Genetics midterm 2

12-10: In a plant, fruit color is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the results shown in the following table. Phenotype | Plant A | Plant B red, long | 46 | 4 yellow, oval | 44 | 6 red, oval | 5 | 43 yellow, long | 5 | 47 Total | 100 | 100

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12-5: The genes for miniature wings (m) and garnet eyes (g) are approximately 8 map units apart on chromosome 1 in Drosophila. Phenotypically wild-type females (m + g / mg +) were mated to miniature-winged males with garnet eyes.

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14-8: Although recombination data can be used to develop chromosome maps for linked genes, genomic techniques have greatly enhanced the speed of mapping. In fact, the genomes of many organisms have already been mapped.

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15-10: The genetic map of the T4 bacteriophage chromosome is now known.

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15-11: You isolate two mutations in a bacteriophage. One causes bumpy plaques (b) and the other produces tiny plaques (t). Previous mapping experiments have established that the genes responsible for these mutations are 10 m.u. apart. You mix phages with genotype b+ t+ and genotype b t and then use the mixture to infect bacterial cells. After collecting and plating the progeny phages, you observe a total of 2000 plaques.

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15-7: Differences in the appearance of plaques can arise from mutations that affect the phage life cycle. For example, phages with the r (rapid) mutation lyse bacteria relatively more rapidly, so give rise to larger plaques. Phages with the h (host range) mutation can lyse both the B and B/2 strains of bacteria (to give rise to clear plaques on a lawn of B and B/2), while the wildtype allele for host range only enables lysis of the B strain (and, thus, results in cloudy plaques on a lawn of B and B/2).

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15-9: The complete DNA sequence of the λ ("lambda") phage genome is now known, and many of the functional genes in the λ genome have been identified.

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16-7: In an experiment similar to the Lederberg and Tatum experiment, researchers grew two strains of E. coli in complete medium. E. coli strain AX has the genotype AmpSbio+ trp+ leu-. E. coli strain AY has the genotype AmpRbio- trp- leu+. The researchers then plated the bacteria on different media types.

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16-9: With respect to F+ and F− bacterial matings,

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17-11: How is horizontal gene transfer different from vertical gene transfer?

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17-4: You study leucine biosynthesis in a strain of Salmonella typhimurium. You have identified four new leucine auxotrophic mutants in a mutagenesis screen. You label them leuA, leuB, leuC, and leuD. Preliminary mapping of the mutant loci indicates that they all group to one side of the ala3 locus. As part of your research, you decide to map all four loci relative to the ala3 locus using bacteriophage cotransduction.

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17-6: Indicate whether generalized transduction, specialized transduction, or both types of transduction are relevant to each of the characteristics/processes outlined below.

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18-10: While working in a genetics lab over the summer, you isolated a true-breeding strain of wingless Drosophila. After sharing your results with your mentor, you learn that six other true-breeding strains of Drosophila with the same mutant phenotype have been isolated independently in your lab.

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18-6: Moths from Mars usually have dark-blue eyes. You can culture these moths by allowing adult females to lay their eggs onto an artificial medium. After the eggs hatch, the larvae eat the medium, grow, enter a pupal stage, and eventually metamorphose into an adult. You obtain two independently-isolated Martian moth mutants cyan-1 and cyan-2. When grown on artificial medium, both mutant strains have light-blue eyes instead of the wildtype dark-blue. You know that wildtype alleles at the cyan-1 and the cyan-2 loci produce enzymes in a single linear biochemical pathway for eye color. Although you do not know the order in which the enzymes act within the pathway, you do know that the enzymes turn a light-blue precursor into an intermediate metabolite, and then turn this intermediate metabolite into the dark-blue final product.

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19-5: Remember that complementation analysis allows you do determine whether or not mutations are located within the same gene. When two mutants complement each other, their offspring will show the wildtype phenotype. This occurs because their mutations are located on different genes, so the information missing from one strain can be supplied by the other.

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20-13: From the information on the translation of mRNA into proteins as well as on protein structure and function what answers would you propose to the following fundamental questions.

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20-5: In the process of transcription, the genetic information encoded in the sequence of bases that makes up a gene is "transcribed," or copied in the same language, into a strand of RNA bases. The enzyme that catalyzes this reaction is called an RNA polymerase. In eukaryotes, before the resulting strand (called pre-mRNA) leaves the nucleus, it is processed in several ways. The product of this processing is the mRNA that functions as the template for protein synthesis outside the nucleus.Before beginning this tutorial, watch the Transcription and RNA Processing animations. Pay particular attention to the base pairing that occurs during transcription and the various steps involved in RNA processing.

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21-10: In studies of frameshift mutations, Crick, Barnett, Brenner, and Watts-Tobin found that either three nucleotide insertions or deletions restored the correct reading frame.

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21-5: Use the following copolymers to decipher the genetic code: UG and UGG.

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22-13: Nondisjunction is a cell division error in which either homologous chromosomes or sister chromatids fail to separate and thus do not migrate to opposite poles. It is responsible for conditions such as monosomy (one member of a chromosome pair is missing) and trisomy (three copies of a chromosome is present rather than two). Nondisjunction can occur during the first or second meiotic division.

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22-17: Human adult hemoglobin is a tetramer containing two alpha (α) and two beta (β) polypeptide chains. The α gene cluster on chromosome 16 and the β gene cluster on chromosome 11 share amino acid similarities such that 61 of the amino acids of the α-globin polypeptide (141 amino acids long) are shared in identical sequence with the β-globin polypeptide (146 amino acids long).

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22-18: A boy with Klinefelter syndrome (47,XXY) is born to a mother who is phenotypically normal and a father who has the X- linked skin condition called anhidrotic ectodermal dysplasia. The mother's skin is completely normal with no signs of the skin abnormality. In contrast, her son has patches of normal skin and patches of abnormal skin.

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23-3: Reactive oxidants within cells are byproducts of the molecular reduction of oxygen (i.e., cellular respiration) and can cause chemical modifications in DNA. For example, electrophilic oxidants are known to create the modified base named 7,8-dihydro-8-oxoguanine (oxoG) in DNA. Whereas guanine base-pairs with cytosine, oxoG base-pairs with either cytosine or adenine.

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22-16: For a species with a diploid number of 32, indicate how many chromosomes will be present in the somatic nuclei of individuals that are haploid, triploid, tetraploid, trisomic, and monosomic.

- 2n = 32

17-4b: Using the cotransduction frequencies from Part A, create a rough map of the leuA, leuB, leuC, and leuD loci relative to the ala3 locus. Place the four loci on the map by dragging the correct label to the appropriate location. Two loci map to approximately the same location

- ala3 - leuB leuD - leuA - leuC - Creating a map from cotransduction frequencies is similar to creating a recombination map, except that a large cotransduction frequency indicates that two loci are close together and thus more likely to move as a unit to the recipient cell. Conversely, a small cotransduction frequency indicates a large map distance between loci because it is relatively unlikely that the two loci will be present on the same transduced fragment. If two loci are too far apart to be packaged together in the bacteriophage, they will have a cotransduction frequency of zero.

17-9b: You perform a cotransformation mapping experiment using two strains of E. coli. The first (donor) strain has the genotype a+ b+ c+ The second (recipient) strain has the genotype a- b- c-. The results of the experiment are shown in this data table. Genotype | % Transformation a+ b- c- 3.4 a- b+ c- 2.8 a- b- c+ 2.3 a+ b+ c- 0.095 a- b+ c+ NA a+ b- c+ 2.1 More information about cotransformation mapping experiments can be found here: https://www.ncbi.nlm.nih.gov/books/NBK21463/ This link also explains how the 'product rule' can be used to test for close gene linkage. Drag the genes to their correct locations on the gene map below.

- c - a - b - Transformation of two genes that are linked or close together in the donor cell chromosome is much more likely to occur than transformation of two genes that are not linked. The cotransformation frequency of genes that are linked is similar to the transformation frequency of either gene alone. The transformation frequency of the c gene alone (transformants a-b-c+) is 2.3, and the transformation of the b gene alone (transformants a-b+c-) is 2.8. The cotransformation frequency of the c and a genes (transformants a+b-c+) is 2.1. The similarity of this cotransformation frequency to the transformation frequency of calone indicates that the c and a genes are close together on the bacterial chromosome. Although the cotransformation frequency of the c and b genes (transformants a-b+c+) is not known, the cotransformation of the a and b genes (transformants a+b+c-) is 0.095, which indicates that the a and b genes are far apart on the bacterial chromosome. Thus, the order of the genes must be c-a-b, with c and a close together, and a and b farther apart.

18-5: You isolate a number of auxotrophic single-gene mutants (1, 2, 3, and 4) of Neurospora. These mutant strains can grow (+) when certain nutritional supplements (A, B, C, D, and/or E) are added to minimal culture medium. The table below shows the growth patterns for your mutant Neurospora strains according to the supplements provided:

-----------------------A--B--C--D--E Mutant strain 1_________+___0____+___+___0 Mutant strain 2________0___0____+___0__0 Mutant strain 3________+____+____+___+___+ Mutant strain 4________+___0_____+___0__0 No growth is indicated by a "0".

13-10: If interference is complete (i.e., I = 1.0), then what is the frequency of double crossovers?

0

15-12g: What is the extent of chromatid interference?

0 Interference = 1 - C.o.C. = 1 - 1 = 0

19-7g: Given your answer to Part F, you calculate that the frequency of recombinants is:

0.001 30 / 30,000 = 0.001

19-7i: In other words, the map distance between rIIx and rIIy is:

0.1 mu

19-7h: This is equivalent to a percent recombination of:

0.1%

19-8a: What is the map distance between mutational sites rIIa and rIId?

0.10 mu 0.05 x 2 = 0.10

13-9c: For linked genes A, B, and C, the map distance A-B is 5 map units and the map distance B-C is 25 map units. If there are 10 double crossover events out of 1000 offspring, what is the interference?

0.2 The coefficient of coincidence is 0.01/0.0125 = 0.8, so the interference is 1 - 0.8 = 0.2.

13-14g: Calculate the interference.

0.273 Interference = 1 - C.o.C. = 1 - 0.727 = 0.273

19-8b: What is the map distance between mutational sites rIIb and rIId?

0.30 mu 0.15 x 2 = 0.30

13-14f: Calculate the coefficient of coincidence.

0.727 Expected double crossover: Frequency of sc - v crossover: 0.33 Frequency of s - v crossover: 0.1 EDCO: 0.33 * 0.1 = 0.033 Observed double crossover: 0.024 C.o.C. = ODCO / EDCO = 0.024 / 0.033 = 0.727

15-12f: What is the coefficient of coincidence?

1 Expected double crossover: Frequency of h - t crossover: 0.06 Frequency of t - r crossover: 0.13 EDCO: 0.06 * 0.13 = 0.0078 Observed double crossover: 11 / 1390 = 0.008 C.o.C. = ODCO / EDCO = 0.008 / 0.0078 = 1

22-7a: Identify the mutagenic effects of deaminating agents, alkylating agents, base analogs, and intercalating agents.

1. Alkylating agents add a methyl or ethyl group to the amino or keto groups of nucleotides, changing base-pair affinities that result in mutation following DNA replication. The following is a specific example: ethylmethane sulfonate (EMS). 2. Base analogs are incorporated as thymine and adenine, but base pair with guanine and cytosine respectively. The following are specific examples: 2-amino purine and 5-bromouracil. 3. Intercalating agents insert themselves between adjacent bases. If such agents inserts themselves into a template DNA strand, an extra base chosen at random is added to the new DNA strand opposite the intercalating agent during DNA replication. If such agents inserts themselves into a new DNA strand in place of a base, a deletion mutation will result once the new DNA strand loses the intercalating agent and replicates. The following are specific examples: proflavin and ethidium bromide. 4. Nitrous acid change cytosine to uracil and adenine to hypoxanthine by converting an amino group to a keto group, thereby resulting in transition mutations. The following is a specific example: deaminating agents.

17-9a: Place the steps involved in the process of bacterial transformation in the correct order.

1. Donor cell lyses, releasing pieces of its chromosome into the environment 2. Donor cell DNA binds to a receptor site on the recipient cell 3. One strand of the donor cell DNA is degraded. 4. Transformed DNA pairs with homologous region on the recipient cell chromosome 5. Transformed DNA recombines with the recipient cell chromosome Bacterial transformation is a process in which a recipient cell takes up DNA from a lysed donor cell. The donor cell DNA can become part of the recipient cell chromosome if homologous recombination occurs in regions where the two types of DNA are similar.

20-10: The central paradigm of biochemistry holds that information flows from DNA to RNA to protein. The process of making protein from the mRNA is called translation. Translation is carried out by the ribosome, which binds to the mRNA and binds tRNA, which recognizes the codons on the mRNA and brings the appropriate amino acid with it. The ribosome forms the peptide bond between the new amino acid and the growing peptide chain. Complete the following vocabulary exercise related to the process of translation of mRNA to protein by the ribosome.

1. Initiation of translation always happens at the start codon of the mRNA. 2. Amino acids are attached to tRNAs by an enzyme called aminoacyl-tRNA synthetase. 3. The process, performed by the ribosome, of reading mRNA and synthesizing a protein is called translation. 4. The RNA that has an amino acid attached to it, and that binds to the codon on the mRNA, is called a tRNA. 5. Termination of translation happens when the ribosome hits a stop codon on the mRNA. The process of translation, or protein synthesis, is a crucial part of the maintenance of living organisms. Proteins are constantly in use and will break down eventually, so new ones must always be available. If protein synthesis breaks down or stops, then the organism dies.

23-9: Identify the various types of DNA repair mechanisms known to counteract the effects of UV radiation.

1. Photoreactivation repair is dependent on a photon-activated enzyme that cleaves thymine dimers. 2. Excision repair is the process by which an endonuclease clips out UV- induced dimers, DNA polymerase III fills in the gap, and DNA ligase rejoins the phosphodiester backbone. 3. Recombinational repair uses the corresponding region on the undamaged parental strand of the same polarity. 4. SOS repair is a process in E. coli that induces error-prone DNA replication in an effort to fill gaps by inserting random nucleotides.

12-7: You've discovered a new plant species in which yellow fruit color (Y) is dominant to green fruit color (y), and round fruit shape (R) is dominant to long fruit shape (r). The genes for fruit color and shape are located on the same chromosome and are 12 m.u. apart. Determine the proportions of progeny of each phenotype produced for this cross. Yr/yR x yr/yr

1. yellow, round: 6% 2. yellow, long: 44% 3. green, round: 44% 4. green, long: 6% To solve this problem, it is important to understand that genes which are 12 map units apart on the same chromosome are expected to undergo recombination 12% of the time. This would result in 12% recombinant gametes and 88% nonrecombinant gametes being produced. In the first parent, Yr/yR, recombinant gametes YR and yr, would be produced 12% of the time (6% each) and nonrecombinant gametes, Yr and yR would be expected 88% of the time (44% each). Since the other parent is double homozygous recessive, the only gamete this parent can produce is yr. This means that the phenotypes of the offspring from this cross will reflect the genotypes of the first parent. Therefore, 44% of offspring are expected to be yellow, long (Yr/yr), 44% green, round (yR/yr), 6% yellow, round (YR/yr) and 6% green, long (yr/yr).

13-3b: Genes X, Y, and Z are linked. Crossover gametes between genes X and Y are observed with a frequency of 25%, and crossover gametes between genes Y and Z are observed with a frequency of 5%. What is the expected frequency of double crossover gametes among these genes?

1.25% The probability of a double crossover is the product of the probabilities of the single crossovers: 0.25 x 0.05 = 0.0125, or 1.25%.

12-10b: Determine the distance between the genes.

10 mu

13-14d: What is the map distance between v and s?

10 mu SCO s-v: + v sc 46 s + + 30 Total: 76 DCO: s + sc 10 + v + 14 Total: 24 Total crossovers: 76 + 24 = 100 mu = (100/1000) x 100 = 10

15-11d: What will be the total number of plaques represented by the plaque type b t+ ?

100 plaques 2 recombinant types: b+ t b t+ Total recombinant frequency: 10% Frequency of each individual recombinant type: 10% / 2 = 5% 2000 plaques * 0.05 = 100 plaques

15-12e: You want to find out whether the formation of a crossover in one region interferes with the formation of a crossover in the other region. To calculate the extent of interference, you will first need to figure out the coefficient of coincidence. To determine the coefficient of coincidence, you will need to divide the observed number of double crossovers by the expected number of double crossovers. How many double crossover events do you expect given your answers above?

11 DCOs

17-4a: To calculate cotransduction frequencies, you use bacteriophage to transduce DNA from wild-type donor bacteria to recipient bacteria of various genotypes. For each experiment, you select for recipient cells that are prototrophic for alanine by plating cells on a master plate containing minimal medium + leucine, which allows all leu genotypes to grow. You then replica plate each master plate on minimal medium without leucine. Use the data shown below to calculate the frequency at which each leu locus cotransduces with ala3 +.

117 / 390 = 0.3 Whereas auxotrophs require supplements to grow, prototrophs will grow in the presence or absence of the supplement. Cotransduction experiments have a selected marker (in this case, the ala3 + allele) and a non-selected marker (in this case, one of the leu + alleles). Replica plating allows you to determine how often the non-selected marker is transduced along with the selected marker: the cotransduction frequency for the two loci.

15-12c: What is the map distance between the genes t and r?

13 mu Total crossovers: 88 + 75 + 6 + 5 = 174 Total progeny: 1390 mu = (174 / 1390) x 100 = 13

15-8d: Estimate the map distance between the y and z genes.

14 mu

15-8c: What is the frequency of recombination between the y and z genes?

14%

15-8b: What percentage of progeny were recombinant?

14% y+ z 480 y z+ 513 Total: 993 Frequency = (993/7000) x 100 = 14

22-16a: haploid

16 2n / n = n 32 / 2 = 16

13-12c: What is the map distance between sp and dsr?

17 mu The map distance between two genes is equal to the percentage of all detectable genetic exchanges that occur between them, which includes both single and double crossovers. Map units are calculated by adding together the observed number of recombinants, in this case SCO + DCO, and dividing by the total number of progeny, which is 850. Using the information provided, a total of 98 offspring belonged to the class of progeny in which a single crossover event occurred between sp and dsr genes (disrupted wings 52 + speck body, cinnabar eyes 46). A total of 47 offspring belonged to the double crossover class (speck body 22 + disrupted wings, cinnabar eyes 25). Map distance is calculated using the following formula: (SCO+DCO)/total number of progeny x 100 (98 + 47)/850 x 100 = 17 mu

15-12d: What is the map distance between the genes h and r?

19 mu 6 mu + 13 mu = 19 mu

18-9c: To determine whether the mutations in the seven strains leading to the wingless phenotype are in the same gene, your lab sets up complementation tests. You obtain the following results ("+" indicates that complementation was observed; "-" indicates that no complementation was observed). How many different genes are contributing to the wingless phenotype in theses mutant fly strains?

2 Recall that complementation analysis can be used to distinguish mutations in the same gene from mutations in different genes. The table lists the results of crosses of each mutant fly strain against each other mutant. When interpreting the results of a complementation test, any given pair of mutants that complement each other by producing wild-type offspring are mutations of different genes, while any given pair of mutants that produce only progeny with the mutant phenotype fail to complement one another and are mutations of the same gene. In genetic complementation analysis, the number of complementation groups equals the number of genes. As two different complementation groups have been identified in this study, two genes are responsible for the mutations observed.

18-9d: Which mutants are in the same complementation group as mutant strain 1? Select all that apply: 2 3 4 5 6 7

2 4 7 To solve this problem, recall that a complementation group consists of mutants whose phenotypes fail to complement one another. Consider the outcome of all crosses involving mutant strain 1. When mutant strain 1 is crossed with strains 2, 4, and 7, mutant progeny are produced ("-"), indicating that these mutants are in the same complementation group. Using similar reasoning, you can assign mutants 3, 5, and 6 to a separate complementation group.

13-4: Phenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous light-straw males: Phenotype Number light-straw: 22 wild-type: 18 light: 990 straw: 970 Total: 2000 Compute the map distance between the light and straw loci.

2 mu

13-9b: In a three‑point mapping experiment for the genes w-y‑ec, the following percentages of events are observed: NCO events: 65% SCO events between y and w: 15% SCO events between y and ec: 17% DCO events: 3% What is the map distance between y and ec?

20 mu The map distance between any two genes is the sum of the percentages of all detectable recombination events between them, so 17 + 3 = 20.

14-8g: Go back to the homepage for Genome Data Viewer (https://www.ncbi.nlm.nih.gov/genome/gdv/). You will see a phylogenetic map for a wide variety of organisms. Look up the karyotype for the zebrafish Danio rerio by clicking on the image of the fish. According to the karyotype (bottom right), what is the haploid number of chromosomes in this species?

25

18-12c: What ratio of speckled flowers to yellow flowers would you expect in the F1? _____ speckled : _____ yellow

3 : 1

23-8a: How many different groups are revealed based on these data?

3 groups

19-7d: The plaques produced in Part C contained wildtype recombinants. But, don't forget, recombination also produced double mutants. Although they are as common as the wildtype recombinats, they won't grow on a lawn of E. coli strain K12(λ). Thus, we can't detect them! How many plaques would have been produced by all the recombinant progeny if we could also detect the double mutants?

30 15 x 2

13-3c: Assume that the genes from the previous example are located along the chromosome in the order X, Y, and Z. What is the probability of recombination between genes X and Z?

30% Recombination frequencies between linked genes along a chromosome are additive, so the recombination frequency between genes X and Z is 25 + 5 = 30.

19-7f: You plate a unit sample of progeny phages on E. coli strain B and observe 30,000 plaques. How many plaques total were produced by progeny phages in a unit sample?

30,000

22-16e: monosomic

31 2n - 1 32 - 1 = 21

12-5b: If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes? __wild type : __ miniature wings : __ garnet eyes : __ miniature wings, garnet eyes Enter your answer as the number of flies of each phenotype separated by a colon (example: 100:300:100:300).

32 : 368 : 368 : 32 If the genes for miniature wings (m) and garnet eyes (g) are linked with 8 map units between them, then the two recombinant classes must add up to 8% of the total (4% each) and the two parental classes must add up to 92% of the total (46% each). As a result, the following distribution would be expected: wild type: 4% x 800 = 32 miniature wings: 46% x 800 = 368 garnet eyes: 46% x 800 = 368 miniature wings, garnet eyes: 4% x 800 = 32

13-12d: What is the map distance between dsr and cn?

32 mu The map distance between two genes is equal to the percentage of all detectable genetic exchanges that occur between them, which includes both single and double crossovers. Map units are calculated by adding together the observed number of recombinants, in this case SCO + DCO, and dividing by the total number of progeny, which is 850. Using the information provided, a total of 229 offspring belonged to the class of progeny in which a single crossover event occurred between dsr and cn genes (wild type 112 + disrupted wings, speck body, cinnabar eyes 117). A total of 47 offspring belonged to the double crossover class (speck body 22 + disrupted wings, cinnabar eyes 25). Map distance is calculated using the following formula: (SCO+DCO)/total number of progeny x 100 (229 + 47)/850 x 100 = 32 mu

22-16d: trisomic

33 2n + 1 32 + 1 = 33

13-14c: What is the map distance between sc and v?

33 mu SCO sc-v: s v + 150 + + sc 156 Total: 306 DCO: s + sc 10 + v + 14 Total: 24 Total crossovers: 306 + 24 = 330 mu = (330/1000) x 100 = 33

14-6a: What is the designation (i.e., the name) of the 10th band from the left in this polytene chromosome?

3A4

14-6d: Based on your answer in Part C, the white mutation must be located in band:

3C2

15-7a: How many plaque phenotypes are possible given the mutations described above?

4

15-11b: What percentage (%) of the total plaques will be represented by the plaque type b+ t+ ?

45% 2 parental types: b+ t+ b t Total parental frequency: 90% Frequency of each individual parental type: 90% / 2 = 45%

22-16b: triploid

48 3n = 3 * 16 = 48

21-2a: Based on this information, what is the minimum size of a codon for these hypothetical Martian life-forms?

5 bases In the most general case of x bases and y bases per codon, the total number of possible codons is equal to xy . In the case of the hypothetical Martian life-forms, is the minimum codon length needed to specify 17 amino acids is 5 (2^5 = 32), with some redundancy (meaning that more than one codon could code for the same amino acid). For life on Earth, x = 4 and y = 3; thus the number of codons is 4^3, or 64. Because there are only 20 amino acids, there is a lot of redundancy in the code (there are several codons for each amino acid).

21-4: Below is a partial DNA sequence used to answer the following questions.

5' - TATGCAGCACATT - 3' 3' - ATACGTCGTGTAA - 5'

13-7: Assume that two genes are 80 map units apart on chromosome II of Drosophila and that a cross is made between a doubly heterozygous female and a homozygous recessive male. What percent recombination would be expected in the offspring of this type of cross?

50%

13-5: What is the theoretical limit of observed recombination due to crossing over?

50% Since two non-sister chromatids in a tetrad remain unchanged, the maximum percentage of recombinant gametes that could be produced is 50%.

21-13a: If the codon specifying glycine is GGA, how many different bases could be substituted into this codon to result in an amino acid substitution at position 210? (A base substitution from G at the 2nd position to A at the 2nd position would count as one change.) Include any changes from an amino acid codon to a different amino acid codon and any changes from an amino acid codon to a stop codon.

6

15-12b: What is the map distance between the genes h and t?

6 mu Total crossovers: 41 + 35 + 6 + 5 = 87 Total progeny: 1390 mu = (87 / 1390) x 100 = 6

22-16c: tertraploid

64 4n = 4 * 16 = 64

14-8d: Notice that the chromosome (shown in a horizontal position at the very top of page) is divided into two sections (arms) based on the location of the centromere (in pink). As you already learned, gene positions can be ascribed to either the short arm (called p) or the long arm (called q). Positions of genes are further located to a region (represented by a distinctive pattern of light and dark bands) and a sub-band within a light or dark area. For example, a chromosomal location of 7q11.21 in the human genome would refer to region 11 within the long arm of chromosome 7, with the sub-band location 21. Without zooming in further on chromosome 7, how would the chromosomal location of a gene on the outermost part of the short arm of chromosome 7 be indicated?

7p22

18-12h: What ratio of speckled flowers to yellow flowers to green flowers would you expect in the F2? _____ speckled: _____ yellow: _____ green (Hint: The classical dihybrid ratio is 9:3:3:1. So, 3/4 × 3/4 = 9/16 of the progeny would be AAB-C- and, thus, speckled. And 3/4 × 1/4 = 3/16 of the progeny would be AAB-cc and, thus green. How many would be yellow?)

9 : 4 : 3

15-11a: What percentage (%) of the total plaques will be represented by the parental plaque types?

90% 10 mu = 10% recombinant type 100% progeny - 10% recombinant type = 90% parental type

15-11c: What will be the total number of plaques represented by the plaque type b+ t+ ?

900 plaques 2000 plaques * 0.45 = 900 plaques

18-4b: Which metabolite(s) would be present in excess in cells that are mutant for enzyme 1? Select all that apply: A B C D

A

18-5a: Take a look at the results for mutant strain 3. According to the table, this strain will be able grow in the presence of which supplements? Select all that apply: A B C D E

A B C D

20-9a: Once elongation is underway, tRNAs involved in the process occupy a series of sites on the complexed ribosome. The occupation of sites occurs in the following order.

A Site, P Site, E Site tRNAs associate with sites on the ribosome in the order listed

22-15: Chromosomal mutations are changes in the normal structure or number of chromosomes. The diagram below shows two normal chromosomes in a cell. Letters represent major segments of the chromosomes. The following table illustrates some structural mutations that involve one or both of these chromosomes. Identify the type of mutation that has led to each result shown.

A deletion is the loss of part of a chromosomal segment. A duplication is the repetition of a segment. The repeated segment may be located next to the original or at a different location, and its orientation may be the same as the original or the reverse. An inversion is the removal of a segment followed by its reinsertion into the same chromosome in the reverse orientation. A translocation is the transfer of a segment to a nonhomologous chromosome. Translocations may be reciprocal (two nonhomologous chromosomes exchange segments) or nonreciprocal (one chromosome transfers a segment without receiving one).

22-9b

A functional sequence of nucleotides, a gene, is likely to be the product of perhaps a billion or so years of evolution. Each gene and its protein product function in an environment that has also evolved. A coordinated output of each gene product is required for life. Deviations from the norm, caused by mutation, are likely to be disruptive because of the complex and interactive environment in which each gene product must function. However, on occasion a beneficial variation occurs.

22-13b: Which of the following offspring genotypes could result from fertilization of a normal haploid sperm with an oocyte that had undergone nondisjunction in meiosis II? Hint: It may be helpful to sketch the outcome of nondisjunction to solve this problem. Select the FIVE answers that apply: AAa Aaa aaa Aa aa a

AAa aaa Aa aa a During meiosis I, the chromosome with gene A and its homolog containing gene a would have separated. During meiosis II, sister chromatids separate. If nondisjunction occurs, both sister chromatids migrate into same daughter cell, instead of different daughter cells. If this occurs in Jill, there are two possible outcomes. It is possible to produce gametes containing two copies of gene A or none at all with the nondisjuction event, with normal division occurring in the other daughter cell, producing gametes with one copy each of gene a. Alternatively, gametes containing two copies of gene a or none at all could be produced by the nondisjunction event, with normal division occurring in the other daughter cell, producing gametes with one copy each of gene A. Considering Jack is giving sperm containing one copy of gene a, all of the following offspring genotypes are possible: AAa, aaa, Aa, aa, and a.

22-13a: Jill is heterozygous for gene A and is going to have a child with Jack, who is homozygous recessive for gene A. Which of the following offspring genotypes could result from fertilization of a normal haploid sperm (from Jack) with an egg (from Jill) that had undergone nondisjunction in meiosis I? To solve this problem, it is very helpful to sketch out the consequences of nondisjunction for Jill's eggs (gametes). Select the TWO answers that apply: AAa Aaa aaa Aa aa a

Aaa a During meiosis I, homologous chromosomes separate. If nondisjunction occurs, both homologous chromosomes migrate into the same daughter cell instead of different daughter cells. If this happens in the mother, Jill, the chromosome containing gene A and the chromosome containing gene a end up in one cell, leaving the other daughter cell without a chromosome containing the gene. After meiosis II, Jill would produce gametes containing either Aa or gametes lacking that gene. If these gametes are fertilized with a normal haploid sperm containing gene a, offspring with Aaa or a genotypes are possible.

22-6: Below is a partial DNA sequence (the original sequence with no mutations); only the coding strand is shown.

Assume the sequence is transcribed and translated from left to right with the reading frame as indicated. 5' - | GGC | GTG | GTA | TTA | GCG | - 3'

20-6: During transcription, RNA polymerase synthesizes RNA from a DNA template with the help of accessory proteins. In this tutorial, you will review the steps of transcription in eukaryotes and bacteria and investigate splicing of mRNAs in eukaryotes. The diagram below shows a length of DNA containing a bacterial gene. Drag the labels to their appropriate locations in the diagram to describe the function or characteristics of each part of the gene.

Bacterial transcription is a four-stage process. 1. Promoter recognition: RNA polymerase is a holoenzyme composed of a five-subunit core enzyme and a sigma ( σ ) subunit. Different types of σ subunits aid in the recognition of different forms of bacterial promoters. The bacterial promoter is located immediately upstream of the starting point of transcription (identified as the +1 nucleotide of the gene). The promoter includes two short sequences, the -10 and -35 consensus sequences, which are recognized by the σ subunit. 2. Chain initiation: The RNA polymerase holoenzyme first binds loosely to the promoter sequence and then binds tightly to it to form the closed promoter complex. An open promoter complex is formed once approximately 18 bp of DNA around the -10 consensus sequence are unwound. The holoenzyme then initiates RNA synthesis at the +1 nucleotide of the template strand. 3. Chain elongation: The RNA-coding region is the portion of the gene that is transcribed into RNA. RNA polymerase synthesizes RNA in the 5′ → 3′ direction as it moves along the template strand of DNA. The nucleotide sequence of the RNA transcript is complementary to that of the template strand and the same as that of the coding (nontemplate) strand, except that the transcript contains U instead of T. 4. Chain termination: Most bacterial genes have a pair of inverted repeats and a polyadenine sequence located downstream of the RNA-coding region. Transcription of the inverted repeats produces an RNA transcript that folds into a stem-loop structure. Transcription of the polyadenine sequence produces a poly-U sequence in the RNA transcript, which facilitates release of the transcript from the DNA.

14-7b: With which band is the u allele associated?

Band 3

14-7a: With which band is the x allele associated?

Band 6

15-9a: Based on the genetic map for this bacteriophage, it is apparent that there is a _____________ distribution of genes along the λ chromosome. non-random random

Based on the genetic map for this bacteriophage, it is apparent that there is a [non-random] distribution of genes along the λ chromosome.

13-6: Why is a 50 percent recovery of single-crossover products the upper limit, even when crossing over always occurs between two linked genes?

Because crossing over occurs at the four-strand stage of the cell cycle, notice that each single crossover involves only two of the four chromatids.

20-11: In eukaryotic cells, the nuclear DNA codes for the synthesis of most of the cell's proteins. Each step of protein synthesis occurs in a specific part of the cell. In addition, various forms of RNA play key roles in the processes of protein synthesis.

Before beginning this tutorial, watch the Protein Synthesis animation. Pay particular attention to the cellular locations where the various steps of protein synthesis occur in eukaryotic cells, as well as the different types of RNA and how they function.

19-10

Benzer used crosses involving deletion mutants to simplify his goal of completing the fine-structure map of the rII region of bacteriophage T4. Because deletions involve the loss of a segment of DNA, deletion mutations cannot revert (back-mutate) to a wild-type state. Further, deletions can affect more than one gene. For example, a deletion may affect both the genes rIIA and rIIB within the rII region of bacteriophage T4. Finally, deleted portions of DNA cannot participate in recombination, so cannot produce wildtype recombinants. To put it simply, cross-over cannot take place when the deletion overlaps onto the area to be crossed. In contrast, a cross between two point mutants can produce wildtype recombinants.

16-14b: Where is the F factor located in Hfr strain 4?

Between the genes b and e

16-9b: How was it established that chromosome transfer was unidirectional?

By treating cells with streptomycin, an antibiotic, it was shown that recombination between two bacterial strains would not occur if one of the strains was initially treated with streptomycin and, thus, inactivated. If the other strain was treated, recombination would occur. Thus, directionality was suggested, with one strain being a donor strain and the other being the recipient.

18-5c: According to the table, which must be the metabolite that is produced at the very end of this biochemical pathway? (Hint: the addition of the final product should result in growth of all the mutant strains.)

C

18-4a: Which of the metabolite(s) would be missing in cells that are mutant for enzyme 2? Select all that apply: A B C D

C D

21-4c: If transcription occurred from left to right, what would be the codon associated with the second amino acid of the growing polypeptide chain? (Hint: the initiation of translation requires a start codon.)

CAG Grouping RNA nucleotides into codons does not start until the AUG start codon. The AUG in this sequence begins with the second nucleotide from the 5' end of the coding strand (top strand). If AUG/Methionine is the first codon, then the second codon is CAG which encodes glutamine (gln).

19-14

Charles Yanofsky examined the relationship between the linear sequence of mutant sites in a gene and the linear sequence of amino acids in the corresponding polypeptide. He worked with the trpA gene in E. coli. The trpA gene codes for the A polypeptide chain within the multimeric enzyme - tryptophan synthase. Yanofsky ordered the mutations in the trpA gene to produce a detailed genetic map. For each mutant strain, he also determined the amino acid sequence of the alteredA polypeptide chain. Yanofsky found an exact match between the sequence of the mutational sites in the trpA gene and the location of the corresponding altered amino acids in the A polypeptide chain. As an example, he found that when two mutational sites were farther apart in map units, there were more amino acids between the two amino acid substitutions in the polypeptide. Therefore, Yanofsky demonstrated colinearity -- the correspondence between the linear sequence of the gene and that of the polypeptide.

12-8: Colored aleurone in the kernels of corn is due to the dominant allele C. The recessive allele c, when homozygous, produces aleurone that lacks pigment. The plant color (not the kernel color) is controlled by another gene with two alleles, G and g. The dominant G allele results in green color, whereas the homozygous presence of the recessive g allele causes the plant to appear yellow. In a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained.

Colored, green : 88 Colored, yellow : 12 Unpigmented, green : 8 Unpigmented, yellow : 92

19-8: You conduct two-factor phage crosses with four rII mutant strains of T4. Because wildtype r+ recombinants will be the only progeny phages to produce plaques on E. coli strain K12(λ), you are able to calculate the % of wildtype recombinants for each two-point cross, as shown below.

Cross | % of wildtype (r+) recombinants rIIa × rIIb_______0.10 rIIa × rIIc_______0.25 rIIa × rIId_______0.05 rIIb × rIIc_______0.35 rIIb × rIId_______0.15 rIIc × rIId_______0.20

20-14: Briefly describe the role of ribosomes in the process of translation in prokaryotes.

During initiation a functional ribosome will contain the following components: mRNA, charged tRNA(fmet), large and small ribosomal subunits, initiation factors, GTP, Mg2+ During elongation a functional ribosome will contain the following components: mRNA, charged tRNA, large and small ribosomal subunits, elongation factors, GTP, Mg2+, peptidyl transferase During termination a functional ribosomes will contain the following components: mRNA, large and small ribosomal subunits, GTP-dependent release factors, GTP, Mg2+

18-5b: The supplements A, B, C, D, and E all happen to be metabolites in the same linear biochemical pathway. The fact that mutant strain 3 is able to grow when either the supplements A, B, C, or D are provided, but NOT when the supplement E is provided, suggests that the first metabolite in this linear pathway must be:

E

18-5d: Write out the entire order of the biochemical pathway.

EBDAC

16-13: Indicate the types of matings that are possible. Select the THREE correct answers: F' × F- F+ × F' F+ × F- F+ × Hfr Hfr × F- Hfr × F'

F' × F- F+ × F- Hfr × F-

16-10: List all major differences between the F+, F−, Hfr, and F′ bacteria.

F+ bacteria -possess the F factor F- bacteria -lack the F factor Hfr bacteria -the F factor is integrated into the bacterial chromosome F' bacteria -the F factor possesses a piece of the bacterial chromosome

15-10a: The physical chromosome of the T4 bacteriophage is circular. TRUE FALSE

FALSE

16-6: The "interrupted mating technique" can be used on Drosophila to produce a genetic map. TRUE FALSE

FALSE

17-11a: In horizontal gene transfer, genetic material is passed down from one generation to another within a species. TRUE FALSE

FALSE

17-11b: Horizontal gene transfer may play a role in the evolution of prokaryotes (e.g., in the spread of bacterial antibiotic resistance), but does not play a role in the evolution of eukaryotes (e.g., in humans). TRUE FALSE

FALSE

18-6Ee: As mentioned above, you don't actually know the order in which the enzyme products of the cyan-1 and cyan-2 loci act in the generation of eye color, but would like to find out! Thus, you grind up cyan-2 adults, mix them with artificial medium, and feed this to cyan-1 larvae. You find that moths with wildtype (dark-blue) eyes are produced. However, if you grind up cyan-1 adults, mix them with artificial medium, and feed this to cyan-2 larvae, only moths with mutant (light-blue) eyes are produced. The gene product of cyan-2 must occur first in the biochemical pathway. (Hint: A cyan-1 mutant will be wildtype for the cyan-2 locus, and vice versa.) TRUE FALSE

FALSE

18-6c: Both a cyan-1 mutant and a cyan-2 mutant would display wildtype dark-blue eyes if the precursor were provided. TRUE FALSE

FALSE

19-2: Different mutational sites within a gene will give rise to only one type of altered polypeptide. TRUE FALSE

FALSE

19-4: Recombination occurs between genes, but never within genes. TRUE FALSE

FALSE

19-5b: In other words, to construct a fine-structure map, one should perform crosses between mutants that complement each other. TRUE FALSE

FALSE

19-7c: You coinfect E. coli strain B with rIIx and rIIy - two separate mutants in the rII locus. You plate progeny phages on E. coli strain K12(λ) and observe that 15 plaques were produced from your unit sample. These plaques contain the double mutant recombinants. TRUE FALSE

FALSE

12-6: If two genes are physically located on the same chromosome, then these two genes will be linked (i.e., they will be inherited together). TRUE FALSE

FALSE Linkage means that two loci are close enough together on one chromosome that they are inherited together (or, at least, are inherited together more frequently than if they were on completely separate chromosomes, i.e., they are inherited together more than 50% of the time). But, what if these two loci were at the opposite ends of ONE very long chromosome? Would they still tend to be inherited together? No! Because the chance for crossover and recombination INCREASES with distance between loci, to the point that any loci that are more than 50 map units apart behave as if they were on separate chromosomes. For purposes of definitions, I suggest the following: 1. Linked genes: genes that are located close to one another on a chromosomes, making them likely to be inherited together, i.e., they are physically linked. Such genes may exhibit complete linkage (are always inherited together) or incomplete linkage (genes tend to be inherited together, but this relationship is sometimes broken up by crossovers). 2. Unlinked genes on separate chromosomes. 3. Genes that are on the same chromosome, but that are so far apart that they behave as if they were on separate chromosomes. We could NOT refer to these types of genes as "linked". Crossovers will completely break up any physical relationship between these two genes.

13-9a: To construct a mapping cross of linked genes, it is important that the genotypes of only SOME of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny. TRUE FALSE

FALSE To construct a mapping cross of linked genes, it is important that the genotypes of all of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny, taking into consideration that the homozygote produced only recessive gametes. Gametes and their genotypes can never be observed directly.

13-11: Two different female Drosophila were isolated, each heterozygous for the autosomally linked genes black body (b), dachs tarsus (d), and curved wings (c). These genes are in the order d−b−c, with b closer to d than to c. Shown in the following table is the genotypic arrangement for each female, along with the various gametes formed by both.

Female A (d b + / + + c) 1. d b c 2. + + + 3. + + c 4. d b + 5. d + + 6. + b c 7. d + c 8. + b + Female B (d + + / + b c) 1. d b + 2. + + c 3. d + c 4. + b + 5. d b c 6. + + + 7. d + + 8. + b c

13-14e: Are there more or fewer double crossovers than expected?

Fewer than expected

15-1: Describe the temporal sequence of the bacteriophage life cycle. Start with binding to the host cell. Rank from first to last.

First event virus binds to the bacterial host cell phage DNA in the head is injected into the bacterial host cell bacterial synthesis of DNA, RNA, and proteins is inhibited; degradation of the bacterial host DNA is initiated; some synthesis of viral molecules begins phage DNA replication occurs and synthesis of the components of head, tail, and tail fibers is completed mature viruses are assembled from phage DNA, head, tail, and tail fiber components Last event

22-2a: If a segment of DNA were replicated without any errors, the replicated strand would have the following sequence of nucleotides: 5' - ACTACGTGA - 3'. Sort the following replicated DNA sequences by the type of point mutation each contains (frameshift, base substitution, or neither), as compared to the correct sequence shown above.

Frameshift mutation 5'-ACTCGTGA-3' 5'-ACTTACGTGA-3' Base substitution mutation 5'-ACTACGTGT-3' 5'-ACTAAGTGA-3' A base substitution mutation can occur if the DNA polymerase inserts the wrong nucleotide base as it synthesizes a new strand of DNA. A frameshift mutation can occur if the DNA polymerase leaves out a nucleotide or adds an extra nucleotide to the sequence. Certain forms of cancer occur because of mutations in DNA sequences that are located in so-called mutational hotspots. These hotspots are locations in the DNA sequence where mutations occur more often than in other places.

14-8e: Notice that the chromosome (top of page) is highlighted within a box. Drag the right side bar of this box to resize the box so that it fits over the very end of the short arm. Chage your magnification to 50% (using the slide bar between magnifying glasses). Then, use the side-arrows (below chromosome) to move away from the end of the short arm (from left to right) until you are about 880 kb away from the end. What is the name of the gene that spans the location 876 kb to 896 kb?

GET4

21-5c: Which codons are present in the UGG copolymer? Select all that apply: GGT GGU UGU UGA GUG UGG

GGU GUG UGG The possible codons are determined by repeating the copolymer. Within a UGGUGGUGG sequence of RNA, it is possible to have a UGG codon, a GGU codon and a GUG codon. It is not possible to have a UGU codon, because the copolymer is repeated UGG.

21-5f: Based on the results of the copolymer experiment, what triplet code can definitely be assigned to valine?

GUG Both copolymers have the GUG codon and only have Valine in common.

21-5a: Which codons are present in the UG copolymer? Select all that apply: UGG GUG GUA TGT UGU

GUG UGU

17-6a: Phages that participate in this type of transduction carry DNA of only bacterial origin; and never of both bacterial and viral origins at the same time.

Generalized transduction When phages that are produced in a host cell package viral DNA correctly, then they carry DNA of only of viral origin. However, sometimes instead of viral DNA, bacterial DNA is accidentally packaged instead; this results in phage heads carrying DNA of only of bacterial origin. Generalized transduction occurs when this bacterial DNA is introduced into another bacterium.

12-10a: Determine the genotypes of the unknown plants A and B.

Genotype of plant A: Ro / rO Genotype of plant B: RO / ro

16-14: You mix 4 different Hfr strains (all with genotype a+ b+ c+ d+ e+ ) with an F- strain that has the genotype: a b c d e. You interrupt conjugation at regular intervals and determine the appearance of chromosomal genes from each Hfr strain in recipient cells. You obtain the following order of gene transfer for the four Hfr strains examined:

Hfr1 b+ e+ a+ c+ d+ Hfr2 a+ c+ d+ b+ e+ Hfr3 a+ e+ b+ d+ c+ Hfr4 b+ d+ c+ a+ e+

16-14c: The polarity of the F factor in Hfr strain 4 is the same as the polarity of the F factor in which other strain?

Hfr3

14-6c: Using the same figure, indicate the deletion mutant Drosophila strains for which the white mutation will be uncovered (i.e., so that pseudodominance can be observed). Select all that apply. I II III IV V VI

II III IV

20-6b: The following eukaryotic structural gene contains two introns and three exons.Eukaryotic structural gene with two introns and three exons The table below shows four possible mRNA products of this gene. Use the labels to explain what may have caused each mRNA.

If a mutation alters a splicing signal sequence of an intron, that intron will not be removed accurately during the splicing reaction. This will result in the production of an abnormally spliced mature mRNA. Mutations in promoter sequences will affect transcription initiation and are likely to result in no mRNA being produced.

18-11: Complementation tests can be used to determine whether two independently-isolated mutants with the same phenotype have mutations within the same gene.

If mutations are on the same gene, then complementation analysis will NOT restore the wildtype phenotype. In other words, the two mutants do not complement each other.

21-2: Life as we know it depends on the genetic code: a set of codons, each made up of three bases in a DNA sequence and corresponding mRNA sequence, that specifies which of the 20 amino acids will be added to the protein during translation.

Imagine that a prokaryote-like organism has been discovered in the polar ice on Mars. Interestingly, these Martian organisms use the same DNA → RNA → protein system as life on Earth, except that... -there are only 2 bases (A and T) in the Martian DNA, and -there are only 17 amino acids found in Martian proteins.

13-11c: What is the relative frequency of each class of gametes? Rank from least frequent to most frequent.

Least frequent Double crossovers Single crossovers from d to b Single crossovers from b to c Noncrossovers Most frequent

19-13

Let's discuss the difference between recombination and complementation. The process of recombination refers to the creation of new combinations of genes through chromosome breakage and rejoining. Recombinant progeny have new genotypes that are different from the original parental genotypes. On the other hand, complementation involves the mixture of gene products. This may occur if two chromosomes are in the same cell and each supply a missing function. Following complementation, chromosomes remain unaltered.

15-2: What are the differences between the lytic cycle and lysogeny when bacteriophage infection occurs?

Lytic cycle: Integration of phage DNA into the bacterial chromosome does not occur. There is no extensive latent period. Lysogeny: Integration of phage DNA into the bacterial chromosome occurs. There is an extensive latent period.

21-8b: During translation, nucleotide base triplets (codons) in mRNA are read in sequence in the 5' → 3' direction along the mRNA. Amino acids are specified by the string of codons. What amino acid sequence does the following mRNA nucleotide sequence specify? 5′− AUG GCA AGA AAA −3′

Met-Ala-Arg-Lys An amino acid sequence is determined by strings of three-letter codons on the mRNA, each of which codes for a specific amino acid or a stop signal. The mRNA is translated in a 5' → 3' direction.

21-8c: Before a molecule of mRNA can be translated into a protein on the ribosome, the mRNA must first be transcribed from a sequence of DNA. What amino acid sequence does the following DNA template sequence specify? 3′− TAC AGA ACG GTA −5′

Met-Ser-Cys-His Before mRNA can be translated into an amino acid sequence, the mRNA must first be synthesized from DNA through transcription. Base pairing in mRNA synthesis follows slightly different rules than in DNA synthesis: uracil (U) replaces thymine (T) in pairing with adenine (A). The codons specified by the mRNA are then translated into a string of amino acids.

21-10b: If the code were a sextuplet (consisting of six nucleotides), would the reading frame be restored by the addition or loss of three nucleotides? YES NO

NO

21-10d: nine nucleotides? YES NO

NO

13-11b: For female B (d + + / + b c), identify which gametes are noncrossovers, single crossovers, or double crossovers.

Noncrossovers (NCOs): d + + + b c Single crossovers (SCOs): + + + d b c d + c + b + Double crossovers (DCOs): d b + + + c

13-11a: For female A (d b + / + + c), identify which gametes are noncrossovers, single crossovers, or double crossovers.

Noncrossovers (NCOs): d b + + + c Single crossovers (SCOs): d + c + + + d b c + b + Double crossovers (DCOs): d + + + b c

14-6e: Which of the 5 genes shown were not uncovered by any of the deletions used in this study? Select all that apply: Notch zeste zw2 white roughest

Notch roughest

16-5b: During conjugation, cells gain genes in the order they occur in the donor cell chromosome. Genes that are near the origin of transfer will be transferred first. Use the graph to help you construct a map of the four genes.

Origin - met - his - pro - bio - The order of the lines on the interrupted mating graph (from highest to lowest) is the same as the order of the genes (from nearest the origin to farthest from the origin). Genes near the origin take less time to transfer during conjugation; genes far from the origin take more time to transfer. When mating is interrupted, most cells will contain the genes nearest the origin. In contrast, genes that are farthest from the origin will be present in fewer cells when mating is interrupted.

13-12: Assume that a female fly that has disrupted wings (dsr) and a speck body (sp) is mated to a male that has cinnabar eyes (cn). Phenotypically wild-type F1 female progeny were mated to males that had speck bodies, disrupted wings and cinnabar eyes, and the following progeny were observed.

Phenotype Number of offspring wild-type 112 disrupted wings 52 speck body 22 cinnabar eyes 235 disrupted wings, speck body 241 disrupted wings, cinnabar eyes 25 speck body, cinnabar eyes 46 disrupted wings, speck body, cinnabar eyes 117

13-14: In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F1, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F2 generation, and 1000 offspring were counted, with the results shown in the table. No determination of sex was made in the data.

Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14

15-8: You have isolated two mutations that affect plaque morphology in phages (y and z). You conduct a genetic cross between the two phage strains by coinfecting bacteria with a wild-type strain and a strain that carries both mutations. Phage chromosomes enter the host and replicate. You collect some of the progeny phages and plate them onto a lawn of bacterial cells. You observe the following number of plaques:

Plaque phenotype | Number y+ z+ 3042 y+ z 480 y z+ 513 y z 2965 Total 7000

22-2b: When a base substitution mutation occurs, one nucleotide in a replicating DNA sequence is substituted for another, which results in the production of a mutant strand of DNA. The result of the mutation depends on how the substituted nucleotide base alters the string of amino acids coded by the mutant DNA. The three types of base substitution mutations are nonsense mutations, missense mutations, and silent mutations. Each type is defined by how it affects protein synthesis. Label the four mutated DNA segments shown below according to the type of point mutation each represents. Use the codon table above to determine how each mutation would affect the amino acid coding for each segment.

Point mutations in DNA sequences can profoundly affect protein synthesis, or they can have no effect at all. Point mutations can be beneficial to an organism but are more commonly neutral or harmful.

13-14h: Does this represent positive or negative interference?

Positive interference

18-4: Here is a metabolic pathway in which precursor A is converted into the final product D. The sequence of steps in this process is as follows:

Precursor A --Enzymes 1--> B --Enzyme 2--> C --Enzymes 3--> D Each of the enzymes shown is the product of a different gene.

15-12: You infect E. coli cells with two strains of phage λ. One strain has a mutant host range, produces larger-than-normal plaques, and is temperature sensitive (genotype: h r t ). The other strain expresses the wild-type phenotype for these characters. Progeny phages from the lysed E. coli cells are collected and plated on a lawn of bacteria. You observe the following number of plaques for each progeny phage phenotype (which directly corresponds to its genotype):

Progeny phage genotype | Number of plaques h+ r+ t+ 562 h r t 578 h+ r t 41 h r+ t+ 35 h+ r t+ 88 h r+ t 75 h+ r+ t 6 h r t+ 5

17-6b: In this type of transduction, a piece of bacterial chromosome is acquired by the phage chromosome, i.e., phages carry viral DNA covalently linked to bacterial DNA.

Specialized transduction Specialized transduction occurs when a prophage incorrectly excises from a bacterial chromosome and takes a part of this bacterial chromosome with it. Thus, this genetic material is now a mix of both phage DNA and bacterial DNA. This is now a transducing phage, because once this phage infects the next bacterium, this bacterial DNA will also enter the infected cell and can recombine with the chromosome of the new host.

21-8a: Use the table to sort the following ten codons into one of the three bins, according to whether they code for a start codon, an in-sequence amino acid, or a stop codon.

Start/Methionine AUG Stop codon UAA UGA UAG Amino acid CAC AUC UGC AAA GCA ACU Nearly every mRNA gene that codes for a protein begins with the start codon, AUG, and thus begins with a methionine. Nearly every protein-coding sequence ends with one of the three stop codons (UAA, UAG, and UGA), which do not code for amino acids but signal the end of translation.

22-4a: Suppose that a transient tautomeric shift occurred in the guanine base to produce a rare tautomer in the partial DNA sequence just prior to a round of DNA replication. Which base would be added opposite this rare tautomer during DNA replication?

T When the replicative polymerase encounters this rare guanine tautomer, thymine is added to the newly synthesized strand instead of cytosine.

12-2: If two gene loci are on nonhomologous chromosomes, genes at these loci are expected to assort independently. TRUE FALSE

TRUE

13-2: The cross GE/ge X ge/ge produces the following progeny: GE/ge 404, ge/ge 396, gE/ge 97, Ge/ge 103. From these data, one can conclude that there are 20 map units between the G and E loci. TRUE FALSE

TRUE

15-6: Viral mutations and variants are often categorized by changes in host range and/or plaque morphology. TRUE FALSE

TRUE

15-9b: Indeed, the genes are clustered by function, so that one area might consist entirely of genes whose products are required for assembly of the structure of the phage (e.g., head and tail proteins). TRUE FALSE

TRUE

18-6a: If the enzyme product of the cyan-2 locus acts before the enzyme product of the cyan-1 locus in the linear pathway, then cyan-2 mutants will accumulate the light-blue precursor. TRUE FALSE

TRUE

18-6b: If the enzyme product of the cyan-2 locus acts before the enzyme product of the cyan-1 locus in the linear pathway, then cyan-1 mutants will accumulate the intermediate metabolite. TRUE FALSE

TRUE

18-6d: A cyan-1 cyan-2 double mutant will express mutant light-blue eyes. TRUE FALSE

TRUE

19-11a: Recombination can occur between the point mutant r1 and the deletion mutant r11. TRUE FALSE

TRUE

19-11b: Your answer to Part A indicates that the deleted segment of deletion mutant r11 does not span the point mutation present in mutant r1. TRUE FALSE

TRUE

19-1: A single gene can have multiple sites at which mutations occur. TRUE FALSE

TRUE

19-5a: To construct a fine-structure map, you want mutations in the strains you are working with to be located within the same gene. TRUE FALSE

TRUE

22-12b: Nondisjunction is viewed as a major cause of aneuploidy. TRUE FALSE

TRUE

22-14: A deletion may set up a genetic circumstance where a recessive allele is expressed in an otherwise heterozygous organism. TRUE FALSE

TRUE

22-1b: Of the two cell lines that can contain a mutation in an organism -- somatic line and germ line -- a mutation in the germ line is more consequential to subsequent generations. TRUE FALSE

TRUE

23-4: Postreplication repair is a system that responds after damaged DNA has escaped repair and failed to be completely replicated. TRUE FALSE

TRUE

16-1: An E. coli nutritional mutant, described as an auxotroph, requires minimal medium with supplementation of an organic component for growth. TRUE FALSE

TRUE An auxotroph is a mutant bacterium that has lost its ability to synthesize one or more organic components, so that component must be added to compensate for the mutation. Wild-type bacteria, described as prototrophs, do not require this supplementation.

23-2c: Thymine dimers can be repaired by Photoreactivation Repair or Nucleotide Excision Repair. TRUE FALSE

TRUE Both Photoreactivation Repair and Nucleotide Excision Repair will target UV-induced pyrimidine dimers in DNA.

20-8c: Each aminoacyl tRNA synthetase is specific for one amino acid and a small number of tRNAs. TRUE FALSE

TRUE Each aminoacyl tRNA synthetase enzyme recognizes only one amino acid, but each enzyme can often recognize several tRNAs because there is usually more than one codon for each amino acid.

13-3a: Two genes that are separated by 10 map units show a recombination percentage of 10%. TRUE FALSE

TRUE One map unit is equal to 1% recombination between two genes; 10 map units would be equal to 10% recombination between the genes.

16-12: Describe the origin of F′ bacteria and merozygotes.

The F+ element can enter the host bacterial chromosome, and upon returning to its independent state, it may pick up a piece of the bacterial chromosome. When transferred to a bacterium with a complete chromosome, a partial diploid, or merozygote, is formed

20-3a: Identify the the following elements on a diagram of a transcription bubble.

The RNA polymerase reads the template strand from the 3' to 5' direction, and synthesizes RNA from 5' to 3'. The template strand is the DNA strand that is based paired to RNA as transcription proceeds. In this example, the RNA is paired with the bottom strand. The bottom strand is oriented with 3' end to the left and 5' end to the right. The coding strand (non-template) is complementary and anti-parallel to the template strand. Therefore the coding strand is oriented with the 5' end to the left and the 3' end to the right.

12-9a: Imagine that you have joined a genetics lab on campus and want to gain experience mapping genes in Drosophila. You conduct a series of two-point mapping crosses involving five genes located on chromosome III and obtain the following percentage of recombinant offspring: h-cu: 23.5% eyg-h: 9.0% eyg-cu: 14.5% rai-eyg: 18.5% rai-h: 9.5 rai-cu: 33.0% th-cu: 6.8% th-h: 16.7% th-eyg: 7.7% th-rai: 26.2% Considering that rai is near the end of chromosome III, construct a map of these genes.

The easiest way to approach this problem is to look for the most distant loci. Given that rai is near the end of chromosome III, the gene furthest from it is cu, as these genes are 33 mu apart. The intermediate loci can then be filled in using the data from the two-point crosses. When determining the location of an unknown gene on the map, it is helpful to reference the recombination frequency of the unknown gene with respect to both rai and cu. To ensure that the location of the unknown gene is actually between rai and cu and not outside of them, the map distance between rai and the unknown gene and between cu and the unknown gene should be approximately the sum of the distance between rai and cu.

22-7c: This is because proflavin can result in a frameshift when the mutation occurs within the protein coding region of a gene. This type of mutation often results in a different set of amino acids being specified by the codons following the site of mutation.

The effect of such mutations can be reversed by inducing a mutation at a second site within the mutant. For example, if the original mutation was a(n) deletion, the reading frame can be restored with an addition mutation. When such a reversion occurs within the same gene, it is referred to as an intragenic suppressor mutation. Although this second mutation has restored the wildtype phenotype, this individual is an example of a double mutant.

16-5a: Using the data in Figure 1, label each line on the results graph below. Drag the phenotype to the appropriate line on the graph.

The graph shows the time in minutes in which each phenotype started to appear. Data in this format are useful for determining gene order during interrupted mating experiments.

19-7b

The rII locus contains multiple mutational sites. Thus, it is possible to isolate separate mutants for this locus, such as rIIx and rIIy. When E. coli strain B is coinfected with rIIx and rIIy, a small proportion of recombinant progeny might arise if a crossover occurs within the rII locus. These progeny could be assayed for by growth on E. coli strain K12(lambda) on which only wildtype phages can grow. Although this would allow for the detection of r+ recombinants, it would not allow for detection of the rIIxy double mutants, which are as likely to occur. Thus, the total number of recombinant phages will always be double the number of the r+ recombinants.

20-12: Identify the the following elements on a diagram of translation.

The small subunit complexes with eukaryotic initiation factors and a charged tRNA Met and is recruited to the 5' cap. Then the initiation complex scans the mRNA for the start codon in the Kozak sequence at which point the large subunit joins the complex. Initiation factors disassociate and are replaced by elongation factors as the ribosome begins to translate the mRNA 5' to 3' generating a polypeptide beginning at the amino terminal end (N-term). As translation proceeds, charged tRNAs enter the A site. Peptidyl transferase forms a peptide bond between amino acids in the P and A sites. Uncharged tRNAs are released to the E site.

23-8: Presented here are hypothetical findings from studies of heterokaryons formed from seven human xeroderma pigmentosum (XP) cell strains.

These data are measurements of the occurrence or nonoccurrence of unscheduled DNA synthesis in the fused heterokaryon. None of the strains alone shows any unscheduled DNA synthesis.

22-6a: This sequence represents a single mutation. Mutation #1: 5' - | GGC | GCG | GTA | TTA | GCG | - 3' Complete these sentences about this mutated DNA sequence.

This mutant DNA sequence is the result of a transition (transition or transversion?) mutation. The effect of this base substitution on the amino acid sequence results in a missense mutation. The second codon changed from GTG to GCG as the result of a transition mutation (pyrimidine to pyrimidine). This point mutation results in a single amino acid change from valine to alanine, causing a missense mutation.

22-6b: This sequence represents a single mutation. Mutation #2: 5' - | GGC | GTG | GTA | TAA | GCG | - 3' Complete these sentences about this mutated DNA sequence.

This mutant DNA sequence is the result of a transversion (transition or transversion?) mutation. The effect of this base substitution on the amino acid sequence results in a nonsense mutation. The fourth codon changed from TTA to TAA as the result of a transversion mutation (pyrimidine to purine). This point mutation results in a change from a codon for leucine to a stop codon, causing a nonsense mutation.

22-6c: This sequence represents a single mutation. Mutation #3: 5' - | GGC | GTG | GTC | TTA | GCG | - 3' Complete these sentences about this mutated DNA sequence.

This mutant DNA sequence is the result of a transversion (transition or transversion?) mutation. The effect of this base substitution on the amino acid sequence results in a silent mutation. The third codon changed from GTA to GTC as the result of a transversion mutation (purine to pyrimidine). Although this point mutation results in a codon change, both codons still code for valine, causing a silent mutation.

19-6: To create a fine-structure map, one should first isolate non-complementary mutants and perform genetic crosses between them. In very rare cases, crossing over in the area between two mutational sites within a gene will occur, leading to recombinant progeny

To construct a fine-structure map of our intragenic mutational sites, all we have to do is collect the mix of parental and recombinant progeny phages and identify the recombinant types. This number can be used to calculate recombination frequency. From the frequency of recombinants, we can estimate the genetic distance between the two mutational sites.

20-11a: RNA is a part of many cellular processes, particularly those associated with protein synthesis: transcription, RNA processing, and translation. Drag the labels to the appropriate bins to identify the step in protein synthesis where each type of RNA is utilized. If an RNA is produced in one process, and then utilized in another process, choose the bin for the process in which it is utilized. If an RNA does not play any role in protein synthesis, drag it to the "not used in protein synthesis" bin.

Transcription/RNA processing: pre-mRNA Translation: rRNA mature mRNA tRNA Not used in protein synthesis: RNA primers In eukaryotes, pre-mRNA is produced by the direct transcription of the DNA sequence of a gene into a sequence of RNA nucleotides. Before this RNA transcript can be used as a template for protein synthesis, it is processed by modification of both the 5' and 3' ends. In addition, introns are removed from the pre-mRNA by a splicing process. The product of RNA processing, mRNA (messenger RNA), exits the nucleus. Outside the nucleus, the mRNA serves as a template for protein synthesis on the ribosomes, which consist of catalytic rRNA (ribosomal RNA) molecules bound to ribosomal proteins. During translation, tRNA (transfer RNA) molecules match a sequence of three nucleotides in the mRNA to a specific amino acid, which is added to the growing polypeptide chain. RNA primers are not used in protein synthesis. RNA primers are only needed to initiate a new strand of DNA during DNA replication.

20-5a: Suppose that a portion of double-stranded DNA in the middle of a large gene is being transcribed by an RNA polymerase. As the polymerase moves through the sequence of six bases shown in the diagram below, what is the corresponding sequence of bases in the RNA that is produced?

UGAGCC There are three principles to keep in mind when predicting the sequence of the mRNA produced by transcription of a particular DNA sequence. 1. The RNA polymerase reads the sequence of DNA bases from only one of the two strands of DNA: the template strand. 2. The RNA polymerase reads the code from the template strand in the 3' to 5' direction and thus produces the mRNA strand in the 5' to 3' direction. 3. In RNA, the base uracil (U) replaces the DNA base thymine (T). Thus the base-pairing rules in transcription are A→U, T→A, C→G, and G→C, where the first base is the coding base in the template strand of the DNA and the second base is the base that is added to the growing mRNA strand.

21-4f: If transcription occurred from right to left, what would be the codon associated with the second amino acid of the growing polypeptide chain?

UGC The RNA sequence is synthesized 5' to 3' by base complementarity with the template strand. If the top strand is the template, then the RNA sequence should be identical to the bottom strand except with uracil (U) instead of thymine (T).Translation then starts at the first AUG from the 5' end of the RNA. RNA nucleotide are not grouped into codons until the AUG. If the first codon is AUG/Methonine, then the second codon is UGC which encodes cysteine.

17-5: You conduct a three-factor transduction cross with trp+ pyr+ qts+ as the donor and trp pyr qts as the recipient. You select for the marker trp+ in recipient cells and determine the number of times that the other (unselected) markers from the donor were transferred. You obtain the following information:

Unselected Markers | Number pyr qts+ 24 pyr+ qts+ 8 pyr+ qts 0

21-12: When the amino acid sequences of insulin isolated from different organisms were determined, some differences were noted. For example, alanine was substituted for threonine, and serine was substituted for glycine at corresponding positions in the protein.

Use Figure 13.7 to answer the following questions.

22-4: Spontaneous mutations can arise following DNA replication when DNA bases undergo a tautomeric shift. Tautomers have the same chemical composition, but the hydrogen atom is in a different position. This change can result in different hydrogen bonding configurations. Stable common tautomeric forms make up the standard base pairing rules (A-T and G-C). Rare unstable tautomers have altered base pairing rules, as shown in this figure.

Use this partial DNA sequence (the original sequence with no mutations) to answer the following questions. 5' - C A A - 3' 3' - G T T - 5'

20-13b: What experimental information verifies that certain codons in mRNA specify chain termination during translation?

When UAA, UGA, or UAG triplets occur at internal sites in genes, premature translation termination occurs and verifies the chain-terminating function of these triplets.

22-3: Shown here are the amino acid sequences of the wild-type and three mutant forms of a short protein.

Wild-type: Met-Trp-Tyr-Arg-Gly-Ser-Pro-Thr Mutant 1: Met-Trp Mutant 2: Met-Trp-His-Arg-Gly-Ser-Pro-Thr Mutant 3: Met-Cys-Ile-Val-Val-Val-Gln-His

19-7a

Wildtype phage T4 can grow on the E. coli strain(s) B and K12(lambda). The rII mutants of T4 can grow on the E. coli strain(s) B only. This means that rII mutants display distinct host range properties. Further, the rII mutants of T4 lyse bacteria more rapidly and, thus, give rise to larger plaques than their wildtype counterparts. These distinct phenotypes associated with rII mutants allow for the differentiation of r+ wildtype phages and rII phages.

23-8b: Which strains fall into the same complementation groups?

XP2 XP1 XP3 XP4 XP6 XP5 XP7

21-10c: six nucleotides? YES NO

YES

21-13c: Is it possible that a single-base substitution in GGA (gly) could result in a stop codon? NO YES

YES

19-12c: Indicate the zone within which point mutant r2 may be located.

Zone A

19-12b: Using the deletion map, indicate the zone in which point mutant r1 may be located.

Zone C

19-12d: Indicate the zone within which point mutant r3 may be located.

Zone D

14-8c: The number of genes on chromosome 7 is shown within the Exon Navigator box (in blue). According to Genome Data Viewer, about how many genes are on chromosome 7? a. 2800 genes b. 28,000 genes c. 280,000 genes

a. 2800 genes

22-4b: Suppose that the transient rare guanine tautomer shifted back to the common guanine tautomer prior to a second round of replication. Which DNA sequence(s) would be present in the sister chromatids after this second round of replication? Select the TWO that apply: a. 5' - C A A - 3' 3' - G T T - 5' b. 5' - T A A - 3' 3' - G T T - 5' c. 5' - C G G - 3' 3' - G C C - 5' d. 5' - U A A - 3' 3' - A U U - 5' e. 5' - T A A - 3' 3' - A T T - 5'

a. 5' - C A A - 3' 3' - G T T - 5' e. 5' - T A A - 3' 3' - A T T - 5' After the first round of DNA replication, the DNA sequence of the sister chromatids will be: (Note: G* is the rare guanine tautomer.) 5' - CAA - 3' 3' - GTT - 5' 5' - TAA - 3' 3' - G*TT - 5' After transient rare guanine tautomer shifted back to the common guanine tautomer prior to a second round of replication, the DNA sequence of the sister chromatids will be: (Note: Italics represent the template strand used during the second DNA replication.) 5' - TAA - 3' 3' - ATT - 5' 5' - CAA - 3' 3' - GTT - 5' (Note that this sister chromatid is present after the first replication and after the second replication.) Now a permanent mutation has been incorporated into one of the sister chromatids as a result of a tautomeric shift.

18-8b: The original one-gene-one-enzyme hypothesis was not supported by data. Which of the following statements best describes the data that refuted the one-gene-one-enzyme hypothesis? a. An enzyme may be made of more than one polypeptide chain b. Genes do not code for enzymes c. Enzymes are not examples of proteins d. Enzymes only exhibit tertiary structure

a. An enzyme may be made of more than one polypeptide chain

20-13a: To carry out its role, each transfer RNA requires at least four specific recognition sites that must be inherent in its tertiary structure. What are they? Check the FOUR correct answers: a. Attachment of the specific amino acid b. Interaction with the ribosome c. Interaction with the aminoacyl tRNA synthetase d. Interaction with the codon (anticodon) e. Interaction with the aminoacyl mRNA synthetase f. Attachment of the specific nucleotide

a. Attachment of the specific amino acid b. Interaction with the ribosome c. Interaction with the aminoacyl tRNA synthetase d. Interaction with the codon (anticodon)

16-11: Why are the recombinants produced from an Hfr×F− cross rarely, if ever, F+? a. Because part of the F factor is the last element to be transferred, the likelihood for complete transfer is low. b. Because the conjugation tube is fragile, no transfer can occur. c. Because the F factor is the first element to be transferred, the likelihood for complete transfer is high. d. Because the F factor within an Hfr cell is free in the cytoplasm, it cannot "infect" an F- cell.

a. Because part of the F factor is the last element to be transferred, the likelihood for complete transfer is low.

14-8a: The NCBI Genes and Disease website allows you to view human chromosome maps and the location of specific disease genes within these chromosomes. Access the Genes and Disease site at: https://www.ncbi.nlm.nih.gov/books/NBK22183/. Under "Contents", click on "Chromosome Map" (bottom of page) to see an image of a karyotype of human chromosomes. Click on chromosome 7. According to the NCBI Genes and Disease website, what is a disease associated with chromosome 7? a. Cystic fibrosis b. Phenylketonuria c. Tangier disease d. Gyrate atrophy

a. Cystic fibrosis

23-5: DNA damage brought on by a variety of natural and artificial agents elicits a wide variety of cellular responses. In addition to the activation of DNA repair mechanisms, there can be activation of pathways leading to apoptosis (programmed cell death) and cell-cycle arrest. Why would apoptosis and cell-cycle arrest often be part of a cellular response to DNA damage? a. DNA repair mechanisms often cannot reduce the impact of mutations. b. Cell-cycle arrest and apoptosis consume less energy. c. Cell-cycle arrest and apoptosis are milder ways to cope with mutations. d. Many cells do not have DNA repair mechanisms.

a. DNA repair mechanisms often cannot reduce the impact of mutations.

18-3c: Which of the following can be inferred from the Beadle and Tatum experiments? a. For a mutation resulting in the production of a defective enzyme involved in a biosynthetic pathway, the compound preceding the corresponding step will accumulate. b. A given Neurospora mutant can be deficient in either a vitamin synthesis pathway or an amino acid synthesis pathway, but not in both. c. Neurospora vitamin auxotrophs do not require amino acids in order to grow. d. X-ray-induced mutations are confined to genes encoding enzymes.

a. For a mutation resulting in the production of a defective enzyme involved in a biosynthetic pathway, the compound preceding the corresponding step will accumulate. The defective enzyme is unable to convert the precursor to the next compound in the pathway. Therefore, the precursor accumulates.

23-7: The graph (modified from Kraemer, 1997. Proc. Natl. Acad. Sci. (USA) 94: 11-14) depicts the age of onset of skin cancers in patients with or without XP, where the cumulative percentage of skin cancer is plotted against age. The non-XP curve is based on 29,757 cancers surveyed by the National Cancer Institute, and the curve representing those with XP is based on 63 skin cancers from the Xeroderma Pigmentosum Registry. Based on the graph, which of the following statements are true? Select the FOUR that apply: a. For non-XP individuals, the chance of skin cancer increases as they age. b. At all ages, the non-XP individuals have half the rate of skin cancer as individuals with XP. c. XP individuals almost always develop skin cancer by age 40. d. By age 20, approximately 80% of the XP population has skin cancer. e. Individuals with XP are more likely to develop skin cancer in their youth than non-XP individuals.

a. For non-XP individuals, the chance of skin cancer increases as they age. c. XP individuals almost always develop skin cancer by age 40. d. By age 20, approximately 80% of the XP population has skin cancer. e. Individuals with XP are more likely to develop skin cancer in their youth than non-XP individuals.

21-12b: Which of the following single base changes could result in serine being substituted for glycine (i.e., in glycine being converted into serine)? a. G changed to A in the first position of the codon. b. G changed to U in the first position of the codon. c. U changed to C in the third position of the codon.

a. G changed to A in the first position of the codon.

22-17b: Which of the following statements describe the Ohno's hypothesis regarding the origin of new genes during evolution? Check the THREE statements that apply: a. Gene duplication is a source of new genes appearing during evolution. b. If an essential gene is present as a single copy, the gene is likely to be eliminated by natural selection. c. Copies of essential genes may undergo multiple mutations, possibly leading to acquisition of a new function. d. If an essential gene is present as a single copy, mutations in it are likely to be eliminated by natural selection.

a. Gene duplication is a source of new genes appearing during evolution. c. Copies of essential genes may undergo multiple mutations, possibly leading to acquisition of a new function. d. If an essential gene is present as a single copy, mutations in it are likely to be eliminated by natural selection.

22-10: Most mutations in a diploid organism are recessive. Why? a. In most cases, the amount of product from one gene of each pair is sufficient for production of a normal phenotype. b. In most cases, the product from the mutated gene is usually degraded by the product from the normal gene. c. In most cases, the product from the mutated gene has a higher affinity to the biological targets of the normal genetic product.

a. In most cases, the amount of product from one gene of each pair is sufficient for production of a normal phenotype.

22-3d: The wild-type RNA consists of nine triplets. What is the role of the ninth triplet in wild-type RNA? a. It is a stop codon. b. It is an amino acid codon. c. It is a start codon.

a. It is a stop codon.

22-9a: Which of these concepts that affect gene mutations in diploid organisms would apply to a haploid organism such as E. coli. Select all that apply: a. Mutations have a wide range of effects on organisms depending on the type of base-pair alteration, the location of the mutation within the chromosome, and the function of the affected gene product. b. Mutations may be homozygous, but it is more likely that most new mutations will be heterozygous or hemizygous. c. Mutations are a source of genetic variation and provide the raw material for natural selection. d. Mutations comprise any change in the base-pair sequence of DNA. e. Mutations can occur spontaneously as a result of natural biological and chemical processes, or they can be induced by external factors, such as chemicals or radiation.

a. Mutations have a wide range of effects on organisms depending on the type of base-pair alteration, the location of the mutation within the chromosome, and the function of the affected gene product. c. Mutations are a source of genetic variation and provide the raw material for natural selection. d. Mutations comprise any change in the base-pair sequence of DNA. e. Mutations can occur spontaneously as a result of natural biological and chemical processes, or they can be induced by external factors, such as chemicals or radiation.

20-5b: After transcription begins in eukaryotes, several steps must be completed before the fully processed mRNA is ready to be used as a template for protein synthesis on the ribosomes. Which THREE statements correctly describe the processing that takes place before a mature mRNA exits the nucleus? a. Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes. b. Coding sequences called exons are spliced out by ribosomes. c. A cap consisting of a modified guanine nucleotide is added to the 5' end of the pre-mRNA. d. A translation stop codon is added at the 3' end of the pre-mRNA. e. A poly-A tail (50-250 adenine nucleotides) is added to the 3' end of the pre-mRNA.

a. Noncoding sequences called introns are spliced out by molecular complexes called spliceosomes. c. A cap consisting of a modified guanine nucleotide is added to the 5' end of the pre-mRNA. e. A poly-A tail (50-250 adenine nucleotides) is added to the 3' end of the pre-mRNA. Once RNA polymerase II is bound to the promoter region of a gene, transcription of the template strand begins. As transcription proceeds, three key steps occur on the RNA transcript: 1. Early in transcription, when the growing transcript is about 20 to 40 nucleotides long, a modified guanine nucleotide is added to the 5' end of the transcript, creating a 5' cap. 2. Introns are spliced out of the RNA transcript by spliceosomes, and the exons are joined together, producing a continuous coding region. 3. A poly-A tail (between 50 and 250 adenine nucleotides) is added to the 3' end of the RNA transcript. Only after all these steps have taken place is the mRNA complete and capable of exiting the nucleus. Once in the cytoplasm, the mRNA can participate in translation.

19-7e: Now that you have estimated the total number of plaques produced by recombinants, you can find out the frequency of recombination by dividing the total number of recombinant plaques by the total number of plaques produced by: a. Progeny phages b. Double mutant phages c. Wildtype phages d. Parental phages

a. Progeny phages

20-4b: An intron is a section of ________. a. RNA that is removed during RNA processing b. protein that is clipped out posttranslationally c. DNA that is removed during DNA processing d. transfer RNA that binds to the anticodon e. carbohydrate that serves as a signal for RNA transport

a. RNA that is removed during RNA processing

16-3: Why is it easier to identify spontaneous mutations in bacteria than in most eukaryotes? a. Spontaneous mutations are expressed directly in descendant cells because bacteria are haploid. b. Spontaneous mutations are visible as color changes within a bacterial plaque. c. Spontaneous mutations can be induced via exposure to the bacteriophage T4. d. Spontaneous mutations occur at a much higher frequency in bacteria than in eukaryotes.

a. Spontaneous mutations are expressed directly in descendant cells because bacteria are haploid. No mutants are hidden as recessive alleles within heterozygote genotypes.

20-2: When considering the initiation of transcription, one often finds consensus sequences located in the region of the DNA where RNA polymerase(s) binds. Which of the following is a common consensus sequence? a. TATAAT b. satellite DNAs c. any trinucleotide repeat d. TTTTAAAA e. GGTTC

a. TATAAT

22-17a: How might one explain the existence of two polypeptides with partially shared function and structure on two different chromosomes? a. The clusters arose from the same ancestral gene. b. Similar function always require similar gene structure. c. Similar genes arose by random chance.

a. The clusters arose from the same ancestral gene.

22-18a: Which parent contributed the abnormal gamete? a. The father must have contributed the abnormal X-linked gene. b. Both parents must have contributed the abnormal X-linked gene. c. The mother must have contributed the abnormal X-linked gene.

a. The father must have contributed the abnormal X-linked gene.

12-8c: Now determine the precise arrangement of the alleles on the homologs in the unknown plant. Choose the correct arrangement. a. The genes are linked, and the arrangement is coupled, i.e., the alleles are in a cis arrangement (one homolog has both dominant alleles; the other homolog has both recessive alleles). b. The genes are linked, and the arrangement is not coupled, i.e., the alleles are in a trans arrangement (each homolog has a dominant and a recessive allele). c. The genes are unlinked (e.g., on different chromosomes).

a. The genes are linked, and the arrangement is coupled, i.e., the alleles are in a cis arrangement (one homolog has both dominant alleles; the other homolog has both recessive alleles).

18-8a: Which of the following hypotheses is most accurate given the current state of knowledge? a. The one-gene-one-polypeptide hypothesis b. The one-gene-one-enzyme hypothesis c. The one-gene-one-protein hypothesis

a. The one-gene-one-polypeptide hypothesis

21-5d: Which amino acids are encoded by the UGG copolymer? Select all that apply: a. Tryptophan b. Valine c. Glycine d. Arginine

a. Tryptophan b. Valine c. Glycine The three codons present in the UGG copolymer are UGG which codes for tryptophan, GGU which codes for glycine and GUG which codes for valine.

22-11a: Why are X rays more potent mutagens than UV radiation? a. X rays are of higher energy and shorter wavelength than UV light. They have greater penetrating ability and can create more disruption of DNA. b. X rays are of higher energy and longer wavelength than UV light. They have greater penetrating ability and can create more disruption of DNA. c. UV light is of higher energy and shorter wavelength than X rays.

a. X rays are of higher energy and shorter wavelength than UV light. They have greater penetrating ability and can create more disruption of DNA.

22-12a: The condition that exists when an organism gains or loses one or more chromosomes but not a complete haploid set is known as ________. a. aneuploidy b. trisomy c. polyploidy d. triploidy e. euploidy

a. aneuploidy

21-3: In 1964, Nirenberg and Leder used the triplet binding assay to determine specific codon assignments. A complex of which of the following components was trapped in the nitrocellulose filter? a. charged tRNA, RNA triplet, and ribosome b. uncharged tRNAs and ribosomes c. free tRNAs d. ribosomes and DNA e. sense and antisense strands of DNA

a. charged tRNA, RNA triplet, and ribosome

12-8b: What is the phenotype of the unknown plant? a. colored, green b. unpigmented, green c. unpigmented, yellow d. colored, yellow

a. colored, green

22-8a: What type(s) of mutations are likely produced by chemical #1? Select all that apply: a. frameshifts b. transversions c. transitions

a. frameshifts Mutations that are induced by chemical #1 could be reversed by mutagens that can create the same type of mutations. If chemical #1 induced frameshift mutations by making small deletions or insertions, then acridine orange (which also induces frameshift mutations) could restore the proper reading frame with a second mutation. Since acridine orange is the only mutagen that reversed the mutations, then it is likely that chemical #1 induces frameshift mutations.

20-12b: What is the direction of translation in this diagram? a. from left to right b. from right to left c. unable to be determined

a. from left to right The ribosome reads the mRNA in the 5' to 3' direction, and synthesizes protein from the N-terminal end to the C-terminal end. The mRNA is oriented with the 5' end on the left and the 3' end on the right. This indicates that translation is proceeding from left (5') to right (3').

20-15b: Choose all enzymes that participate in the translation processes. Select all that apply: a. large subunit ribozyme b. DNA polymerase III c. release factors d. DNA polymerase I e. aminoacyl tRNA synthetases f. DNA polymerase II

a. large subunit ribozyme c. release factors e. aminoacyl tRNA synthetases

20-3c: Where would you expect to find the promoter for this gene? a. unable to be determined a. left side b. right side

a. left side The promoter is located in the DNA that is adjacent to the start site of transcription. The start site corresponds with the 5' end of the RNA (or the 3' end of the DNA template strand).

20-5c: In which cellular organelle do the three posttranscriptional modifications often seen in the maturation of mRNA in eukaryotes occur? a. nucleus b. lysosome c. Golgi d. mitochondrion e. cytoplasm

a. nucleus

22-7b: Assuming one mutational event in a gene, on average, which of the following mutagens or mutational conditions would be expected to cause the most damage to a protein synthesized by such a mutagenized gene? a. proflavin b. 5-bromouracil c. ethylmethane sulfomate (EMS) d. nitrous acid

a. proflavin

23-10a: The recA gene product is part of ______________ that is activated when damaged DNA has escaped repair and the distortion disrupts the process of replication. a. recombinational DNA repair b. nucleotide excision repair

a. recombinational DNA repair

21-10a: Assuming the code is a triplet, what effect would the addition or loss of six nucleotides have on the reading frame? a. reestablishing of the subsequent reading frame b. frame shift c. the effect is unpredictable

a. reestablishing of the subsequent reading frame

16-4a: In an interrupted mating experiment, the purpose of plating cells on a selective medium is _______. a. to ensure that only recombinant genotypes are recovered b. to ensure that conjugation has been completed c. to eliminate all recipient cells d. to determine the genes present in the Hfr

a. to ensure that only recombinant genotypes are recovered Selective media in these experiments allow elimination of parental genotypes and recovery of only those whose genotypes result from transfer of donor genes.

22-8d: What type(s) of mutations are likely produced by chemical #4? Select all that apply: a. transversions b. transitions c. frameshifts

a. transversions Mutations that are induced by chemical #4 could be reversed by mutagens that can create the same type of mutations. Since none of three mutagens reversed mutations created by chemical #4, it is likely that chemical #4 does not induce transition mutations or frameshift mutations. Chemical #4 does in fact induce mutations since exposing E. coli to chemical #4 resulted in observed mutant phenotypes. Because these mutant phenotypes were not reversed by any of the three mutagens, it is likely that chemical #4 induces a different type of mutation. The only remaining choice listed as a mutation type is transversions. In the future, a mutagen that induces transversion mutations could be used to determine if the chemical #4 mutations could be reversed.

18-3a: The Beadle and Tatum experiments were based on all of the following assumptions except that _______. a. two strains of auxotrophic Neurospora that grow on minimal medium supplemented with biotin have mutations in the same gene b. supplemented media permit growth of auxotrophic strains of Neurospora c. X-irradiation can induce mutations. d. auxotrophs fail to grow on minimal media

a. two strains of auxotrophic Neurospora that grow on minimal medium supplemented with biotin have mutations in the same gene Since biosynthetic pathways have multiple steps, each catalyzed by a separate enzyme, these two strains would not necessarily have the same mutation.

21-9: The genetic code is fairly consistent among all organisms. The term often used to describe such consistency in the code is ________. a. universal b. trans-specific c. exceptional d. overlapping e. none of the answers listed is correct

a. universal

12-5a: Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event? Select all that apply. a. wild type b. miniature wings, garnet eyes c. miniature wings d. garnet eyes

a. wild type b. miniature wings, garnet eyes When a single genetic exchange occurs between two nonsister chromatids during the tetrad stage, two noncrossover (parental) and two crossover (recombinant) gametes are produced. Therefore, a female with the m+ g / m g+ genotype would produce two parental gametes which were not involved in genetic exchange -- m+ g and m g+ -- and two recombinant gametes from the single genetic exchange between the m and g genes -- m+ g+ and m g. When crossed with a male with miniature wings and garnet eyes (mg / mg), the phenotypes observed in the resulting offspring would reflect the genotypes of the female gametes.

18-12g: In a cross of the F1 plants from Part F, what phenotype(s) would result in the F2 generation? P1: speckled (AABbCc) × speckled (AABbCc) Select all that apply: a. yellow b. speckled c. green d. colorless

a. yellow b. speckled c. green

16-14a: What is the order of genes on the bacterial chromosome? Please start with gene a.

acdbe

21-13b: Which of the following amino acids could result from a single-base substitution in GGA (gly)? Select the FOUR that apply: alanine (ala) arginine (arg) aspartic acid (asp) cysteine (cys) glutamic acid (glu) leucine (leu) proline (pro) serine (ser) valine (val)

alanine (ala) arginine (arg) glutamic acid (glu) valine (val)

21-4e: If transcription occurred from right to left, what RNA sequence would be produced? a. 5' - AUGCAGCACAUU - 3' b. 3' - AUACGUCGUGUAA - 5' c. 5' - UAUGCAGCACAUU - 3' d. 3' - ATACGTCGTGTAA - 5' e. 5' - TATGCAGCACATT - 3'

b. 3' - AUACGUCGUGUAA - 5' The RNA sequence is synthesized 5' to 3' by base complementarity with the template strand. If the top strand is the template, then the RNA sequence should be identical to the bottom strand except with uracil (U) instead of thymine (T).

17-8: Which of the following statements about cotransformation is true? See Section 8.4 (pp. 153-154). a. Non-linked genes are passed simultaneously from one cell to another. b. A bacterial cell receives two adjacent genes on a single piece of DNA from the medium. c. Two bacterial cells within a culture are transformed by the same bacteriophage. d. A few chromosomal genes are passed from one cell to another through a cytoplasmic bridge.

b. A bacterial cell receives two adjacent genes on a single piece of DNA from the medium. Cotransformation makes it possible to map pairs of genes that are located near each other.

21-12a: Which of the following single base changes would result in alanine being substituted for threonine (i.e., threonine being converted into alanine)? a. C changed to G in the second position of the codon. b. A changed to G in the first position of the codon. c. A changed to C in the third position of the codon.

b. A changed to G in the first position of the codon.

21-11: In studies using repeating copolymers, AC . . . incorporates the amino acids 1 and 2, and CAACAA . . . incorporates the amino acids 1, 3 and 4. What triplet code can definitely be assigned to amino acid 1? a. CAC b. ACA c. CAA d. ACC

b. ACA

15-12h: What does your answer in Part G suggest? a. Some of the crossovers that were expected, did not occur. b. All of the crossovers that were expected, occurred. c. None of the crossovers that were expected, occurred.

b. All of the crossovers that were expected, occurred.

18-9b: What would be the outcome of crossing two strains of wingless flies that belong to the same complementation group? a. All offspring will be wild-type. b. All offspring will be wingless. c. Some offspring will be wild-type, and some will be wingless.

b. All offspring will be wingless. A cross of mutants belonging to the same complementation group will produce only the mutant phenotype in progeny because both parents possess mutations in the same gene. As a result, all progeny will be homozygous for two mutant alleles and display the wingless phenotype because they do not possess a wild-type copy of the gene important for wing development. The only way that offspring could display the wild-type phenotype and develop normal wings would be if the mutant parental flies belonged to different complementation groups and had mutations in separate genes that affected wing development. In that case, a cross of flies with mutations in two different genes would result in offspring heterozygous for both genes. Because they would have one wild-type allele at each locus, the offspring would develop normal wings.

18-5e: Indicate where in the pathway Mutant Strain 1 is blocked. a. Between E and B b. Between B and D c. Between D and A d. Between A and C

b. Between B and D

21-5b: Which amino acids are encoded by the UG copolymer? Select all that apply: a. Tryptophan b. Cysteine c. Arginine d. Valine

b. Cysteine d. Valine The two codons present in the UG copolymer are UGU which codes for cysteine and GUG which codes for valine.

23-8c: What can you conclude about the genetic basis of XP from these data? a. Each complementation group represents a group of genes in the same biochemical pathway. Each strain is defective in a different protein. b. Each complementation group represents a gene. All strains within a group are defective in the same protein.

b. Each complementation group represents a gene. All strains within a group are defective in the same protein.

18-8c: However, even the updated hypothesis (i.e., the one-gene-one-polypeptide hypothesis) is not entirely accurate. Select the THREE reasons why even the one-gene-polypeptide hypothesis is not entirely supported by data: a. One gene can only code for a single enzyme product b. Expression of one gene can result in more than one type of polypeptide c. Transcription of some genes does not result in a polypeptide d. Alternative splicing can lead to the formation of different polypeptide products from a single RNA transcript

b. Expression of one gene can result in more than one type of polypeptide c. Transcription of some genes does not result in a polypeptide d. Alternative splicing can lead to the formation of different polypeptide products from a single RNA transcript

23-3a: Assume that an unrepaired oxoG lesion is present in the helix of DNA opposite cytosine. Predict the type of mutation that will occur following several rounds of replication. a. A:T pair will be replaced with C:G pair b. G:C pair will be replaced with T:A pair c. insertion of an adenine base d. deletion of an adenine base

b. G:C pair will be replaced with T:A pair

16-4b: Mapping bacterial genes by conjugation is based on which of the following assumptions? a. F minus bacterial strains are capable of transferring genes to Hfr strains. b. Genes are transferred from donor to recipient in a linear fashion. c. Conjugation is not disrupted when a bacterial culture is mixed in a blender. d. Bacterial cells are capable of growing in the presence of sodium azide.

b. Genes are transferred from donor to recipient in a linear fashion. If this were not true, the distance between genes could not be measured as a function of time.

17-10: Bacterial recombination sometimes produces heteroduplex DNA. What is true with respect to heteroduplex DNA? Select the TWO statements that are true: a. Heteroduplex DNA is single stranded b. Heteroduplex DNA contains DNA strands that originate from different sources c. Heteroduplex DNA is composed of strands that are always exactly complementary d. Heteroduplex DNA is formed during the process of transduction e. Heteroduplex DNA is formed during the process of transformation f. Heteroduplex DNA is formed during the process of conjugation

b. Heteroduplex DNA contains DNA strands that originate from different sources e. Heteroduplex DNA is formed during the process of transformation

18-7b: Which mutant strain is deficient in enzyme 1 activity? a. Mutation A b. Mutation B c. Impossible to tell

b. Mutation B

22-18b: Using the appropriate genetic terminology, describe the meiotic mistake that occurred. Be sure to indicate in which division the mistake occurred. a. Non-disjunction must have occurred during meiosis II. b. Non-disjunction must have occurred during meiosis I.

b. Non-disjunction must have occurred during meiosis I.

18-7a: What flower color would you expect in the mutant strain defective for enzyme 1? a. Blue b. Red c. Purple

b. Red

12-3: For the linked gene loci shown above, indicate which of the statements is correct: a. The a and c alleles are in a cis arrangement, while the b allele is in a trans arrangement with respect to the other two alleles. b. The a and b alleles are in a cis arrangement, while the c allele is in a trans arrangement with respect to the other two alleles. c. The b and c alleles are in a cis arrangement, while the a allele is in a trans arrangement with respect to the other two alleles.

b. The a and b alleles are in a cis arrangement, while the c allele is in a trans arrangement with respect to the other two alleles.

15-10b: Although T4 actually has a linear chromosome, its chromosome map is drawn as a circle because: a. Genes are clustered at the ends b. The ends of DNA molecules of different T4 bacteriophages can occur at random positions c. The genes at one end of the chromosome have duplicated d. Gene locations are random

b. The ends of DNA molecules of different T4 bacteriophages can occur at random positions

15-4: The bacteriophage genome consists primarily of genes encoding proteins that make up the head, collar and tail, and tail fibers. When these genes are transcribed following phage infection, how are these proteins synthesized, since the phage genome lacks genes essential to ribosome structure? a. The proteins are synthesized by a special mechanism, which does not require ribosomes. b. The proteins are synthesized using bacterial ribosomes. c. The phage uses the bacterial genome to synthesize its own ribosomes. d. None of the above

b. The proteins are synthesized using bacterial ribosomes.

18-10a: Your mentor asks you to determine if the mutants belong to the same complementation group. What is true about flies that belong to the same complementation group? a. They all have the same mutation in the same wing- development gene. All flies in a complementation group have identical DNA sequences for this gene. b. They all have some mutation in the same wing-development gene. Each strain may have a different mutation, but the same gene is mutated in all strains in a complementation group. c. They all have some mutation in some wing-development gene. Each strain may have a different mutation in a different gene, but all strains within a complementation group have the same phenotype.

b. They all have some mutation in the same wing-development gene. Each strain may have a different mutation, but the same gene is mutated in all strains in a complementation group. All mutations found to be present in a single gene are in the same complementation group, and they will complement mutations in other complementation groups. When large numbers of mutations affecting the same trait are available and studied using complementation analysis, it is possible to predict the total number of genes involved in the determination of a trait.

22-18c: Using the appropriate genetic terminology, explain the son's skin phenotype. a. This son's mosaic phenotype is caused by Y-chromosome inactivation, a form of dosage compensation in mammals. b. This son's mosaic phenotype is caused by X-chromosome inactivation, a form of dosage compensation in mammals.

b. This son's mosaic phenotype is caused by X-chromosome inactivation, a form of dosage compensation in mammals.

17-3: Which of the following statements about using transduction for mapping genes on the bacterial chromosome is FALSE? a. Transduction mapping can be used to determine the order of several genes. b. Transduction studies require the direct contact between bacterial cells. c. Transduction mapping is based on the fact that phages can transfer two genes simultaneously. d. Genes that are located close to one another on the chromosome have a higher probability of being transduced together.

b. Transduction studies require the direct contact between bacterial cells.

22-1c: Why would a mutation in a somatic cell of a multicellular organism escape detection? a. A somatic mutation is a type of mutation that can be easily repaired. Therefore, the mutation is often corrected before the genetic screen takes place. b. When conducting genetic screens, it is assumed that all the cells are genetically identical. Since a somatic mutation first appears in a single cell, it is highly unlikely that the organism will be sufficiently altered to respond to a screen because none of the other cells will have the mutation. c. Most somatic mutations are dominant and do not cause a phenotypic change in the cell. As a result, it is highly unlikely that the organism will be sufficiently altered to respond to a screen.

b. When conducting genetic screens, it is assumed that all the cells are genetically identical. Since a somatic mutation first appears in a single cell, it is highly unlikely that the organism will be sufficiently altered to respond to a screen because none of the other cells will have the mutation.

14-5a: Each of the banded protrusions originating from the chromocenter can best be described as: a. a heterochromatic duplication b. a pair of synapsed homologous chromosomes c. a single chromosome d. a single chromatid

b. a pair of synapsed homologous chromosomes

16-5c: The distance between genes on the gene map is measured in minutes because the distance between genes is measured by the time it takes for the gene to be transferred from the host cell into the recipient cell. This distance can be determined by looking at the first appearance of each phenotype on the graph. What is the distance between the met and his genes? a. less than 1 minute b. approximately 1 minute c. between 1 and 2 minutes d. between 2 and 3 minutes

b. approximately 1 minute The distance between genes on a bacterial chromosome can be measured by the time in minutes that it takes to transfer the genes from a host cell to a recipient cell. From the graph, you can see that the met + phenotype first appears around the 9-minute mark. The his + phenotype first appears at the 10-minute mark. Therefore, you can conclude that met and his are about 1 minute apart.

23-6b: How do these defects create the phenotypes associated with the disorder? a. by decreasing the rate of DNA synthesis in skin cells b. by decreasing the rate of DNA mutation repair in skin cells c. by increasing the protein longevity in skin cells d. by decreasing the mRNA longevity in skin cells

b. by decreasing the rate of DNA mutation repair in skin cells

14-5b: Within a polytene chromosome, each homologous tightly-synapsed pair of chromosomes has a large volume because it is made up of hundreds of: a. unwound chromatin fibers b. chromatids c. single-stranded DNA molecules d. chromosomes

b. chromatids

16-2: What is a form of recombination in bacteria that involves the F plasmid? a. transduction b. conjugation c. meiosis d. vertical transfer e. transformation

b. conjugation

14-6h: The method of mapping described in this question is used to construct what type of a map of Drosophila polytene chromosomes? a. cDNA map b. cytological map c. genetic map d. geographic map

b. cytological map

22-3c: What type of mutation led to Mutant 3? a. nonsense b. frameshift c. silent d. missense e. sense

b. frameshift

22-8c: What type(s) of mutations are likely produced by chemical #3? Select all that apply: a. transversions b. frameshifts c. transitions

b. frameshifts c. transitions Mutations that are induced by chemical #3 could be reversed by mutagens that can create the same type of mutations. Since all three mutagens reversed mutations created by chemical #3, it is likely that chemical #3 induces transition mutations as well as small deletions or insertions that result in frameshift mutations.

20-3b: What is the direction of transcription in this diagram? a. from right to left b. from left to right c. unable to be determined

b. from left to right The RNA polymerase reads the template strand from the 3' to 5' direction, and synthesizes RNA from 5' to 3'. The template strand is the DNA strand that is based paired to RNA as transcription proceeds. In this example, the RNA is paired with the bottom strand. The bottom strand is oriented with 3' end to the left and 5' end to the right. This means transcription is proceeding from left (3') to right (5').

22-4c: Suppose that the top strand is the coding (nontemplate) strand and the three bases shown represent a single "in-frame" codon in a gene. What will be the effect of the tautomeric shift-induced mutation on the amino acid sequence? a. missense mutation b. nonsense mutation c. silent mutation d. not enough information to determine

b. nonsense mutation The codon changed from CAA to TAA as a result of a tautomeric shift-induced mutation. This point mutation results in a change from a glutamine codon to a stop codon, causing a nonsense mutation.

23-10b: The uvr gene products are part of ______________ repair in which many nucleotides are removed at once, including the damaged region. a. recombinational DNA repair b. nucleotide excision repair

b. nucleotide excision repair

22-3e: Another mutation (Mutant 4) is isolated. Its amino acid sequence is unchanged, but the mutant cells produce abnormally low amounts of the wild-type proteins. Which of the following mutations would explain this result? a. missense mutation b. promoter mutation c. frameshift mutation d. nonsense mutation

b. promoter mutation

15-3: Viral DNA that has integrated into a bacterial chromosome is referred to as a ________. a. virulent phage b. prophage c. prephage d. plaque e. temperate phage

b. prophage

20-8a: All EXCEPT which of the following are involved in the process of tRNA "charging"? a. aminoacyl tRNA synthase b. rRNA c. amino acids d. ATP

b. rRNA

18-12a: In a cross of the following true-breeding plants (P1), what phenotype would result in the F1 generation? P1: speckled (AABBCC) × yellow (AAbbCC) a. yellow b. speckled c. green d. colorless

b. speckled

18-12f: In a cross of the following true-breeding plants (P1), what phenotype would result in the F1 generation? P1: yellow (AAbbCC) × green (AABBcc) a. yellow b. speckled c. green d. colorless

b. speckled

18-12b: In a cross of the following plants (P1), what phenotype would result in the F1 generation? P1: speckled (AABbCC) × speckled (AABbCC) Select all that apply: a. colorless b. speckled c. green d. yellow

b. speckled d. yellow

22-1a: Mutations that arise in nature, from no particular artificial agent, are called ________. a. induced mutations b. spontaneous mutations c. chromosomal aberrations d. oblique mutations e. cosmic mutations

b. spontaneous mutations

21-1a: When scientists were attempting to determine the structure of the genetic code, Crick and coworkers found that when three base additions or three base deletions occurred in a single gene, the wild-type phenotype was sometimes restored. These data supported the hypothesis that ________. a. AUG is the initiating triplet b. the code is triplet c. the code is overlapping d. there are three amino acids per base e. the code contains internal punctuation

b. the code is triplet

21-6: When examining the genetic code, it is apparent that ________. a. there can be more than one amino acid for a particular codon b. there can be more than one codon for a particular amino acid c. there are 44 stop codons because there are only 20 amino acids d. AUG is a terminating codon e. the code is ambiguous in that the same codon can code for two or more amino acids

b. there can be more than one codon for a particular amino acid

22-11b: Pyrimidine dimers that distort the normal DNA conformation, thereby leading to errors during DNA replication, are caused by: a. ionizing radiation from X rays b. ultra-violet radiation

b. ultra-violet radiation

23-2b: Ultraviolet light causes pyrimidine dimers to form in DNA. Some individuals are genetically incapable of repairing such dimers at "normal" rates. Such individuals are likely to suffer from ________. a. Huntington disease b. xeroderma pigmentosum c. SCID d. phenylketonuria e. muscular dystrophy

b. xeroderma pigmentosum

16-7c: Next, the researchers mixed equal amounts of the AX and AY E. coli strains. On which type(s) of media would growth provide evidence that genetic transfer had occurred between the two strains of bacteria? Select the THREE that apply: bio- leu- bio- trp- trp- leu- leu- trp- bio- none of the above

bio- leu- trp- leu- leu- trp- bio-

14-8b: For a more detailed view of the genes located on chromosome 7 visit Genome Data Viewer at: https://www.ncbi.nlm.nih.gov/genome/gdv/. On the home page for this genome browser, you're shown the human karyotype (bottom right; note that this browser sometimes goes offline - just check back later). Click on chromosome 7. You will be directed to an overview page for this chromosome. The number of base pairs in chromosome 7 is shown on the top right, immediately after "Chr 7 (NC_000007.14)" According to Genome Data Viewer, about how many base pairs does chromosome 7 contain? a. 159 bp b. 159,000 bp c. 159,000,000 bp

c. 159,000,000 bp

23-2d: Which of the following statements regarding Nucleotide Excision Repair (NER) and Base Excision Repair (BER) is true? a. Both NER and BER involve the creation of an apyrimidinic (AP) site. b. Only NER involves the action of DNA ligase to seal nicks in the DNA backbone. c. Both NER and BER involve the removal of one or more nucleotides. d. Both NER and BER can be activated by exposure to visible light

c. Both NER and BER involve the removal of one or more nucleotides. This statement is true. In both NER and BER a nuclease will target damaged or distorted regions of DNA.

14-8f: Hover over the name of this gene. This gene codes for a _____________ . a. Transfer RNA b. T cell receptor c. Golgi to ER traffic protein d. Myosin heavy chain

c. Golgi to ER traffic protein

12-8a: Based on the data, what can you say about the genotype of the unknown plant? a. It was heterozygous for one gene, and homozygous for the other. b. It was homozygous for both genes. c. It was heterozygous for both genes.

c. It was heterozygous for both genes.

14-2: Which of the following are NOT commonly used as DNA landmarks for mapping human genes? a. Restriction fragment length polymorphisms (RFLPs) b. Single-nucleotide polymorphisms (SNPs) c. Mobile genetic elements (which include elements such as transposons) d. Microsatellites (also known as short tandem repeats, or STRs)

c. Mobile genetic elements (which include elements such as transposons)

20-16: The concept of colinearity states that the linear sequence of triplet codons corresponds precisely with the linear sequence of amino acids in each protein. Review the concept of colinearity and consider the following question: Certain mutations called amber in bacteria and viruses result in premature termination of polypeptide chains during translation. Many amber mutations have been detected at different points along the gene that codes for a head protein in phage T4. How might this system be further investigated to demonstrate and support the concept of colinearity? a. The nearer to the 5′ end of mRNA the amber mutation occurs, the longer the resulting polypeptide. b. The nearer to the Shine-Dalgarno sequence of mRNA the amber mutation occurs, the longer the resulting polypeptide. c. The nearer to the 5′ end of mRNA the amber mutation occurs, the shorter the resulting polypeptide. d. The position of the occurring amber mutation cannot be used to support the concept of colinearity.

c. The nearer to the 5′ end of mRNA the amber mutation occurs, the shorter the resulting polypeptide.

22-2c: Generally speaking, which of the following mutations would most severely affect the protein coded for by a gene? a. a base substitution at the beginning of the gene b. a base substitution at the end of the gene c. a frameshift deletion at the beginning of the gene d. a frameshift deletion at the end of the gene

c. a frameshift deletion at the beginning of the gene A frameshift mutation at the beginning of a gene would affect every codon after the point where the mutation occurred. During protein synthesis, incorrect amino acids would be inserted from the point where the frameshift mutation occurred on; the resulting protein would most probably be nonfunctional. For this reason, a frameshift mutation at the beginning of a gene is generally the most severe type of mutation.

23-10c: At a UV fluence of 3 J m^-2, cells that exhibit a mutation in ___________ show the relatively lowest rate of survival. a. the recA gene only b. the uvrA gene only c. both uvrA and recA (i.e., the double mutant)

c. both uvrA and recA (i.e., the double mutant)

18-3b: In their first round of screening, Beadle and Tatum plated spores on minimal medium. The purpose of this screen was to _______. a. identify which mutants were deficient in vitamin synthesis pathways and which were deficient in amino acid synthesis pathways b. determine which of the mutants generated were auxotrophic for thiamine c. determine whether any auxotrophic mutants had been generated d. eliminate all nonmutant Neurospora from the population

c. determine whether any auxotrophic mutants had been generated

23-10d: The fact that the double mutant shows the lowest survival suggests that nucleotide excision and recombinational DNA repair systems ____________ . a. are both required for any DNA repair to occur b. do not increase cell survival following DNA damage c. function independently of one another d. do not function in wildtype cells e. function in the same pathway of repair

c. function independently of one another

12-12c: The results of the Stern experiment supported the general idea that _______. a. only females exhibit incomplete dominance b. crossing over does not occur in males c. genetic recombination is a result of physical exchange between homologous chromosomes d. alleles of different genes assort independently

c. genetic recombination is a result of physical exchange between homologous chromosomes Whenever recombinant phenotypes occurred, the cytological markers indicated that a physical exchange between the X chromosomes had also occurred.

12-12a: The discernible difference in length between the two X chromosomes of the female fruit fly was important in Stern's experiments because _______. a. it allowed him to predict the phenotypic ratios among offspring b. only the longer of the two X chromosomes could be passed on to offspring c. it allowed cytological detection of physical exchange between the chromosomes d. the carnation eye color phenotype was difficult to detect

c. it allowed cytological detection of physical exchange between the chromosomes The differences in structure between the two chromosomes allowed Stern to track the inheritance of recombinant and nonrecombinant X chromosomes.

12-1: When two genes fail to assort independently, the term normally applied is ________. a. tetrad analysis b. Mendelian inheritance c. linkage d. dominance and/or recessiveness e. discontinuous inheritance

c. linkage

13-13: Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are most consistent with ________. a. 100 percent recombination b. independent assortment c. linkage with approximately 33.3 map units between the two gene loci d. sex-linked inheritance with 30 percent crossing over e. linkage with 50 percent crossing over

c. linkage with approximately 33.3 map units between the two gene loci

12-4: Crossing over normally occurs between all EXCEPT which of the following? a. homologous chromosomes b. non-sister chromatids c. nonhomologous chromosomes d. sex chromosomes

c. nonhomologous chromosomes Normally, crossing over does not occur between nonhomologous chromosomes.

12-12b: Stern observed all of the following results EXCEPT _______ in his experiment. a. one of the recombinant phenotypes was associated with an X chromosome of normal length b. the number of car, B+ male offspring was roughly equal to the number of car+, B male offspring c. offspring with car+, B+ eyes resulted from fertilization of eggs containing recombinant X chromosomes d. the number of males was roughly equal to the number of females

c. offspring with car+, B+ eyes resulted from fertilization of eggs containing recombinant X chromosomes

13-8: When analyzing three genes that reside on the same chromosome, the expected frequency of double-crossover events can be determined by multiplying the frequency of single crossovers between each pair of genes. What would cause the number of observed double-crossover events to be less than the expected value? See Section 7.4 a. the product law b. the coefficient of coincidence (C) c. positive interference

c. positive interference Positive interference occurs when the number of observed double-crossover events is less than the number expected.

14-5c: One benefit of using polytene chromosomes is that: a. genetic mapping through the use of three-point crosses can be greatly facilitated b. the degree to which the formation of one crossover interferes with the formation of a second crossover can be more easily calculated c. the genetic position of linked genes can be connected with their physical location on the chromosome d. double crossover events can be more easily detected

c. the genetic position of linked genes can be connected with their physical location on the chromosome

14-1b: What advantage do they confer? a. they are used to suppress crossing over in the predetermined region of the chromosome b. they are used to induce crossing over in the predetermined point of the chromosome c. they are used as landmarks d. they are recognized by enzymes used in artificial enzymatic crossing over

c. they are used as landmarks

22-8b: What type(s) of mutations are likely produced by chemical #2? Select all that apply: a. transversions b. frameshifts c. transitions

c. transitions Mutations that are induced by chemical #2 could be reversed by mutagens that can create the same type of mutations. Since both EMS and 5-Bromouracil reversed the mutations created by chemical #2, and these mutagens create transition mutations, it is likely that chemical #2 also induces transition mutations. For example if chemical #2 induced an adenine-to-guanine transition mutation that gave rise to a mutant phenotype, then EMS could induce a guanine-to-adenine transition mutation to reverse the mutant phenotype.

18-9: You obtain two independently-isolated auxotrophs of Neurospora, both of which are unable to synthesize the amino acid valine. In order to determine whether the mutations in these two auxotrophs are on the same gene, you could conduct a _________________ test.

complementation

14-8h: Click on chromosome 15. About how many genes are on this chromosome? a. 100 genes b. 400 genes c. 800 genes d. 1200 genes

d. 1200 genes

21-4a: If transcription occurred from left to right, which strand is the template strand? a. 5' - TATGCAGCACATT - 3' b. 5' - ATACGTCGTGTAA - 3' c. 3' - TATGCAGCACATT - 5' d. 3' - ATACGTCGTGTAA - 5'

d. 3' - ATACGTCGTGTAA - 5' RNA polymerase travels along the template strand in the 3' to 5' direction, synthesizing an RNA molecule from 5' to 3'. If the polymerase is moving from left to right, the strand that is oriented from 3' to 5' left to right is the bottom strand. This is the template strand. The top strand is called the coding strand (or sometimes the non-template) strand. The template strand is complementary to the mRNA sequence; the coding strand is the same sequence with uracil (U) instead of thymine (T).

21-4d: If transcription occurred from right to left, which strand is the template strand? a. 3' - TATGCAGCACATT - 5' b. 3' - ATACGTCGTGTAA - 5' c. 5' - ATACGTCGTGTAA - 3' d. 5' - TATGCAGCACATT - 3'

d. 5' - TATGCAGCACATT - 3' Regardless of whether transcription is being carried out from left to right, or right to left, the RNA polymerase moves in the 3' to 5' direction along the template strand. If transcription is proceeding from right to left, the top strand is the template strand as it is oriented from 3' to 5' going from right to left.

21-4b: If transcription occurred from left to right, what RNA sequence would be produced? a. 5' - TATGCAGCACATT - 3' b. 3' - ATACGTCGTGTAA - 5' c. 5' - AUGCAGCACAUU - 3' d. 5' - UAUGCAGCACAUU - 3' e. 3' - AUACGUCGUGUAA - 5'

d. 5' - UAUGCAGCACAUU - 3' The RNA sequence is produced 5' to 3' by base complementarity with the template strand. This produces an RNA sequence identical to the coding strand of DNA (top strand in this example) except with uracil (U) instead of thymine (T).

23-3b: Which DNA repair mechanism might work to counteract an oxoG lesion? a. DNA double-strand break repair b. Photoreactivation by photolyase c. Homologous recombination repair d. Base excision repair

d. Base excision repair

18-5f: Indicate where in the pathway Mutant Strain 2 is blocked. a. Between E and B b. Between B and D c. Between D and A d. Between A and C

d. Between A and C

16-9a: How was it established that physical contact was necessary? a. By placing bacterial cells in a U-tube. Under this condition, conjugation occurs. b. By placing bacterial cells in a U-tube. Under this condition, conjugation does not occur. c. By placing a filter in a U-tube between bacterial cells. Under this condition, conjugation occurs. d. By placing a filter in a U-tube between bacterial cells. Under this condition, conjugation does not occur.

d. By placing a filter in a U-tube between bacterial cells. Under this condition, conjugation does not occur.

14-1a: DNA markers have greatly enhanced the mapping of genes in humans. What are DNA markers? a. DNA markers are the regions of DNA that, due to specific base sequence, have a very low rate of random mutations b. DNA markers are well studied DNA molecules for which the data on exact base sequence are available c. DNA markers are chemically modified nucleotides that can be incorporated into DNA during replication d. DNA markers are unique DNA sequences whose sequence and chromosomal location are known

d. DNA markers are unique DNA sequences whose sequence and chromosomal location are known

21-1b: Which type of mutation helped lead to the understanding that the genetic code is based on triplets? a. Nonsense (generation of a stop codon) b. Deletion or insertion mutation within the promoter c. Base substitution (substitution of one base for another) d. Frameshift

d. Frameshift

21-7: Introns are known to contain termination codons (UAA, UGA, or UAG), yet these codons do not interrupt the coding of a particular protein. Why? a. UAA, UGA, and UAG are initiator codons, not termination codons. b. These triplets cause frameshift mutations, but not termination. c. More than one termination codon is needed to stop translation. d. Introns are removed from mRNA before translation. e. Exons are spliced out of mRNA before translation.

d. Introns are removed from mRNA before translation.

20-8b: Which of the following statements best describes the function of aminoacyl tRNA synthetase? a. It provides the energy required to attach a specific amino acid to a tRNA molecule. b. It synthesizes tRNA molecules. c. It helps tRNA synthesize proteins. d. It attaches a specific amino acid to a tRNA molecule.

d. It attaches a specific amino acid to a tRNA molecule. Aminoacyl tRNA synthetase catalyzes the charging reaction that links a specific amino acid to a tRNA molecule.

18-2: Which of the following statements is true of Neurospora carrying a mutation that affects tyrosine synthesis (tyr-)? a. The cells can grow on minimal medium + histidine. b. The cells can grow on minimal medium. c. The cells cannot grow on complete medium. d. The cells can grow on minimal medium + tyrosine.

d. The cells can grow on minimal medium + tyrosine. The cells can grow if the nutrient that the mutant cannot produce itself (i.e., tyrosine) is supplied in the medium.

18-12d: How would the results in the F1 generation differ if genes A and B were linked with no crossing over between them? a. Additional colors would be represented in the F1. b. There would be more speckled flowers in the F1. c. There would be more yellow flowers in the F1. d. There would be no change.

d. There would be no change.

18-12e: How would the results in the F1 generation differ if genes A and B were linked and 20 mu apart? a. Additional colors would be represented in the F1. b. There would be more speckled flowers in the F1. c. There would be more yellow flowers in the F1. d. There would be no change.

d. There would be no change.

23-2a: What are the consequences of having pyrimidine dimers in DNA? a. They form an extra phosphodiester bond between them. b. They create an apyrimidinic site c. They prevent the transcription of the DNA into RNA. d. These dimers distort the DNA structure and result in errors during DNA replication.

d. These dimers distort the DNA structure and result in errors during DNA replication.

17-2: What are the observations that led Zinder and Lederberg to conclude that the prototrophs recovered in their transduction experiments were not the result of HFr-mediated conjugation? a. They observed that the two auxotrophic strains required cell-to-cell contact in order for genetic exchange to occur and for prototrophs to form. b. They plated two auxotrophic strains on a minimal medium to recover prototrophic cells. c. They marked auxotrophs with labels visible under UV-light and then checked to see if recovered prototrophs were labeled. d. They placed a filter between the two auxotrophic strains to prevent contact. The following treatment with DNase showed that the filterable agent was not naked DNA.

d. They placed a filter between the two auxotrophic strains to prevent contact. The following treatment with DNase showed that the filterable agent was not naked DNA.

22-17c: How can duplications arise? a. Spindle failure during chromosome separation b. The loss of a telomere c. A chromosome break and exchange d. Uneven crossing over during meiotic prophase

d. Uneven crossing over during meiotic prophase

21-5e: Which one of the following amino acids is encoded by both copolymers? a. Tryptophan b. Arginine c. Glycine d. Valine

d. Valine The UG copolymer has UGU and GUG codons. The UGG copolymer has UGG, GGU and GUG codons. Both copolymers have the GUG codon in them, which encodes valine.

22-5: All except which of the following mutagenic events occur spontaneously? a. depurination b. tautomerization c. deamination d. alkylation

d. alkylation A chemical mutagen called an alkylating agent is required to alkylate a base.

14-6g: A new mutant Drosophila strain VII is found that has a deleted chromosome part that spans bands 3A1 through 3C3. Which mutation(s) will be uncovered when using this new strain? a. zeste b. zw2 c. white d. all of the above e. none of the above

d. all of the above

12-9b: In another set of experiments, a sixth gene, jv, was tested against h and th. You obtain the following percentage of recombinant offspring: jv-h: 7.3% jv-th: 24.0% Based on this information, where is jv located on the map you constructed? a. between h and eyg b. between th and cu c. between eyg and th d. between rai and h

d. between rai and h To solve this problem, it is helpful to review the two-point mapping data generated for all three genes involved. Considering that the map distances between jv-h (7.3 mu) and h-th (16.7 mu) equal the sum of the distance between jv-th (24 mu), and given the distance between th-rai is greater (26.2 mu), the jv gene is located between rai and and h.

17-7: Bacterial cells are capable of being transformed only under certain conditions. Bacteria that are in the particular physiological state to become transformed are called ________. a. linked b. conjugated c. recombinant d. competent

d. competent

23-6a: What genetic defects result in the disorder xeroderma pigmentosum (XP) in humans? a. defects in the mRNA translation system b. defects in the mRNA transport system c. defects in the pre-mRNA splicing system d. defects in the DNA repair system

d. defects in the DNA repair system

20-9b: When a peptide bond is formed between two amino acids, one is attached to the tRNA occupying the P site and the other _______. a. is attached through hydrogen bonds to the mRNA b. is free in the cytoplasm c. is attached to the tRNA occupying the E site d. is attached to the tRNA occupying the A site

d. is attached to the tRNA occupying the A site

16-9c: What is the genetic basis of a bacterium being F+? a. it lacks the F− sequence in its genome b. it contains the F+ sequence in its genome c. it lacks the F− plasmid d. it contains the F+ plasmid

d. it contains the F+ plasmid

20-1: The relationship between a gene and a messenger RNA is that ________. a. mRNAs make proteins, which then code for genes b. mRNA is directly responsible for making Okazaki fragments c. genes are structurally identical to mRNAs d. mRNAs are made from genes e. genes are made from mRNAs

d. mRNAs are made from genes

22-3a: What type of mutation led to Mutant 1? a. sense b. missense c. frameshift d. nonsense e. silent

d. nonsense

12-11: A linkage group __________. a. is a chromosomal region in which crossing over cannot occur b. is a collection of genes that never produce recombinant gametes c. can occur only on an autosome d. represents some of the genes located on the same chromosome that are physically close together

d. represents some of the genes located on the same chromosome that are physically close together Genes located on the same chromosome that are physically close together will show some evidence of linkage to each other.

14-6b: If a deletion mutation occurs for a dominant allele that spans bands 3C4 through 3C6, pseudominance will be observed for which recessive allele? a. N (Notch) b. zw2 c. w (white) d. rst (roughest) e. z (zeste)

d. rst (roughest)

18-1: By their experimentation using the Neurospora fungus, Beadle and Tatum were able to propose the far-reaching hypothesis that ________. a. more than one codon can specify a given amino acid b. genetic recombination occurred in Neurospora c. several different enzymes may be involved in the same step in a biochemical pathway d. the role of a specific gene is to produce a specific enzyme e. prototrophs will grow only if provided with nutritional supplements

d. the role of a specific gene is to produce a specific enzyme

13-1: The genes dumpy wings (dp), clot eyes (cl), and apterous wings (ap) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the genetic distances shown below were determined. dp−ap: 42 dp−cl: 3 ap−cl: 39 What is the sequence of the three genes?

dp−cl−ap

13-12a: From this information, determine which gene is in the middle.

dsr To determine the sequence of genes, identify the parental, noncrossover classes and the double crossover classes. The noncrossover classes will be those that appear most frequently: cinnabar eyes, sp + dsr + cn (235) and disrupted wings, speck body, sp dsr cn + (241). The double crossover classes will be those that appear least frequently: speck body, sp dsr + cn + (22) and disrupted body, cinnabar eyes, sp + dsr cn (25). Next, identify the gene that switches places between the groups. Because the dsr gene switches places between the noncrossover and double crossover groups, it is in the middle.

23-1: The process of error correction of mismatched bases from the end of a growing polynucleotide that is carried out by DNA polymerases is called ________. a. base-pair substitution b. base excision repair c. photoreactivation d. recombinational DNA repair e. DNA proofreading

e. DNA proofreading

14-6f: Which is the only mutation shown that would result in the same mutant phenotype in an intact heterozygous fly and in a heterozygous fly with a deletion that abolishes the wildtype allele? In other words, for which gene would deletion mapping not be able to indicate location? a. zw2 b. white c. roughest d. zeste e. Notch

e. Notch

20-4a: It has been recently determined that the gene for Duchenne muscular dystrophy (DMD) is more than 2000 kb (kilobases) in length; however, the mRNA produced by this gene is only about 14 kb long. What is a likely cause of this discrepancy? a. The exons have been spliced out during mRNA processing. b. When the mRNA is produced, it is highly folded and therefore less long. c. More amino acids are coded for by the DNA than by the mRNA. d. The DNA represents a double-stranded structure, whereas the RNA is single-stranded. e. The introns have been spliced out during mRNA processing.

e. The introns have been spliced out during mRNA processing.

19-9: Experiments that determined genetic fine-structure, such as the genetic experiments performed by Benzer, revealed that the ___________ is the unit of mutation and the unit of recombination, i.e., that this is the unit that is indivisible by the processes of mutation and recombination. a. chromatid b. allele c. gene d. chromosome e. base pair

e. base pair

22-3b: What type of mutation led to Mutant 2? a. frameshift b. silent c. sense d. nonsense e. missense

e. missense

16-7d: Next the researchers repeated the experiment in Part C but instead of mixing the AX and AY strains, they put one strain on each side of a glass U-tube. The U-tube contains a glass filter between the two strains of bacteria that prevents the cells from interacting directly. On which type of media would you expect cells from the AY side of the U-tube to grow? a. bio- leu- b. bio- trp- c. trp- leu- d. leu- trp- bio- e. none of the above

e. none of the above Bacterial gene transfer by conjugation requires physical contact between cells of the two strains. The glass filter in this experiment prevents cell-cell contact. Without gene transfer, the AY strain remains bio- trp- and cannot grow on any medium that lacks bio and/or trp.

16-8: Name the general category into which double-stranded circular extrachromosomal DNA elements such as F factors, ColE1, and R would fall. a. plaque b. capsid c. r-determinant d. partial diploid e. plasmid

e. plasmid

15-5: A bacteriophage that is capable of entering either a lytic or lysogenic cycle is called a(n) ________. a. plasmid b. virulent bacteriophage c. episome d. plaque-forming unit e. temperate bacteriophage

e. temperate bacteriophage

17-1: Viral-mediated transfer of bacterial DNA is referred to as ________. a. endosymbiosis b. lysogeny c. transformation d. conjugation e. transduction

e. transduction

20-7a: List all molecular constituents present in a functional polyribosome. Select the TEN that apply: elongation factors (optionally) PABP Mg2+ charged tRNA release factors poly-A binding proteins large ribosomal subunit Ca2+ mRNA growing polypeptide chain GTP small ribosomal subunit initiation factors (optionally)

elongation factors (optionally) Mg2+ charged tRNA release factors large ribosomal subunit mRNA growing polypeptide chain GTP small ribosomal subunit initiation factors (optionally)

18-12i: The ratio you entered in Part H should immediately suggest to you that the phenomenon of ________ is present, i.e., that the alleles of these genes are interacting in the control of a single phenotype.

epistasis

19-3: The distance between mutational sites within a gene is known as the __________________ distance.

intragenetic

16-7b: On which type(s) of media will the AY strain (AmpR bio- trp- leu+) grow? Select the THREE that apply: leu- trp- bio- leu- trp- bio- + ampicillin complete medium complete medium + ampicillin

leu- complete medium complete medium + ampicillin Bactria that are AmpR are resistant to ampicillin and can grow in the presence of ampicillin. If a strain of bacteria is wild-type for a gene that is necessary to make an essential nutrient (for example, leu+), then the strain can grow in media that lacks that nutrient.

20-7b: Identify the role of all molecular constituents present in a functional polyribosome.

mRNA carries the information from the DNA specifying the order of amino acids Charged tRNA carries an amino acid into the ribosome. Large ribosomal subunit is responsible for the catalytic activity of the ribosome Small ribosomal subunit is responsible for recognizing the translational start signal Initiation factors bind to the small ribosomal subunit to ensure proper interactions between the mRNA and the initiator tRNA Elongation factors, one of which escorts incoming charged tRNAs into the ribosome and the other of which is responsible for ribosomal translocation GTP provides the energy to release the amino acid from a charged tRNA as well as to induce a conformational change needed for translocation Mg2+ is essential for the activity of several enzymes Release factors stimulate hydrolysis of the completed polypeptide from the peptidyl tRNA and dissociate the ribosomal subunits

14-3: Many dipteran (two-winged) flies have giant banded chromosomes in certain tissues, for example, in salivary glands. These chromosomes begin as normal chromosomes, but then become larger and larger due to repeated DNA replication without cell division. The characteristic banding patterns of these chromosomes are easily detectable under the microscope, making them an effective system in which to observe and localize chromosomal mutations. These types of chromosomes are called:

polytene chromosomes

14-4: When the dominant allele of a heterozygote is deleted, the recessive phenotype will unexpectedly appear. This phenomenon (i.e., the unexpected expression of a recessive trait) is known as:

pseudodominance

15-7b: Indicate the phage genotype(s) that would result in large plaques. Select all that apply: r+ h+ r+ h r h+ r h

r h+ r h

15-7d: Indicate the phage genotype(s) that would result in small, clear plaques.

r+ h

15-7c: Indicate the phage genotype(s) that would result in cloudy (vs. clear) plaques. Select all that apply: r+ h+ r+ h r h+ r h

r+ h+ r h+

19-11c: Indicate all the point mutations that are not within the boundaries of the deletion present in mutant r11. Select all that apply: r1 r2 r3 r4 r5 r6 r7

r1 r2 r3 r4

19-12e: Which is the least useful deletion mutant for the purposes of the questions above? In other words, which deletion mutant could you have excluded from the pair-wise crosses to still obtain correct answers in Parts B - D?

r21

19-12a: You conduct pair-wise crosses between the deletion mutants from above and the point mutants r1, r2, r3, and r4. Your results are shown in the table below, with a "+" indicating the presence of recombinants. When crossed with point mutant r1, which deletion mutant(s) produce recombinant progeny? Select all that apply: r21 r22 r23 r24 r25 r26

r25 r26

19-11d: You isolate a new point mutant r8 and want to locate it within the linkage map described above. You conduct point crosses between r8 and your three deletion mutants. You observe r+ recombinants only in the cross of r8 and r12. Nearest to which point mutant is r8 located?

r6

19-8c: What is the order of these mutational sites within the rII region?

rIIc rIId rIIa rIIb

13-14b: What is the correct sequence of these genes on the X chromosome?

sc v s (v in the middle)

13-12b: With respect to the three genes mentioned in the problem, what are the genotypes of the parents used in making the phenotypically wild-type F1 heterozygote?

sp dsr cn+ / sp dsr cn+ and sp+ dsr+ cn / sp+ dsr+ cn The parental female fly displayed the mutant traits of disrupted wings (dsr) and speck body (sp), while the parental male fly had cinnabar eyes (cn). This would imply that the female fly had a wild-type copy of the cinnabar eye gene and that the male fly had wild-type copies of the disrupted wings and speck body genes. Recall that wild-type versions of genes are indicated as "+". Therefore the female fly would have the sp dsr cn+ / sp dsr cn+ genotype while the male fly would have the sp+ dsr+ cn / sp+ dsr+ cn genotype.

15-12a: Which gene is located between the other two on the λ phage chromosome?

t Parental type: h+ r+ t+ h r t DCOs: h+ r+ t h r t+ Gene that switched: t

20-15a: Contrast the roles of tRNA and mRNA during translation.

tRNA -contains anticodon -carries amino acid mRNA -contains a copy of DNA triplet codes -contains codon

17-5a: Choose the correct gene order:

trp qts pyr

16-7a: On which type(s) of media will the AX strain (AmpS bio+ trp+ leu-) grow? Select the THREE that apply: leu- trp- bio- leu- trp- bio- + ampicillin complete medium complete medium + ampicillin

trp- bio- complete medium Bactria that are AmpS cannot grow in the presence of ampicillin. If a strain of bacteria has a mutation in a gene that is necessary to make an essential nutrient (for example, leu-), then the strain requires that nutrient to be provided in the medium.

14-7c: What is the order of the genes, starting from band 1?

wzuvyx

15-8a: Which are examples of recombinant progeny? Select all that apply: y+ z+ y+ z y z+ y z

y+ z y z+

14-7: Here, six bands in a polytene chromosome of Drosophila are shown, along with the extent of five deletions. You know that the genes corresponding to the recessive alleles u, v, w, x, y, and z are in the region, but do not know their order. When individual flies are heterozygous for a deletion, then the following results are obtained:

| u | v | w | x | y | z Del 1 - + - + + - Del 2 - + + + + + Del 3 - - + + + - Del 4 - - + + - + Del 5 + - + - - + In this table, the minus sign means that the deletion uncovers the recessive allele. Use these data to determine the following: