This sequence is complementary and in the correct orientation.

A2Q3 - Which of the following outcomes would be most likely if the Hershey-Chase experiments were repeated without the step involving the blender? a. Both preparations of infected bacteria would contain both P32 and S35. b. The phage would fail to infect bacteria. c. Neither preparation of infected bacteria would exhibit radioactivity. d. Both preparations of infected bacteria would exhibit radioactivity.

d. Both preparations of infected bacteria would exhibit radioactivity. Instead of being removed from the preparation, the "ghosts" would be retained. Because both bacterial preparations would include ghosts as well as viral DNA, both would be radioactive, one with P32, one with S35.

A4Q6 - What is the first order of chromatin packing? a. Formation of a solenoid b. Formation of a 300‑nm fiber c. Looping of 300‑nm fibers d. Coiling around nucleosomes

d. Coiling around nucleosomes The first order of chromatin packing occurs when DNA coils around nucleosomes, whereby DNA is reduced to about one‑third its original length

A2Q3 - The Hershey and Chase experiments involved the preparation of two different types of radioactively labeled phage. Which of the following best explains why two preparations were required? a. Each scientist had his own method for labeling phage, so each conducted the same experiment using a different isotope. b. Establishing the identity of the genetic material required observation of two phage generations. c. The bacteriophage used in the experiments was a T2 phage. d. It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment.

d. It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment. Because it was concluded that the component associated with bacteria at the end of the experiment must be the genetic material, it was critical that the component be identifiable as either DNA or protein.

A2Q7 - Which of the following statements about DNA structure is true? a. Nucleic acids are formed through phosphodiester bonds that link nucleosides together. b. Hydrogen bonds formed between the sugar‑phosphate backbones of the two DNA chains help to stabilize DNA structure. c. The pentose sugar in DNA is ribose. d. The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.

d. The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions. This statement is true; the 5′-3′ orientation of each chain runs in opposite directions.

A5Q6 - Assume that an organism exists in which crossing over does not occur, but that all other processes associated with meiosis occur normally. Consider how the absence of crossing over would affect the outcome of a single meiotic event.Which of the following statements would be true if crossing over did not occur? a. Independent assortment of chromosomes would not occur. b. The daughter cells of meiosis I would be diploid, but the daughter cells of meiosis II would be haploid. c. The four daughter cells produced in meiosis II would all be different. d. There would be less genetic variation among gametes. e. The two daughter cells produced in meiosis I would be identical. f. The two sister chromatids of each replicated chromosome would no longer be identical.

d. There would be less genetic variation among gametes. Crossing over contributes significantly to the genetic variation seen in gametes. This is because the exchange of maternal and paternal genes between the nonsister chromatids of a homologous chromosome pair creates recombinant chromosomes with unique combinations of alleles. However, crossing over is not the only process that introduces genetic variation in meiosis I. The independent assortment of homologous chromosomes (which are never identical) in meiosis I produces daughter cells that differ from each other. The effect of crossing over on genetic variation is shown below. Without crossing over, sister chromatids remain identical and thus, pairs of daughter cells would be identical. With crossing over, however, all four daughter cells are genetically unique.

A2Q15 - What is this gene's function? a. This gene is part of a noncoding sequence b. This gene codes for an enzyme involved in the Krebs cycle c. This gene is involved in the synthesis of growth hormone d. This gene is involved in maintaining genomic stability and in tumor suppression

d. This gene is involved in maintaining genomic stability and in tumor suppression

A4Q6 - What makes up the protein component of a nucleosome core? a. Eight different histone proteins b. Histone H1 protein c. One tetramer of histone proteins d. Two tetramers of histone proteins

d. Two tetramers of histone proteins The protein component of a nucleosome is composed of two tetramers of histone proteins. One tetramer is composed of two units each of histones H2A and H2B, and the other is composed of two units each of histones H3 and H4.

A5Q4 - Separation of sister chromatids occurs _______. a. in meiosis, but not in mitosis b. at anaphase II in mitosis and anaphase in meiosis c. in mitosis, but not in meiosis d. at anaphase in mitosis and anaphase II in meiosis

d. at anaphase in mitosis and anaphase II in meiosis

A4Q7 - In addition to highly repetitive and unique DNA sequences, a third category of DNA sequences exists. What is it called, and which types of elements are involved in it? a. permissive DNA; centromeres and heterochromatin b. dominant DNA; euchromatin and heterochromatin c. multiple gene family DNA; hemoglobin and 5.0S RNA d. moderately repetitive DNA; SINEs, LINEs, and VNTRs e. composite DNA; telomeres and heterochromatin

d. moderately repetitive DNA; SINEs, LINEs, and VNTRs

A5Q4 - Novel combinations of genes can arise from _______. a. reciprocal exchange of DNA between homologs during prophase II b. reciprocal exchange of DNA between sister chromatids during prophase c. the alignment of chromosomes at Metaphase II d. reciprocal exchange of DNA between homologs during prophase I

d. reciprocal exchange of DNA between homologs during prophase I

A2Q13 - When double-stranded DNA is heated to 100°C, the two strands separate because the hydrogen bonds between the strands break. When the solution is cooled, the two strands can find each other to re-form the double helix, a process called renaturation or reannealing. For example, consider the following DNA double helix 5' - GCGCGCGCGCGCGC - 3' 3' - CGCGCGCGCGCGCG - 5' Now, imagine that this DNA is heated to 100°C and then cooled, but the two separated strands never find each other. The local secondary structure of each single strand will most likely assume the shape of a ___________ . a. double helix b. bowtie c. ligand d. ribose zipper e. hairpin (stem) loop

e. hairpin (stem) loop

A4Q1 - Structures located at the ends of eukaryotic chromosomes are called ________. a. recessive inversions b. permissive mutations c. telomerases d. centromeres e. telomeres

e. telomeres

A3Q6 - The discontinuous aspect of replication of DNA in vivo is caused by ________. a. polymerase slippage b. trinucleotide repeats c. topoisomerases cutting the DNA in a random fashion d. sister-chromatid exchanges e. the 5' to 3' polarity restriction.

e. the 5' to 3' polarity restriction.

A3Q3 - DNA polymerase III adds nucleotides ________. a. to the 5' end of the primer b. in the place of the primer after it is removed c. to both ends of the primer d. to internal sites in the DNA template e. to the 3' end of the primer

e. to the 3' end of the primer

A3Q12 - You are trying to sequence a piece of DNA. You use chain terminating tags to obtain the complementary DNA fragments for your unknown DNA sequence. What technique will you use to separate your fragments according to size?

gel electrophoresis