Genetics Mastering and LC: Test 2

b (After several cell divisions, only the larger of the two translocated chromosomes remains. Its reciprocal is lost.)

A Robertsonian translocation is considered non-reciprocal because _______. a. an uneven number of gametes is produced in each meiosis b. the smaller of the two reciprocal products of translocated chromosomes is lost c. for every viable gamete formed, there are two inviable gametes formed d. trisomies of chromosome 21 are viable, whereas monosomies of the same chromosome are not

e (A broken fragment of a chromosome is referred to as acentric because it lacks a centromere altogether. This means it can easily be lost from a cell during mitosis because it cannot be attached to the mitotic spindle.)

A broken fragment of a chromosome that lacks a centromere is referred to as _________. a. submetacentric b. telocentric c. metacentric d. acrocentric e. acentric

b (During DNA replication, DNA polymerase I removes the RNA primers provided by primase after DNA polymerase III has used these primers to synthesize the rest of the DNA. If a cell cannot replace the RNA with DNA prior to DNA ligase re-joining the DNA fragments, DNA polymerase III is most likely defective.)

A cell is able to replicate DNA, but it is unable to remove and replace the RNA primers. Which enzyme is likely defective? a. Helicase b. DNA polymerase I c. Ligase d. DNA polymerase III e. Topoisomerase f. Primase

ABAB⋅CDEFG - Tandem Duplication AB⋅CDEABFG - Displaced Duplication AB⋅CFG - Interstitial Deletion AB⋅CDEF - Terminal Deletion

A chromosome has the following segments AB⋅CDEFG, where ⋅ represents the centromere. Match each new chromosome configuration with the resulting type of structural change. 1st choice Interstitial Deletion 2nd choice Displaced Duplication 3rd choice Terminal Deletion 4th choice Tandem Duplication AB⋅CFG AB⋅CDEABFG AB⋅CDEF ABAB⋅CDEFG

a, b, c, d (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, both X chromosomes end up in one cell, with no X chromosome in the other cell. If this happens in the father, both the X and Y chromosome end up in one cell, with no sex chromosome in the other cell. In both cases it is possible for one parent to not give a sex chromosome to the child. If the other parent gives an X chromosome, the child will be XO and have Turner syndrome. During meiosis II, sister chromatids separate. If nondisjunction occurs, both sister chromatids migrate into same daughter cell, instead of different daughter cells. If this happens in the mother, two identical copies of an X chromosome end up in one cell, with no X chromosome in the other cell. If this happens in the father, either two identical copies of the X chromosome end up in one cell, with no sex chromosome in the other cell - or - two identical copies of the Y chromosome end up in one cell, with no sex chromosome in the other cell. In all cases, it is possible for one parent to not give a sex chromosome to the child. If the other parent gives an X chromosome, the child will be XO and have Turner syndrome)

A couple has a daughter with Turner syndrome, a condition in which only a single copy of the X chromosome is present. This results from nondisjunction, the failure of the X chromosome to segregate properly during meiosis. During which meiotic division, and in which parent, could nondisjunction have occurred to produce a child with this condition? Hint: It may be helpful to sketch a diagram depicting the outcome of nondisjunction in meiosis I and II of both parents to solve this problem. Select all that apply. a. Meiosis I in the mother b. Meiosis II in the mother c. Meiosis I in the father d. Meiosis II in the father

a, b, c (To produce a child with XXY, either mom gave two X chromosomes and dad gave Y OR mom gave one X chromosome and dad gave both X and Y. There are two ways mom could give two X chromosomes - either she gave both of her X's (meiosis I nondisjunction), or she gave two copies of one of her X's (meiosis II nondisjunction). In both these cases if dad gives a Y chromosome, the child is XXY and has Klinefelter's. There is only one way dad could give X and Y, and that's if X and Y did not separate in meiosis I. Then if mom gives an X chromosome, the child is XXY and has Klinefelter's. Nondisjunction in meiosis II in dad would give both sister chromatids - either XX or YY, which would produce either a XXX girl or a XYY boy... not a XXY boy.)

A couple has a son with Klinefelter syndrome, a condition in which an extra copy of the X chromosome is present. This condition also results from nondisjunction. During which meiotic division, and in which parent, could nondisjunction have occurred to produce a child with this condition? Select all that apply. a. Meiosis I in the mother b. Meiosis II in the mother c. Meiosis I in the father d. Meiosis II in the father

d (The key to solving this problem is to realize that the child got two Y chromosomes. Both of those Y chromosomes had to have come from the father, as the mother has no Y chromosomes. In order to get two Y chromosomes, nondisjunction would have to have occurred in meiosis II - identical Y sister chromatids migrated together into one cell. During meiosis II, sister chromatids should separate. If nondisjunction occurs, both sister chromatids migrate into same daughter cell, instead of different daughter cells. If this happens in the father, either two identical copies of the X chromosome end up in one cell, with no sex chromosome in the other cell - or as in this case two identical copies of the Y chromosome end up in one cell, with no sex chromosome in the other cell. If the sperm containing two copies of the Y chromosome fertilizes an egg containing an X chromosome, the child will be XYY)

A couple has a son with XYY syndrome, a condition in which an extra copy of the Y chromosome is present. This condition also results from nondisjunction. During which meiotic division, and in which parent, could nondisjunction have occurred to produce a child with this condition? Select all that apply. a. Meiosis I in the mother b. Meiosis II in the mother c. Meiosis I in the father d. Meiosis II in the father

37

A diploid organism has 2n=36 chromosomes. How many chromosomes will be found in a trisomic member of this species?

d

A female has the genotype A1B1C / A2B1c, and a male has the genotype A1B2c / A2B2c. Suppose gene C represents a dominantly inherited disease-causing allele. If these individuals have children, what haplotype would you be looking for to predict whether their child will be affected? a. A2B2 b. A1B1 or A2B2 c. A2B1 d. A1B1C e. A1B2

a (A human germ cell that experiences a meiotic disjunction of chromosome 21 in meiosis I will result in gametocytes in which two cells will have 24 chromosomes, and two cells will have 22 chromosomes. A nondisjunction during meiosis I would result in all four gametocytes being aneuploid.)

A human germ cell experiences a meiotic disjunction of chromosome 21 in meiosis I. Which of the following statements best describes the pattern you would expect to observe in the resulting gametocytes? a. Two cells will have 24 chromosomes, and two cells will have 22 chromosomes. b. Two cells will have 24 chromosomes, and two cells will have 23 chromosomes. c. Two cells will have 23 chromosomes, and two cells will have 22 chromosomes. d. One cell will have 24 chromosomes, one cell will have 22 chromosomes, and two cells will have 23 chromosomes. e. All cells will have 23 chromosomes.

c (A B C G F - centromere - E D H I J is a pericentric inversion, since the chromosome broke between genes C and D, and between genes G and H. The chromosome segment rotated, and the inversion included the centromere.)

A metacentric chromosome has the following sequence of genes: A B C D E - centromere - F G H I J Which of the following gene sequences would result from a pericentric inversion? a. A B C D E - centromere - J I H G F b. A B C D E C D E - centromere - F G H I J c. A B C G F - centromere - E D H I J d. A D C B E - centromere - F G H I J e. A B C D E - centromere - G H I J

d (A partial deletion heterozygote is characterized by one wild-type chromosome and one homologous chromosome with a terminal deletion. The terminal deletion results from a single break at the end of a chromosome, while an interstitial deletion is the loss of an internal segment of a chromosome that results from two breaks and the rejoining of the ends. Cri-du-chat syndrome is an example of a partial deletion heterozygote.)

A partial deletion heterozygote is characterized by what type of mutation? a. Two homologous chromosomes with terminal deletions. b. Two homologous chromosomes with interstitial deletions. c. One wild-type chromosome and one homologous chromosome with a deletion of an internal segment. d. One wild-type chromosome and one homologous chromosome with a terminal deletion. e. One wild-type chromosome and one homologous chromosome with a chromosomal rearrangement.

c (The parental haplotype A2B1D occurs when there is no crossing over between genes A, B, and D. So, if the probability of crossing over between A and B is 0.2, the probability of no crossover is 0.80. If the probability between B and D is 0.15, the probability of no crossover is 0.85. So, (0.5)(0.80)(0.85) = 0.34)

A patient has a disease-causing allele, D, and the genotype A1B1D / A2B2d. You are trying to decide if this haplotype can be used to predict the inheritance of the allele, D, in this patient's children. First, you need to decide how often crossing over occurs between these genetic markers to decide if it will occur. The distance between A and B is 20 cM, and between B and D is 15 cM. What is the probability of an affected individual inheriting the parental haplotype with the disease-causing allele (A1B1D)? a. 17% b. 68% c. 34% d. 12% e. 16%

d (To calculate this frequency, you have two options. First, you know the probability of this individual passing the disease-causing allele is 50%. If 34% of their affected offspring have the parental haplotype, then 50-34 = 16%. Or, you can calculate the total probability of recombinant offspring (single crossovers and double crossover), then divide that number by half. Both methods will give you 16%.)

A patient has a disease-causing allele, D, and the genotype A1B1D / A2B2d. You are trying to decide if this haplotype can be used to predict the inheritance of the allele, D, in this patient's children. First, you need to decide how often crossing over occurs between these genetic markers to decide if it will work. The distance between A and B is 20 cM, and between B and D is 15 cM. What is the probability of an affected individual inheriting a recombinant haplotype with the disease-causing allele? a. 68% b. 34% c. 17% d. 16% e. 12%

c (For a paracentric inversion heterozygote who is attempting to complete meiosis, the sequence of the acentric fragment lost during anaphase I separation is Z Y X W V S T U V W X Y Z. The inversion loop would allow for crossover between the homologs, so the acentric chromosome would be the two sequences on either side of the point of the inversion. In this case, the inversion rotates around the T, so one half of the acentric fragment is the correct order (T U V W X Y Z), and the other side is the inversion (T S V W X Y Z). If you put those two fragments together into one complete acentric fragment with T in the middle, you would get this complete fragment.)

A patient is a paracentric inversion heterozygote, so their two chromosomes are the following sequences: Q·R S T U V W X Y Z and Q·R U T S V W X Y Z. What is the sequence of the acentric fragment lost during anaphase I separation? a. Q R b. Q R U T S c. Z Y X W V S T U V W X Y Z d. T U V W X Y Z e. Z Y X W V S T

d

A researcher interested in studying a human gene on chromosome 21 and another gene on the X chromosome uses FISH probes to locate each gene. The chromosome 21 probe produces green fluorescent color, and the X chromosome probe produces red fluorescent color. If the subject studied is female, how many green and red spots will be detected? Explain your answer. a. Females contain two copies each of chromosome 21 and the X chromosome one of which is inactivated, thus there will be either two green spots and one red spot or four green spots and two red spots in cells from a female depending on the cell cycle phase. b. Females contain two copies each of chromosome 21 and the X chromosome one of which is inactivated, thus there will be two green spots and one red spot in cells from a female. c. Females contain two copies each of chromosome 21 and the X chromosome, thus there will be either two green spots and two red spots or four green spots and four red spots in cells from a female depending on the cell cycle phase. d. Females contain two copies each of chromosome 21 and the X chromosome, thus there will be two green spots and two red spots in cells from a female.

c

A researcher interested in studying a human gene on chromosome 21 and another gene on the X chromosome uses FISH probes to locate each gene. The chromosome 21 probe produces green fluorescent color, and the X chromosome probe produces red fluorescent color. If the subject studied is male, how many green and red spots will be detected? Explain your answer. a. Males contain two copies of chromosome 21 but only one copy of the X chromosome and one copy of the Y chromosome, thus there will be two green spots and two red spots in cells from a male. b. Males contain two copies of chromosome 21 and one copy of the X chromosome which is inactivated, thus there will be two green spots and no red spots in cells from a male. c. Males contain two copies of chromosome 21 but only one copy of the X chromosome, thus there will be two green spots and one red spot in cells from a male.

b

After a cross, you collect the following progeny: 14 AB / ab 16 ab / ab 34 Ab / ab 36 aB / ab What are the parental genotypes? a. ab and Ab b. Ab and aB c. AB and aB d. AB, ab, Ab, and aB e. AB and ab

daughter DNA

After replication is complete, the strand of new DNA, called_____, is complementary to the parental strand.

b (When the centromere is contained within the inversion, it is neither duplicated nor deleted)

All of the following could result from meiosis in a pericentric inversion heterozygote in which a single crossover occurred within the inversion loop except a _______. a. chromosome with some duplicated regions b. chromosome with two centromeres c. completely normal chromosome d. chromosome with some deleted regions

c (Fusion of haploid gametes occurs after meiosis)

All of the following events occur during normal meiosis except _______. a. sister chromatids separate from one another during meiosis II b. homologous chromosomes separate from one another during meiosis I c. two haploid gametes fuse to form a diploid cell d. one diploid cell produces four haploid cells

a

Allele A has the sequence 5'-ATCCGA-3', while allele B has the sequence 5'-ATCGGA-3'. You have designed a PCR-based test to screen patients for the presence of either allele. Which type of common genetic marker is seen in these variants? a. SNP b. Hotspot c. VNTR d. RFLP e. GWAS

c (1/2*1/2*1/2=1/8)

An individual has a genotype AaBbCc, where ABC and abc are the parental gametes. What is the predicted frequency of the AbC nonparental gamete, assuming no linkage? a. <<<50% b. >>>50% c. 12.5% d. 25% e. 50%

c (The expected number of double recombinants is equal to the product of the two single crossovers. (0.2)(0.22) = 0.044, or 4.4%. The observed number of double crossovers is 3/100 = 0.03, or 3%. Thus, c = 3/4.4 = 0.68. So, I = 1-c, or 0.32. This means that the observed number of double recombinants is 32% lower than expected.)

An organism of the genotype a+b+c+/abc was testcrossed to a triply recessive organism, abc/abc. The phenotypic classes of the progeny are as follows: 28 a+b+c+ 27 abc 10 a+bc 12 ab+c+ 9 a+b+c 11 abc+ 1 a+bc+ 2 ab+c What is the coefficient of coincidence? a. 0.044 b. 1.46 c. 0.68 d. 0.03 e. 0.32

d (Approximately 146 bp of DNA is wrapped around the nucleosome core particle. Therefore 200 bp - 146 bp = 54 bp of linker DNA. This is the amount of DNA present in the "string.")

Assume that the DNA associated with a nucleosome core particle plus the DNA in the linker adds up to 200 bp. Approximately how many base pairs are found in the linker region? a. ~200 b. ~146 c. ~34 d. ~54 e. ~11

48 (12% of the gametes would be recombinant if the loci are 12 m.u. apart. But a single crossover event only produces 50% recombinant gametes. So 24% of oocytes would have a chiasma event (resulting in 12% recombinant gametes))

Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above?

d

Assuming no genetic linkage, how many different gametic combinations can be produced by an individual with the genotype AaBbCcDd? a. 32 b. 2 c. 8 d. 16 e. 4

c (Polyploidy refers to the presence of three (or more) sets of chromosomes in the nucleus. Bread wheat, Triticum aestivum, is a cultivated wheat species with two sets of chromosomes from three different species, making it allohexaploid. "Allo" refers to the fact that chromosomes arise from two or more species, while "hexa" refers to the presence of six sets of chromosomes in the nucleus.)

Bread wheat, Triticum aestivum, is a cultivated wheat species with two sets of chromosomes from three different species. What term describes this type of polyploidy? a. Autotriploid b. Allodiploid c. Allohexaploid d. Allotriploid e. Autohexaploid

c

Certain classes of drugs target the enzyme responsible for relaxing DNA supercoiling ahead of the replication bubble. By preventing this enzyme from reducing the strain, the DNA often experiences random breaks and fragments, ultimately killing the cell. Which enzyme is being targeted by this type of drug? a. DNA polymerase I b. DNA polymerase III c. Topoisomerase d. Primase e. Helicase f. Ligase

a, b, e (All cells within the eye have the same genotype, so differences are attributed to epigenetic changes. White patches in the eye are the result of silencing the w + allele by spreading of heterochromatin. On the other hand, when the heterochromatin does not spread into the w + allele, gene expression is normal, resulting in red patches in the eye. This results in a variegated eye phenotype.)

Chromatin structure is dynamic. In regions of highly condensed chromatin, such as the centromere, the boundary between heterochromatin and euchromatin is variable. Genes that are near this boundary region can be influenced by either type of chromatin in what is referred to as position effects. Recall the early Drosophila melanogaster experiments by Hermann Muller where the repositioning of the w + allele (normal activity of the w + allele produces red eye pigment) by translocation or inversion near this boundary of chromatin produced intermittent w + activity. In the heterozygous state (w +/w), a variegated eye is produced, with white and red patches. Which statements are true in relation to this experiment? Select all that apply. a. When heterochromatin spreading encompasses the new location of w + allele, the gene is not transcribed, producing white eye patches. b. When heterochromatin spreading does not reach the new location of the w + allele, the gene will be transcribed, producing red eye patches. c. Since the genotype is heterozygous, half of the cells will receive the w + allele producing red eye patches, while the other half will receive the w allele producing white patches. d. When heterochromatin spreading encompasses the new location of w + allele, the gene is transcribed, producing white eye patches. e. Even though eye color phenotypic variegation exists within the eye, all cells have the same genotype.

a

Chromosome duplications often result in abnormal phenotypes because: a. developmental processes depend on the relative amounts of proteins encoded by different genes b. extra copies of the genes within the duplicated region do not pair in meiosis c. the chromosome is more likely to break when it loops in meiosis d. extra DNA must be replicated, which slows down cell division

a, c, d (During meiosis I in an individual with Klinefelter syndrome, two sex chromosomes will migrate to one daughter cell and one sex chromosome will migrate to the other daughter cell. At meiosis II, sister chromatids separate. There are three possible ways in which the sex chromosomes could separate in an XXY individual.)

Consider the distribution of sex chromosomes that would occur when a male with Klinefelter syndrome (X1X2Y) undergoes spermatogenesis. Which chromosome combinations are possible? Select all that apply. a. X1X2 and Y b. X1X2Y and O c. X1Y and X2 d. X2Y and X1

b

Crossing over cannot increase genetic variety if: a. homologous non-sister chromatids exchange segments b. the exchanged chromosome segments are identical c. it takes place during gamete formation d. kinetochore microtubules pull the sister chromatids away from each other very fast

d

DNA polymerase I (DNA pol I) possesses which of the following enzymatic activities? a. 5'-to-3' exonuclease activity only b. 3'-to-5' exonuclease activity and 3'-to-5' polymerase activity c. 3'-to-5' exonuclease activity only d. 5'-to-3' exonuclease activity and 5'-to-3' polymerase activity e. 3'-to-5' exonuclease activity and 5'-to-3' polymerase activity f. 5'-to-3' polymerase activity only

a

Dark G bands that appear along chromosomes contain ... a. heterochromatic regions. b. euchromatic regions. c. telomeres. d. centromeres.

d

Deoxyribose is a five-carbon sugar. Which carbon is attached to the phosphate? a. 2' b. 1' c. 3' d. 5' e. 4'

c

Describe what is meant by the Down syndrome critical region (DSCR). a. The Down syndrome critical region (DSCR) refers to the minimal segment of chromosome 12 that can be present in three copies and cause the morphological and physiological symptoms seen in individuals with trisomy 12. b. The Down syndrome critical region (DSCR) refers to the average segment of chromosome 21 that can be present in three copies and cause the morphological and physiological symptoms seen in individuals with trisomy 21. c. The Down syndrome critical region (DSCR) refers to the minimal segment of chromosome 21 that can be present in three copies and cause the morphological and physiological symptoms seen in individuals with trisomy 21. d. The Down syndrome critical region (DSCR) refers to the average segment of chromosome 12 that can be present in three copies and cause the morphological and physiological symptoms seen in individuals with trisomy 12.

c

Discontinuous replication is a result of which property of DNA? a. Complementary bases b. Charged phosphate group c. Antiparallel nucleotide strands d. Five-carbon sugar

b

Discuss the following concerning interpretation of the COVID-19 PCR-based test. If you were a research medical doctor in Seattle testing your patients for CORVID-19 and you one of your patients tested positive. In talking with the patient you learn: -> his brother had been mildly sick 2 weeks earlier (but now feels fine), and -> his mother just started a fever yesterday. What would you expect from their test results? a. Brother but NOT mother would test positive b. Mother but NOT brother would test positive c. Neither brother or mother would test positive d. Both brother and mother would test positive

replication fork

During DNA replication, an open section of DNA, in which a DNA polymerase can replicate DNA, is called a_____.

f

During DNA replication, which enzyme breaks hydrogen bonds between the strands? a. DNA polymerase I b. Primase c. Topoisomerase d. DNA polymerase III e. Ligase f. Helicase

d

Extract from the heat-killed SIII bacteria is treated with the following enzymes, and then mixed with type RII bacteria. Which enzyme treatment results in no recovery of living type SIII bacteria? a. RNase and Protease b. Protease c. RNase d. DNase e. Lipase

e (The frequency of the gamete XyZ is equivalent to half of the number of double crossovers, or (0.5)(0.2)(0.3) = 0.03)

For a chromosome XYZ / xyz, the distance between X and Y is 0.2 and the distance between Y and Z is 0.3. What is the frequency of the gamete XyZ? a. 0.12 b. 0.07 c. 0.1 d. 0.15 e. 0.03 f. 0.06

c

For a given DNA sequence 5'-ATGCAGCGATAT-3', what is the sequence of the complementary strand? a. 3'-ATGCAGCGATAT-5' b. 5'-TACGTCGCTATA-3' c. 3'-TACGTCGCTATA-5' d. 3'-CGTACTATCGCG-5' e. 3'-GCATGATAGCGC-5'

c (A-T: 2 hydrogen bonds; G-C: 3 hydrogen bonds)

For a segment of double-stranded DNA with the following sequence, how many hydrogen bonds would you predict to observe between these two strands? 5'-ACAGGCATCG-3' 3'-TGTCCGTAGC-5' a. 16 b. 10 c. 26 d. 24 e. 20

e

Genes C and D are 15 map units apart. For a heterozygous organism with the genotype CD / cd that is test-crossed to a recessive individual, what percentage of their progeny would be Cd / cd? a. 85% b. 42.5% c. 30% d. 15% e. 7.5%

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

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%. 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? a. 50% b. 20% c. 30% d. 5%

c (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%.)

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? a. 5% b. 30% c. 1.25% d. 50%

b (In the dispersive model, each DNA strand is a composite of the parental and daughter strands in duplex segments, so you would predict only one band if the dispersive model was correct.)

Growing bacteria in the lighter 14N isotope, Meselson and Stahl observed two bands after two rounds of DNA replication. This observation eliminated the possibility of which of the three models of DNA replication? a. Semi-conservative b. Dispersive c. Dispersive and conservative d. Conservative e. Semiconservative and conservative

c (Growing bacteriophage in media containing 32P would label phage DNA because DNA contains a large amount of phosphorus.)

Growing bacteriophage in media containing ³²P would label which component of the phage? a. Lipids b. Proteins c. DNA d. RNA e. Both DNA and protein

c

Haploinsufficiency of which gene is responsible for the symptoms associated with Turner syndrome? a. FISH b. 15q11.12 c. SHOX d. DSCR e. TERT

b (Histones are basic proteins that interact with negatively charged DNA. The strength of this interaction is modulated by epigenetic modifications. One of these modifications is acetylation. Acetylation adds an acetyl group to the positively charged amino group present on the side chain of the amino acid lysine effectively changing the net charge of the protein by neutralizing the positive charge. When the positive charge is reduced, the histones loosen their grip on the negatively charged DNA.)

Histone acetyltransferases (HATs) are capable of remodeling chromatin by adding acetyl groups to various lysine residues in histones that comprise the nucleosome. Following this modification, the lysine residue no longer has a positive charge. Which statement is true? a. Histones in general have a net negative charge that allow them to bind to DNA. Acetylation of histones, decreases their positive charge and strengthens the histone-DNA interaction b. Histones in general have a net positive charge that allow them to bind to DNA. Acetylation of histones, decreases their positive charge and weakens the histone-DNA interaction. c. Histones in general have a net positive charge that allow them to bind to DNA. Acetylation of histones, decreases their positive charge and strengthens the histone-DNA interaction. d. Histones in general have a net negative charge that allow them to bind to DNA. Acetylation of histones, decreases their positive charge and weakens the histone-DNA interaction.

d (If one gamete is aneuploid, with two copies of the chromosome, and the other gamete is normal, then trisomy rescue will randomly eject an extra copy of the chromosome to allow the zygote to develop with normal chromosome content. This permits the survival of a fertilized egg that was initially trisomic, but which of the three copies that is ejected is random. Thus, trisomy rescue can also result in uniparental disomy.)

How can aneuploidy result in uniparental disomy? a. Mosaicism b. Gene dosage c. Chromatin condensation d. Trisomy rescue e. Nondisjunction during mitosis

a

How many chromosomes would be found in an allopolyploid plant if its parents had diploid numbers of 4 and 6 respectively? a. Diploid number of 5 b. Haploid number of 12 c. Haploid number of 5 d. Diploid number of 10

8 (Two molecules each of four histones - H2A, H2B, H3, and H4 - join together to form an octameric (eight-member) nucleosome core particle.)

How many total polypeptide chains would be present in a single nucleosome core particle?

d

How might those genes lead to the main symptoms of Down syndrome? a. Three copies of chromosome 21 lead to a decrease in expression of at least some of the genes on it, which disrupts the balance of gene expression during postnatal development, leading to physical abnormalities. b. Three copies of chromosome 21 lead to a decrease in expression of at least some of the genes on it, which disrupts the balance of gene expression during embryonic development, leading to developmental abnormalities. c. Three copies of chromosome 21 lead to an increase in expression of at least some of the genes on it, which disrupts the balance of gene expression during postnatal development, leading to physical abnormalities. However, there will be no developmental abnormalities. d. Three copies of chromosome 21 lead to an increase in expression of at least some of the genes on it, which disrupts the balance of gene expression during embryonic development, leading to developmental abnormalities.

b (In a pulse-chase, the DNA is exposed to radioactive compounds temporarily, and then the radioactivity is removed. If the DNA replication is bidirectional, they predicted a pattern of labeled and unlabeled DNA segments moving outward from the origin of replication, which they observed.)

Huberman and Riggs used pulse-chase labeling experiments to study which aspect of DNA? a. Polarity of the DNA strand b. Bidirectionality of DNA replication c. Base stacking of the DNA helix d. Semi-conservative model of DNA replication e. Width of the DNA helix

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

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

b (Failure to transform suggests that the chemical degraded in that preparation is the one responsible for transformation, in this case polysaccharides.)

If Avery had observed transformation using only the extracts containing degraded DNA, degraded RNA, and degraded protein, but NOT the extract containing degraded polysaccharides, he would have concluded that _______. a. RNA is the genetic material b. polysaccharides are the genetic material c. the preparations were contaminated d. mice with diets rich in polysaccharides are resistant to bacterial infection

c

If a female (XX) zygote experienced mitotic nondisjunction and the female now has cells that are XO, XX, or XXX, then this is an example of what genetic phenomenon? a. Gene dosage b. Trisomy rescue c. Mosaicism d. Uniparental disomy e. Haploinsufficiency

b and c (During meiosis I in an individual with Klinefelter syndrome, two sex chromosomes will migrate to one daughter cell and one sex chromosome will migrate to the other daughter cell. As a result, following meiosis II, two gametes will contain 2n+1 chromosomes, and the other two gametes will contain n chromosomes.)

If a male with Klinefelter syndrome undergoes spermatogenesis (without nondisjunction), how many chromosomes would be present in the sperm? Select all that apply. a. 22 b. 23 c. 24 d. 45 e. 46 f. 47

5% (Because the opsin gene and the clotting factor gene are 10 map units apart, recombination would take place between these genes 10% of the time, producing crossover gametes. There are two possible crossover gametes: one with mutant alleles for both the opsin and clotting factor genes, and another with the wild-type alleles for both genes. Each of these two types of gametes would be produced 5% (10/2) of the time.)

If the opsin gene and the clotting factor gene are 10 map units apart, what percentage of Maria's children would inherit both a mutant allele for opsin and a mutant allele for clotting factor?

c (Each parent produces two types of gametes, a + b + and ab, giving the simple Mendelian ratio of 3 a + b + : 1 ab.)

If two genes on the same chromosome exhibit complete linkage, what is the expected F2 phenotypic ratio from a selfed heterozygote with the genotype a + b + ⁄⁄ ab? a. 1:1 b. 9:3:3:1 c. 1:1:1:1 d. 3:1

a (Because some of the nonvirulent bacteria acquired properties of the virulent bacteria, instructions for this transformation must be carried by the virulent bacteria.)

In 1928, Frederick Griffith established that _______. a. heat-killed bacteria harbor the constituent(s) necessary to convey genetic properties to living bacteria b. proteases have no effect on DNA c. mouse DNA could be transferred into bacterial cells d. mice could be infected with bacteria

a

In Drosophila, the genes for eye color, wing shape, and wing length are located on chromosome II. Purple eyes (pr), arc bent wings (a), and vestigial wings (vg) are the mutant forms of the wild type traits red eyes, straight wings, and long wings, respectively. You've discovered some data in your genetics laboratory which indicates that the distance between vg and pr is 12.5 m.u., the distance between a and pr is 44.7 m.u., and the distance between a and vg is 32.2 m.u. From this information, deduce the order of these genes on chromosome II and identify which gene is in the middle. a. vg b. pr c. a d. not enough information to tell

106 (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 1250. Using the information provided, we would have expected to obtain 0.322 x 0.125 x 1250 = 50 offspring with double crossover phenotypes. In that case, ((SCO+DCO)/total number of progeny) x 100 = m.u. (SCO + 50)/1250 x 100 = 12.5 m.u., where SCO = 106 offspring.)

In Drosophila, the genes for eye color, wing shape, and wing length are located on chromosome II. Purple eyes (pr), arc bent wings (a), and vestigial wings (vg) are the mutant forms of the wild type traits red eyes, straight wings, and long wings, respectively. You've discovered some data in your genetics laboratory which indicates that the distance between vg and pr is 12.5 m.u., the distance between a and pr is 44.7 m.u., and the distance between a and vg is 32.2 m.u. If a total of 1250 offspring were obtained from your cross, determine the number of offspring that you would expect to obtain that represent a single crossover event between pr and vg if interference does not occur.

c (Knowing that vg is in the middle, it is helpful to rewrite the female's genotype as a vg pr/+ + + or pr vg a/+ + + before attempting this problem. Once you've done this, consider that when a chiasma is formed between genes pr and vg during meiosis, that two recombinant gametes would be produced: + pr +/a + vg. When these gametes are involved in a fertilization event with gametes from a homozygous recessive male (a pr vg), offspring would phenotypically reflect the genotype of the maternal gametes.)

In Drosophila, the genes for eye color, wing shape, and wing length are located on chromosome II. Purple eyes (pr), arc bent wings (a), and vestigial wings (vg) are the mutant forms of the wild type traits red eyes, straight wings, and long wings, respectively. You've discovered some data in your genetics laboratory which indicates that the distance between vg and pr is 12.5 m.u., the distance between a and pr is 44.7 m.u., and the distance between a and vg is 32.2 m.u. You begin studying heterozygous females (a pr vg/+ + +) and homozygous recessive males as part of your thesis project. Based on the information from Part A, if you set up a cross between the female and male flies, which of the following represents the phenotypic class of offspring resulting from a single crossover event between pr and vg? a. + + vg/a pr + b. + pr vg/a + + c. + pr +/a + vg d. a pr +/+ + vg

a

In Griffith's experiment with mice, what did he observe when injecting heat-killed type SIII and living type RII bacteria into a mouse? a. The mouse died, and live type SIII bacteria were recovered. b. The mouse lived, and live type RII bacteria were recovered. c. The mouse lived, and live type RII and SIII bacteria were recovered. d. The mouse lived, and live type SIII bacteria were recovered. e. The mouse died, and live type RII bacteria were recovered.

c

In a diploid species of plant, the genes for plant height and fruit shape are syntenic and separated by 18 m.u. Allele D produces tall plants and is dominant to d for short plants, and allele R produces round fruit and is dominant to r for oval fruit. A plant with the genotype DR/dr produces gametes. Identify gamete genotypes, label parental and recombinant gametes, and give the frequency of each gamete genotype. a. Parental: 9% Dr, 9%dR; recombinant: 41% DR, 41%dr b. Parental: 41% Dr, 41%dR; recombinant: 9% DR, 9%dr c. Parental: 41% DR, 41%dr; recombinant: 9% Dr, 9%dR d. Parental: 9% DR, 9%dr; recombinant: 41% Dr, 41%dR

a

In a diploid species of plant, the genes for plant height and fruit shape are syntenic and separated by 18 m.u. Allele D produces tall plants and is dominant to d for short plants, and allele R produces round fruit and is dominant to r for oval fruit. Give the same information for a plant with the genotype Dr/dR. Identify gamete genotypes, label parental and recombinant gametes, and give the frequency of each gamete genotype. a. Parental: 41% Dr, 41%dR; recombinant: 9% DR, 9%dr b. Parental: 9% DR, 9%dr; recombinant: 41% Dr, 41%dR c. Parental: 41% DR, 41%dr; recombinant: 9% Dr, 9%dR d. Parental: 9% Dr, 9%dR; recombinant: 41% DR, 41%dr

e (In a sampling of 500 gametes with a crossover frequency of 32%, you would expect that 80 gametes would have the Cd genotype. To calculate, you can assume that a frequency of 32% recombination would mean that 16% would be Cd, and 16% would be cD, since those are the two possible recombinant genotypes. 16% of 500 = (0.16) x (500) = 80.)

In a sample of 500 gametes from an organism with the genotype CD/cd, how many gametes would you expect to have the genotype Cd, assuming crossover is 32%? a. 320 b. 160 c. 340 d. 170 e. 80

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

In a three‑point mapping experiment for the genes y‑w‑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? a. 15 map units b. 17 map units c. 18 map units d. 20 map units

d (In a trihybrid chromosome d+e+f+ / def, chromosome de+f represents a double crossover. In this case, one crossover occurred between d and e, and one between e and f.)

In a trihybrid chromosome d+e+f+ / def, which of the following represents a double crossover? a. de+f+ b. d+e+f c. def d. de+f e. d+e+f+

a (In a cross between pr+ vg+ / pr vg females and recessive males, collecting only 12 offspring with red eye, vestigial wing is indicative of genetic linkage. If the genes assort independently, you would expect to observe equal numbers of all four possible genotypes (25 of each). The red eye, full wing and the purple eye, vestigial wing represent the two parental genotypes, whereas the red eye, vestigial wing or the purple eye, full wing represent recombinant genotypes.)

In one of Morgan's test-crosses, he crossed pr+ vg+ / pr vg females with recessive males. Assuming you collect 100 offspring, which of the following results would indicate genetic linkage between these two genes? a. 12 red eye, vestigial wing b. 25 red eye, full wing c. 12 red eye, full wing d. 25 red eye, vestigial wing e. 12 purple eye, vestigial wing f. 25 purple eye, vestigial wing

b

In the Hershey and Chase experiment in 1952, the P-32 was associated with the bacterial pellet and the S-35 was associated with the viral capsule. Why? a. P-32 is labeling the amino acid proline (protein) b. P-32 in labeling the phosphate of the backbone of DNA c. S-35 is labeling the sulfur of nitrogenous of the DNA d. S-35 is labeling the RNAs

e (If wild potatoes have a diploid number of 24, tetraploid cultivated potatoes would be 2n = 4x = 48. The pollen from cultivated potatoes would go through a reductional division during meiosis, so the pollen would contain 24 chromosomes.)

In the wild, potatoes have a diploid number of 24. Cultivated potatoes that we grow for food are tetraploid. How many chromosomes would you expect to find in pollen from cultivated potatoes? a. 96 chromosomes b. 12 chromosomes c. 36 chromosomes d. 48 chromosomes e. 24 chromosomes

d

In which of the following organisms would you expect to find autonomously replicating sequences (ARSs)? a. E. coli b. Drosophila c. Archaea d. Yeast e. Mammals

d

Light G bands that appear along chromosomes contain ... a. heterochromatic regions. b. centromeres. c. telomeres. d. euchromatic regions.

c

Linkage disequilibrium describes a nonrandom relationship between alleles of closely linked genes, which means that specific alleles are linked to a specific haplotype. What type of visual graph can be used to illustrate locations of chromosomes where linkage disequilibrium has been detected? a. RFLP b. LOD score curve c. Manhattan plot d. P value e. Dot plot

a, c, d, e

List the main symptoms of Down syndrome. Select all that apply. a. short stature b. vision loss c. flattened facial profile and back of skull d. mental retardation e. eyes that are up-slanting with thick, epicanthal folds f. malformations of the respiratory system g. webbed feet

c

Mice have 40 chromosomes (2n = 40). Suppose a cell undergoing mitosis experiences a chromosome nondisjunction of a single chromosome. What are the chromosome numbers in the resulting daughter cells? a. One cell will be diploid (2n = 40), and the other will be monosomic (2n = 38). b. One cell will be diploid (2n = 40), and the other will be trisomic (2n = 42). c. One cell will be monosomic (2n-1 or 2n= 39), and the other will be trisomic (2n+1 or 2n= 41). d. One cell will be monosomic (n = 19), and the other will be trisomic (n = 21). e. One cell will be diploid (2n = 40), and the other will be triploid (3n = 60).

b

Mutations in which enzyme lead to human progeroid conditions? a. Topoisomerases b. Helicases c. Telomerases d. DNA polymerase e. Ligases

b (There is only one way that Jack and Jill could have a child with an AAa genotype -- if nondisjuction occurs in meiosis II during gametogenesis in Jill. She is the only parent with A to begin with. This would result in sister chromatids failing to separate, creating eggs with AA or aa. Upon fertilization with sperm containing a, offpsring with the AAa genotype could be produced. If a nondisjunction occurred in maternal meiosis I, the only gametes that would be produced would contain both A and a, or lack gene a altogether, resulting in offspring with Aaa or a genotypes upon fertilization - not AAa.)

Nondisjuction 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. When nondisjunction occurs in meiosis I, homologous chromosomes fail to separate. When nondisjunction occurs in meiosis II, sister chromatids fail to separate. Jill is heterozygous for gene A and is going to have a child with Jack, who is homozygous recessive for gene A. If Jack and Jill have a child with an AAa genotype, during which meiotic division, and in which parent, could nondisjunction have occurred? Select all that apply. a. Meiosis I in the mother b. Meiosis II in the mother c. Meiosis I in the father d. Meiosis II in the father

a, c, d (To produce a child with an Aaa genotype, either Jack gave two copies of gene a and Jill gave A OR Jack gave a and Jill gave both A and a. There are two ways Jack could give two copies of a -- from either meiosis I or II nondisjunction. In both of these cases, if Jill's egg contains the chromosome with gene A, the child is Aaa. There is only one way Jill could give both A and a, and that is if the chromosomes containing A and a did not separate in meiosis I. Then if Jack's sperm contains a chromosome with gene a, the child is Aaa.Nondisjunction in meiosis II in Jill would give both sister chromatids -- either AA or aa, which would produce either AAa or aaa ... not Aaa.)

Nondisjuction 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. When nondisjunction occurs in meiosis I, homologous chromosomes fail to separate. When nondisjunction occurs in meiosis II, sister chromatids fail to separate. Jill is heterozygous for gene A and is going to have a child with Jack, who is homozygous recessive for gene A. If Jack and Jill have a child with an Aaa genotype, during which meiotic division, and in which parent, could nondisjunction have occurred? Select all that apply. a. Meiosis I in the mother b. Meiosis II in the mother c. Meiosis I in the father d. Meiosis II in the father

b, c, d (Recall that if a nondisjunction event occurred in maternal meiosis I, the gametes produced would contain both A and a, or lack gene a altogether, resulting in offspring with Aaa or a genotypes upon fertilization - not aaa. However, nondisjucntion events in maternal meiosis II or paternal meiosis I or II could produce gametes containing aa that could result in offspring having an aaa genotype upon fertilization with a normal haploid gamete containing gene a.)

Nondisjuction 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. When nondisjunction occurs in meiosis I, homologous chromosomes fail to separate. When nondisjunction occurs in meiosis II, sister chromatids fail to separate. Jill is heterozygous for gene A and is going to have a child with Jack, who is homozygous recessive for gene A. If Jack and Jill have a child with an aaa genotype, during which meiotic division, and in which parent, could nondisjunction have occurred? Select all that apply. a. Meiosis I in the mother b. Meiosis II in the mother c. Meiosis I in the father d. Meiosis II in the father

a, c, d, e, f (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 daugher 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.)

Nondisjuction 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. When nondisjunction occurs in meiosis I, homologous chromosomes fail to separate. When nondisjunction occurs in meiosis II, sister chromatids fail to separate. 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 the fertilization of an egg that had undergone nondisjunction in meiosis II with a normal haploid sperm? (Because a nondisjunction event in meiosis II could involve the chromosome containing gene A or the chromosome containing gene a, be sure to consider both possibilities in your final answer.) Select all that apply. a. AAa b. Aaa c. aaa d. Aa e. aa f. a

b and f (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.)

Nondisjuction 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. When nondisjunction occurs in meiosis I, homologous chromosomes fail to separate. When nondisjunction occurs in meiosis II, sister chromatids fail to separate. 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 the fertilization of an egg that had undergone nondisjunction in meiosis I with a normal haploid sperm? Hint: It may be helpful to sketch the outcome of nondisjunction to solve this problem. Select all that apply. a. AAa b. Aaa c. aaa d. Aa e. aa f. a

b

Parental chromosomes have an arrangement with DE on one chromosome and de on the other. Assuming the genes exhibit incomplete linkage and the percentage of the parental gamete DE is 38%, what is the expected frequency of the De gamete? a. 76% b. 12% c. Not enough information to calculate d. 38% e. 24%

d (Patau syndrome is caused by a trisomy of chromosome 13. Humans are extremely sensitive to gene dosage, so many trisomies are incompatible with life. Only trisomies of autosomal chromosomes 13, 18, and 21 are seen in newborns, but the syndromes associated with these trisomies are severe. Patau syndrome affects 1 in 15,000 births, and typically results in severe developmental abnormalities and early death.)

Patau syndrome is a human aneuploidy that results in mental retardation, major organ abnormalities, possible deafness, and early death. Which chromosomal abnormality causes Patau syndrome? a. Trisomy 18 b. Monosomy X c. Trisomy X d. Trisomy 13 e. Trisomy 21

b (If there are more G/Cs, each of them require 3 hydrogen bonds instead of the 2 hydrogen bonds for A/T pairs. Thus, a higher annealing temperature is needed for primers with more G/Cs, but it also ensures that the primer is more tightly annealed to the template DNA.)

Primer annealing temperature can vary anywhere from 45-68°. Some of this variability is due to the length of the primer, but the base composition can also affect the primer's optimal annealing temperature. Why? a. Primers with more purines require a lower annealing temperature. b. Primers with a higher G/C content require a higher annealing temperature than primers with a high A/T content. c. Primers with more pyrimidines require a lower annealing temperature. d. Primers with a higher G/C content require a lower annealing temperature than primers with a high A/T content. e. Primers with more repetitive sequences don't anneal as well, so they require higher annealing temperatures.

e

Red-green color blindness is a human X-linked recessive disorder. A young woman with Turner's syndrome is color blind. Her brother is normal 46, XY but is also color blind. Both parents have normal color vision. Where did the nondisjunction that give rise to the young woman with Turner's syndrome take place? a. Dad, Meiosis I b. Dad, Meiosis II c. Mom, Meiosis I d. Mom, Meiosis II e. Dad, either Meiosis I or Meiosis II f. Mom, either Meiosis I or Meiosis II

c (Sanger sequencing works by limiting the amount of ddNTPs in the sequencing reaction. By including small concentrations of ddNTPs, this ensures that they are incorporated randomly, producing fragments of multiple lengths.)

Sanger sequencing generates shorter fragments of the PCR products by limiting which component of the PCR sequencing reaction? a. dNTPs b. Polymerase c. ddNTPs d. Template DNA e. DNA Primer

a

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

b

Studies of infants born with Down syndrome have established that over 90% of cases can be attributed to what chromosomal abnormality? a. Paternal meiosis II nondisjunction of chromosome 21 b. Maternal meiosis I nondisjunction of chromosome 21 c. Paternal meiosis I nondisjunction of chromosome 21 d. Mosaicism of trisomy 21 e. . Maternal meiosis II nondisjunction of chromosome 21

false (Offspring from this cross would be triploid and produce gametes with an uneven number of homologous chromosomes, making sexual reproduction unlikely)

T/F: A cross between a tetraploid and a diploid member of the same species will produce offspring that can undergo sexual reproduction.

false (Crossing over during meiosis occurs between alleles on nonsister chromatids.)

T/F: Crossing over during prophase I of meiosis occurs between alleles on sister chromatids.

true (During meiosis, the presence of a chromosomal loop signals that a duplication or deletion has occurred)

T/F: Duplications and deletions can be detected during meiosis by the presence of extrachromosomal loops that do not pair properly with their homolog.

false (Females with only one X chromosome are viable but have Turner syndrome, which is characterized by underdeveloped ovaries. Males that lack an X chromosome do not develop; this condition is lethal)

T/F: Females with only one X chromosome do not develop; this condition is lethal.

false (Nondisjunction during either meiosis I or meiosis II creates gametes that will generate trisomies if fertilized)

T/F: In order to create the possibility of generating a trisomy, nondisjunction must occur during meiosis II.

false (Loop formation allows pairing within inverted regions of homologous chromosomes regardless of whether the centromere is included in the inversion)

T/F: Inversion loops do NOT form during meiosis in paracentric inversion heterozygotes.

false (The data obtained from the Meselson-Stahl experiment after one generation was consistent with both the semiconservative and the dispersive model of DNA replication. The conservative model of DNA replication was eliminated because it predicted that there would be two bands representing the original DNA at one density and the newly replicated DNA at a different density.)

T/F: The data obtained from the Meselson-Stahl experiment after one generation of replication eliminated the dispersive model of DNA replication.

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

T/F: The second order of chromatin packing occurs when nucleosomes coil together to form a fiber that is 300 nm in diameter.

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.)

T/F: To construct a mapping cross of linked genes, it is important that the genotypes of some of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny.

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

T/F: Two genes that are separated by 10 map units show a recombination percentage of 10%.

lack, heterochromatic, inhibit, silenced

Telomeric regions of chromosomes generally_____genes because these chromosomal regions are_____and therefore_____gene expression. Thus, any gene located at telomeres would be_____unless something altered telomeric heterochromatin structure.

c (Because it was concluded that the component associated with bacteria at the end of the experiment must be the genetic material, it was critical that the component be identifiable as either DNA or protein)

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

36, 36, 14, 14

The boss in your laboratory has just heard of a proposal by another laboratory that genes for eye color and the length of body bristles may be linked in Drosophila. Your lab has numerous pure-breeding stocks of Drosophila that could be used to verify or refute genetic linkage. In Drosophila, red eyes (c +) are dominant to brown eyes (c), and long bristles (d +) are dominant to short bristles (d). Your lab boss asks you to design an experiment to test the genetic linkage of eye color and bristle-length genes, and to begin by crossing a pure-breeding line homozygous for red eyes and short bristles to a pure-breeding line that has brown eyes and long bristles. Assume the eye color and bristle-length genes are separated by 28 m.u.. What are the approximate frequencies of phenotypes expected from the cross you proposed in part (b)? Parental: c+d/c+d, cd+/cd+ ; progeny: cd+ /c+d. Cross F1 with ccdd Approximate frequencies of phenotypes: cd+____%, c + d____%, cd____%, ____ c + d +

b

The boss in your laboratory has just heard of a proposal by another laboratory that genes for eye color and the length of body bristles may be linked in Drosophila. Your lab has numerous pure-breeding stocks of Drosophila that could be used to verify or refute genetic linkage. In Drosophila, red eyes (c +) are dominant to brown eyes (c), and long bristles (d +) are dominant to short bristles (d). Your lab boss asks you to design an experiment to test the genetic linkage of eye color and bristle-length genes, and to begin by crossing a pure-breeding line homozygous for red eyes and short bristles to a pure-breeding line that has brown eyes and long bristles. Give the genotypes of the pure-breeding parental flies, and the genotype(s) and phenotype(s) of the F1 progeny they produce. a. Parental: cd+ /c+d , cd+ /c+d ; progeny: c+d/c+d b. Parental: c+d/c+d, cd+/cd+ ; progeny: cd+ /c+d c. Parental: cd+ /c+d , cd+ /c+d ; progeny: cd+/cd+ d. Parental: cd+ /c+d , cd+/cd+ ; progeny: cd+ /c+d

25, 25, 25, 25

The boss in your laboratory has just heard of a proposal by another laboratory that genes for eye color and the length of body bristles may be linked in Drosophila. Your lab has numerous pure-breeding stocks of Drosophila that could be used to verify or refute genetic linkage. In Drosophila, red eyes (c +) are dominant to brown eyes (c), and long bristles (d +) are dominant to short bristles (d). Your lab boss asks you to design an experiment to test the genetic linkage of eye color and bristle-length genes, and to begin by crossing a pure-breeding line homozygous for red eyes and short bristles to a pure-breeding line that has brown eyes and long bristles. How would the results of the cross differ if the genes are not linked? Parental: c+d/c+d, cd+/cd+ ; progeny: cd+ /c+d. Cross F1 with ccdd Approximate frequencies of phenotypes: cd+____%, c + d____%, cd____%, ____ c + d +

c

The boss in your laboratory has just heard of a proposal by another laboratory that genes for eye color and the length of body bristles may be linked in Drosophila. Your lab has numerous pure-breeding stocks of Drosophila that could be used to verify or refute genetic linkage. In Drosophila, red eyes (c +) are dominant to brown eyes (c), and long bristles (d +) are dominant to short bristles (d). Your lab boss asks you to design an experiment to test the genetic linkage of eye color and bristle-length genes, and to begin by crossing a pure-breeding line homozygous for red eyes and short bristles to a pure-breeding line that has brown eyes and long bristles. In your experimental design, what is the genotype of the line you propose to cross to the F1 to obtain the most useful information about genetic linkage between the eye color and bristle-length genes? a. c+c d+d b. cc +d +d c. ccdd d. c+ c+ dd

d

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

DNA polymerase

The enzyme that can replicate DNA is called_____.

b (If you collect 35 AB, 37 ab, 15 Ab, and 13 aB gametes, the recombination frequency is 28%. The recombinant gametes in this instance are Ab and aB. Thus, adding the frequency of the two recombinant classes would give you 15% + 13% = 28%.)

The gametes below were collected from an individual with genotype AB/ab: 35 AB 37 ab 15 Ab 13 aB What is the recombination frequency? a. 37% b. 28% c. 13% d. 72% e. 15% f. 35%

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)

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. 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

b and d (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 / mg + genotype would produce two parental gametes which were not involved in genetic exchange -- m + g and mg + -- and two recombinant gametes from the single genetic exchange between the m and g genes -- m + g + and mg. 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.)

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

e (If a mouse in the population (n=12) experiences a Robertsonian translocation, the mouse will have n=11 chromosomes in a gamete. The Robertsonian translocation causes a fusion of two nonhomologous chromosomes, so the resulting mice within the population will be 2n-2. A gamete is haploid, though, so the new species will be n-1.)

The haploid number of chromosomes in one population of mice is 12 (n=12). If a Robertsonian translocation occurs in the germ cells of a mouse from this population, how many chromosomes would be present in a gamete? Suppose a mouse in this population experiences a Robertsonian translocation, resulting in a new species. a. 24 b. 22 c. 12 d. 23 e. 11

b (Nondisjunction during either meiosis I or II can produce a gamete that will result in a trisomic zygote. Since the incidence of Down syndrome increases with maternal age, it is likely that this error occurs in the female gamete)

The incidence of Down syndrome, also known as trisomy 21, increases with increasing maternal age. Which of the following errors most likely produces this condition? a. Nondisjunction during meiosis I in either the male and female gamete b. Nondisjunction during either meiosis I or II in the female gamete c. Nondisjunction during meiosis II in either the male and female gamete d. Nondisjunction during either meiosis I or II in the male gamete

leading strand

The new DNA strand that grows continuously in the 5' to 3' direction is called the_____.

b (There are a total of 1000 progeny. There are 92+108 = 200 recombinant progeny between the two genes. With 200/1000 or 20% recombinants, the genes are 20 cM apart on the chromosome.)

The recessive A and B genes are located on the same chromosome, and a test cross is set up with a double heterozygous individual to determine the relative distance between these two genes. The test cross results in 396 offspring that are phenotypically wild type; 404 offspring that express both recessive traits; 92 offspring that express the recessive trait for A gene but are wild type for B; and 108 offspring that express the recessive trait for B gene but are wild type for A. What is the relative distance of A and B genes on the chromosome in centimorgans? a. 25 cM b. 20 cM c. 10 cM d. 12.5 cM

a (The recombination frequency between genes A and B is 30%, and between B and C is 25%. If you collect 70 double recombinants, the interference is 0.067.)

The recombination frequency between genes A and B is 30%, and between B and C is 25%. You collect 1,000 offspring, including 70 double recombinants. What is the interference? a. 0.067 b. 0.933 c. -0.07 d. 0.075 e. 1.07

a

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

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

The virulent form of the bacteria S. pneumoniae is called the S strain because it is surrounded by a polysaccharide coat that makes it appear smooth under a microscope. Sometimes the S strain mutates into a non-virulent form (called the R strain), which lacks the polysaccharide coat and appears rough. Frederick Griffith performed an experiment in which he injected mice with different combinations of these bacterial strains. What conclusion(s) could Griffith draw from his experiment? Select all that apply. a. A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent. b. The transforming factor is DNA. c. The transforming factor is RNA. d. The transforming factor is protein.

21q22, full spectrum of the clinical, DSCR3, DSCR8

The_____region of chromosome 21 has been identified as containing the DSCR, although the exact location of a subset of chromosome 21 that is responsible for the_____features of Down syndrome has not yet been proven. Many genes are located in this region, including RCAN1, PSGM1,_____, DSCR4, PIGP, DSCR6 and_____.

d (The complete extract possessed the same ability to induce transformation in IIR bacteria as whole heat-killed IIS bacteria.)

To be certain that the extract prepared from virulent cells still contained the transforming principle that was present prior to lysis, Avery _______. a. destroyed proteins, polysaccharides, DNA, and RNA contained in the extract b. incubated virulent cells with the complete extract c. injected mice with the extract d. incubated nonvirulent cells with the complete extract

c (Chargaff's rules state that there is a 1:1 ratio of purines to pyrimidines. More specifically the amount of adenine is equal to the amount of thymine, and the amount of cytosine is equal to the amount of guanine. Two purines would place the bases too far apart for hydrogen bonds to form. Two pyrimidines would place the bases too close together, making the bond unstable due to steric repulsion.)

To help achieve proper base pairing and hence form a double helix, which condition must be met? a. A purine base must pair with a purine base. b. A pyrimidine base must pair with pyrimidine base. c. A purine base must pair with a pyrimidine base.

a (Chargaff's rule confirms the base-pairing rules of A-T and C-G)

Watson and Crick derived their model of DNA structure not from their own laboratory research but rather through the analysis and interpretation of experimental data generated by others, such as Erwin Chargaff, Rosalind Franklin, and Maurice Wilkins. Which of the following provides evidence that DNA nucleotides are arranged as complementary base pairs? a. Percentages of adenine and thymine are approximately equal to one another, and the percentages of cytosine and guanine are approximately equal to one another. b. Complementary strands of DNA are antiparallel. c. DNA nucleotides contain a five-carbon deoxyribose sugar, a phosphate group, and one of four possible nitrogenous bases. d. The nucleotides forming a single strand of DNA are linked together by covalent phosphodiester bonds.

a (A ddNTP is used for Sanger sequencing because it is deoxygenated at both the 2' and 3' carbons, while a normal dNTP is deoxygenated at the 2' carbon only. The lack of a hydroxyl at the 3' carbon prevents the formation of a phosphodiester bond. That means the DNA cannot be extended, so the addition of a ddNTP will terminate strand synthesis, resulting in a large number of partial replication products.)

What is the difference between the structure of a dNTP and a ddNTP? a. A ddNTP is deoxygenated at both the 2' and 3' carbons, while a dNTP is deoxygenated at the 2' carbon only. b. A ddNTP is oxygenated at both the 2' and 3' carbons, while a dNTP is deoxygenated at the 2' carbon. c. A ddNTP is deoxygenated at both the 2' and 3' carbons, while a dNTP is oxygenated at both the 2' and 3' carbons. d. A ddNTP is oxygenated at both the 2' and 3' carbons, while a dNTP is deoxygenated at both the 2' and 3' carbons. e. A ddNTP lacks the phosphate on the 5' carbon, whereas the dNTPs use that phosphate to elongate the DNA chain.

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

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

b

What is the function of DNA polymerase? a. To catalyze the breaking of a phosphodiester bond between the 3' hydroxyl of one nucleotide and the 5' triphosphate group of an adjacent nucleotide. b. To catalyze the formation of a phosphodiester bond between the 3' hydroxyl of one nucleotide and the 5' triphosphate group of an adjacent nucleotide. c. To break the hydrogen bonds between the nucleotide bases of adjacent DNA strands. d. To catalyze the addition of a phosphate on the 5' carbon of deoxyribose. e. To catalyze the formation of a hydrogen bond between the 3' hydroxyl of one nucleotide and the 5' triphosphate group of an adjacent nucleotide.

a

What is the name given to a chromosome with a centrally located centromere and chromosome arms of essentially equal length? a. metacentric b. submetacentric c. acrocentric d. telocentric

e

What is the purpose of the DNA sequences called "telomeres"? a. To eliminate the need for RNA primers by providing a location for DNA pol III to bind at the ends of the chromosomes. b. To provide a DNA sequence for the topoisomerase to bind ahead of the replication fork. c. To stabilize the replication fork. d. To circularize the linear chromosomes during replication. e. To prevent the loss of genes as linear chromosomes shorten during replication.

c (The purpose of an adaptor in a NGS reaction is to bind to the end of a PCR product so it can be anchored to a substrate. These adaptors can also contain a primer to aid in the amplification of the template. However, the DNA must first be bound to the adaptors, then denatured before the amplification step can proceed.)

What is the purpose of the adaptor in the NGS protocol? a. To fragment the DNA b. To denature the DNA c. To anchor the DNA strand to a substrate d. To record the emission wavelength and measure intensity e. To allow the DNA to fluoresce

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

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

c (Balanced translocation heterozygotes have a normal amount of genetic material, but it is in a translocated configuration. As long as there are no position effects, these individuals can be phenotypically normal)

What phenotype would be expected in balanced translocation heterozygotes in the absence of position effects? a. Normal, even though they have too much genetic material b. Abnormal, because they have too much genetic material c. Normal, because they have a normal amount of genetic material d. Abnormal, because they lack some genetic material

a (A-T base pairs are stabilized by two hydrogen bonds, and C-G base pairs are stabilized by three hydrogen bonds)

What type of bond holds the two strands of DNA together through the formation of complementary base pairs? a. Hydrogen bonds b. Phosphodiester bonds c. Covalent bonds d. Ionic bonds

b; b

Where is it more likely to find the sequence encoding the digestive enzyme amylase? a. telomeric region b. euchromatic region c. heterochromatic region d. centromeric region Explain your reasoning. a. Expression of the amylase gene would be repressed if it was located in a centromeric chromosomal region. b. Expression of the amylase gene would be repressed if it was located in a heterochromatic chromosomal region, which includes centromeric and telomeric regions. c. Expression of the amylase gene would be repressed if it was located in a heterochromatic chromosomal region, which excludes centromeric and telomeric regions. d. Expression of the amylase gene would be repressed if it was located in a telomeric chromosomal region.

b

Where would you expect to see nonhistone proteins anchoring chromatin loops to the scaffold? a. Centromere b. MARs c. Nucleosome d. Solenoid e. PEV

d

Where would you expect to see the highly repetitive regions of DNA known as "satellites"? a. On the long or short ends of either arms of metacentric chromosomes b. On the long arm of acrocentric chromosomes c. On the short arm of submetacentric chromosomes d. On the short arm of acrocentric chromosomes e. On the long arm of submetacentric chromosomes

a

Which enzyme catalyzes the addition of nucleotides to a growing DNA chain? a. DNA polymerase b. Primase c. Telomerase d. Helicase

b (During DNA replication, DNA ligase joins the Okazaki fragments generated on the lagging strand by re-connecting the phosphodiester bonds on the DNA backbone between these fragments. It also helps to repair the phosphodiester bonds around the RNA primers after DNA pol I removes and replaces them with DNA, resulting in a continuous DNA backbone at the end of replication.)

Which enzyme catalyzes the formation of a phosphodiester bond between Okazaki fragments? a. Helicase b. Ligase c. DNA polymerase I d. Primase e. Topoisomerase f. DNA polymerase III

d

Which form of DNA exhibits a left-handed helical twist and is often found at the start sites for gene transcription? a. A-form and B-form b. B-form c. Z-form and B-form d. Z-form e. A-form

a

Which gene is responsible for aniridia, one of the symptoms of WAGR syndrome? a. PAX6 b. SHOX c. BDNF d. WT1 e. DSCAM

e

Which genetic markers consist of short sequences of DNA repeated end-to-end in a non-coding region of a chromosome? a. SNP b. Hotspot c. RFLP d. GWAS e. VNTR

a (The histone protein H1 plays a key role in stabilizing the 30-nm solenoid structure. The long N-terminal and C-terminal ends of the H1 protein attach to adjacent nucleosome core particles pulling them into an orderly solenoid array.)

Which histone helps stabilize the solenoid structure? a. H1 b. H2A c. H2B d. H3 e. H4

e

Which histone protein is not part of the nucleosome core particle? a. H2A b. H4 c. H2B d. H3 e. H1

d

Which human condition is the result of uniparental disomy? a. Edward syndrome b. Cri-du-chat syndrome c. Williams-Beuren syndrome d. Prader-Wili syndrome e. WAGR syndrome

c (Ordinarily, homologs separate during meiosis I. Failure of this separation is one of the ways in which nondisjunction can occur)

Which of the following accurately describes a possible meiotic nondisjunction event? a. Sister chromatids fail to separate during meiosis I. b. Fusion of gametes results in trisomy. c. Homologs fail to separate during meiosis I. d. Meiosis fails to proceed to completion.

a, b, c, d

Which of the following are examples of heterochromatin? Select all that apply. a. Barr body b. telomeric DNA c. centromeric DNA d. repetitive DNA

b (A large heterozygous inversion will result in greater numbers of genes that are duplicated or deleted when crossing over occurs)

Which of the following arrangements would yield the greatest reduction in gamete viability? a. An inversion heterozygote for a small inversion b. An inversion heterozygote for a large inversion c. An inversion homozygote for a small inversion d. An inversion homozygote for a large inversion

a

Which of the following best describes a "haplotype"? a. Syntenic SNPs close together within a small region of a chromosome b. A pattern of SNPs used to detect genes that act in a coordinated pattern to establish phenotype c. A collection of SNPs spread across multiple chromosomes d. Syntenic VNTRs located within a single chromosome e. RFLP patterns from autosomal chromosomes

e

Which of the following best describes a submetacentric chromosome? a. A chromosome whose centromere is in the center of the chromosome. b. A chromosome whose centromere is nearly at the end of the chromosome. c. A chromosome that lacks a centromere. d. A chromosome with a terminal centromere and no short arm. e. A chromosome whose centromere is nearer to one end of the chromosome, producing one arm that is shorter than the other.

d

Which of the following best describes the inheritance pattern of VNTRs? a. VNTRs exhibit incomplete dominance. b. VNTRs exhibit pleiotropy. c. VNTRs exhibit complete dominance. d. VNTRs exhibit codominance. e. VNTRs exhibit variable expressivity.

c

Which of the following choices describes the function of DNA polymerase? a. Opens up double-stranded DNA to make it single stranded b. Cuts DNA segments into pieces of one or two nucleotides c. Catalyzes the linking of dATP, dCTP, dGTP, and dTTP in a specific order, using single-stranded DNA as a template d. Fuses intermediate-length DNA segments into longer segments

b

Which of the following choices would be an immediate effect of neutralizing positive charges on histone tails? a. Histones would bind DNA tighter. b. Histones would bind less tightly to the DNA. c. Histones would no longer be attracted to each other. d. Histones would cause supercoiling of the DNA.

d

Which of the following is a common consequence of polyploidy? a. increased cell size b. increased flower and fruit size c. decreased recessive homozygosity d. all of the above

b

Which of the following is a common consequence of unequal crossover between homologs in repetitive DNA regions of the chromosome? a. Inversion b. Partial duplication and deletion c. Trisomy d. Reciprocal translocation

c

Which of the following outcomes is not a consequence of polyploidy? a. Hybrid vigor b. Increased fruit and flower size c. Increased fertility d. Increased cell and nuclei size e. Increased disease resistance

a (Instead of being removed from the preparation, the "ghosts" would be retained. Because both bacterial preparations would include ghosts as well as viral DNA, both would be radioactive, one with P32, one with S35.)

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

c

Which of the following results would represent significant evidence of genetic linkage? a. A positive LOD score b. A negative LOD score c. A LOD score greater than 3 d. A LOD score between 3.0 to -2.0 e. A LOD score of 0

b (Once helicase unwinds the double helix, single‑strand binding proteins bind to the open DNA and prevent it from winding together again.)

Which of the following statements about DNA replication is true? a. DNA gyrase unwinds the DNA double helix. b. Single‑strand binding proteins stabilize the open conformation of the unwound DNA. c. Okazaki fragments are DNA fragments synthesized on the leading strand. d. DNA polymerase adds dNTP monomers in the 3′-5′ direction.

c (Polyploid individuals are sterile if they produce genetically unbalanced gametes; both autopolyploid and allopolyploid individuals may be sterile)

Which of the following statements about allopolyploid individuals is true? a. They result from a mating between individuals of the same species. b. They are generated from exposure to colchicine. c. They may be sterile and unable to produce offspring. d. They can be formed when two sperm simultaneously fertilize an ovum within the same species.

e

Which of the following statements about gene linkage is true? a. If syntenic genes exhibit linkage, parental gametes occur with equal frequency as nonparental gametes. b. If syntenic genes exhibit linkage, nonparental gametes occur more frequently than parental gametes. c. If syntenic genes exhibit independent assortment (no linkage), parental gametes occur more frequently than nonparental gametes. d. If syntenic genes exhibit independent assortment (no linkage), nonparental gametes occur more frequently than parental gametes. e. If syntenic genes exhibit linkage, parental gametes occur more frequently than nonparental gametes.

b

Which of the following statements about nondisjunction is true? a. Nondisjunction only results in gametes with n+1 or n-1 chromosomes. b. Nondisjunction occurring during meiosis II results in 50 percent normal gametes. c. Nondisjunction during meiosis I results in 50 percent normal gametes. d. Nondisjunction always results in four different kinds of gametes.

b (When inversion heterozygotes undergo recombination during meiosis, inversion suppresses the production of recombinant chromosomes. The viable gametes produced by inversion heterozygotes contain either the normalorder chromosome or the inversion-order chromosome, but no recombinant chromosomes are viable, due to duplications and deletions of chromosome segment. This is known as crossover suppression.)

Which of the following statements about recombination in inversion heterozygotes is true? a. Large inversions produce fewer recombinant gametes. b. Inversion suppresses the production of recombinant chromosomes. c. Small inversions have a high frequency of crossovers. d. Inversions result in a higher proportion of recombinant chromosomes. e. Fertility is adversely affected by both small and large inversions.

a (Nondisjunction during meiosis I yields two gametes that are missing a particular chromosome after meiosis II)

Which of the following statements about the products produced when nondisjunction occurs during meiosis I is true? a. Two products of the second meiotic division have both the maternal and paternal chromosomes of a set and the other two products have none for that set. b. Three products of the second meiotic division have too many chromosomes, while one is missing a chromosome for a particular set. c. Both products of the first meiotic division contain dyads representing one chromosome of each homologous pair. d. Both products of the first meiotic division have too many chromosomes.

d

Which of the following statements best describes a heterochromatic region of DNA? a. A chromosome region with less chromatin compaction, few expressed genes, repetitive DNA sequences, and lightly stained regions of G-banded chromosomes. b. A chromosome region with less chromatin compaction, many actively expressed genes, and lightly stained regions of G-banded chromosomes. c. A chromosome region with few expressed genes, repetitive DNA sequences, and lightly stained regions of G-banded chromosomes. d. A chromosome region with few expressed genes, repetitive DNA sequences, and darkly stained regions of G-banded chromosomes. e. A chromosome region with less chromatin compaction, many actively expressed genes, and darkly stained regions of G-banded chromosomes.

a (Because phage DNA and not protein was associated with bacteria at the end of the experiment, it could be concluded that DNA - not protein - must be the genetic material)

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

e

Which of the following statements describes the relationship between the calculated map distances and the actual physical distances? a. The calculated map distance usually underestimates the physical distance between genes because of undetected single crossovers. b. The calculated map distance usually overestimates the physical distance between genes because of negative interference. c. The calculated map distance usually overestimates the physical distance between genes because the expected number of double crossovers is too low. d. The calculated map distance usually underestimates the physical distance between genes because of interference. e. The calculated map distance usually underestimates the physical distance between genes because of undetected double crossovers.

a (Three viable zygote types are possible when fertilization occurs between one normal and one heterozygous parent. One-third has a normal chromosome complement, one-third is a carrier, and one-third has Down syndrome.)

Which of the following statements regarding familial Down syndrome is false? a. If one parent is a translocation heterozygote for chromosomes 14/21, a child born to this parent is as likely to have Down syndrome as he is to be normal. b. It accounts for approximately 5% of all Down syndrome cases. c. Heterozygous carriers of the Robertsonian translocation that cause familial Down syndrome can have children who are also carriers of the translocation. d. Heterozygous carriers of the Robertsonian translocation that causes familial Down syndrome can have children who are phenotypically normal and are not carriers of the translocation.

c (Fragile X syndrome is an example of a trinucleotide repeat disorder, which shows increasing severity with increased copy number of a specific three‑base sequence)

Which of the following syndromes is not paired with its causative chromosomal aberration? a. Huntington disease - duplication b. Down syndrome - Robertsonian translocation c. Fragile X syndrome - deletion d. Cri du chat syndrome - deletion

e

Which of the following techniques could be used to identify a microdeletion in a chromosome? a. Southern blot b. Deletion mapping c. Chromosome band analysis d. Pulse-chase analysis e. FISH

c (The DNA strand must be heated to 95° to denature it and break the hydrogen bonds, but then the temperature is dropped anywhere from 45-68° to allow the small DNA primers to anneal. The temperature is raised to give the polymerase time to extend the primers and generate the PCR product, a small piece of double-stranded DNA.)

Which of these correctly matches the order of the steps of PCR with the correct temperature? a. Denature - 60° Anneal - 95° Extend - 72° b. Extend - 95° Denature - 60° Anneal - 72° c. Denature - 95° Anneal - 60° Extend - 72° d. Denature - 95° Anneal - 72° Extend - 60° e. Anneal - 60° Denature - 95° Extend - 72°

e (Shelterin is the protein complex that interacts with the T loop, protecting the telomere from enzymatic degradation. The telomeric repeat regions and the shelterin-protected T loops will protect the telomeres from extensive shortening for at least several dozen cycles of replication. Despite shelterin, eventually the telomere length becomes too short, triggering apoptosis.)

Which protein complex interacts with the T loop, protecting the telomere from enzymatic degradation? a. preRC b. TERT c. Telomerase d. ORC e. Shelterin

d (In E. coli, DnaB acts as the helicase for DNA replication. DnaA binds to the 9-mer region of oriC, the origin sequence. Single-stranded binding proteins bind to the DNA at the replication bubble, preventing it from closing. DnaC helps to shuttle the DnaB (helicase) to the open site so that replication can proceed when the helix is unwound.)

Which protein functions as a helicase during DNA replication in E. coli? a. DnaC b. SSB c. DnaA d. DnaB e. oriC

b (Nucleotides are joined together by a phosphodiester linkage, which requires a free 3' OH group on the sugar molecule, and an alpha phosphate of an entering dNTP.)

Which statement about the polarity of DNA strands is true? a. The 3' end has a free phosphate group. b. The 3' end has a free OH group. c. The 5' end has a free OH group.

e

While many of the traits we've used to create gene maps involve one gene linked to one trait, we can also look for genes across multiple chromosomes that influence traits or phenotypes associated with alleles in populations of organisms. Which type of analysis is best suited to mapping associations of alleles in populations? a. RFLP b. LOD c. Haplotype d. VNTR e. GWAS

b (During meiosis, the Robertsonian chromosome may sort to one pole of the cell alone, with the normal copy of chromosome 14, or the normal copy of chromosome 21. The other chromosomes migrate to the opposite pole.)

With respect to the chromosomes involved in the translocation, 14 and 21, what is the total number of different gametes possible for a heterozygous carrier parent to produce? a. 4 b. 6 c. 3 d. 2

3′− GCTATAACTCGATTCGAA −5′

Write the complementary sequence for the following DNA sequence, in order from 3' to 5': 5′−CGATATTGAGCTAAGCTT−3′

16% A major driving force of having double stranded DNA is the hydrogen bonding between base pairs (A and T; C and G). Therefore, whatever percent thymine, you would have the same percentage of adenine (34%T + 34%A = 68%A:T) When subtracting this value from 100% you get 32%. This is the percentage of G and C in your DNA (100% - 68% = 32%GC). To be able to determine the percentage of each G and C, you would divide 32% by 2. This would result in 16% guanine and 16% cytosine in the DNA sample.

You are isolating a previously unknown virus and analysis of its genome reveals that it is composed of a double stranded DNA molecule containing 34% thymine. Based on this information, what would you predict the percentage of guanine to be?

e

You have a patient diagnosed with Williams-Beuren syndrome. What would you expect to see in their karyotype analysis? a. Chromosome 13 trisomy b. Chromosome 7 with a nonfunctional hybrid PMSA-PMSB gene, and a duplication of the PMSA, PMSB, and the 17 genes between them. c. Chromosome 11 with an interstitial deletion affecting chromosome bands 11p1.3 and the adjoining band, 11p2. d. Chromosome 15 with a partial deletion of the 15q11.12 portion. e. Chromosome 7 with a nonfunctional hybrid PMSA-PMSB gene, and a deletion of the PMSA, PMSB, and the 17 genes between them.

a

You want to create a chromosome map that reveals the relationships between four different genes: A, B, G, and H. Through controlled crosses, you learn that A and H are 17.9 cM apart; A and B are 10 cM apart; B and H are 7.9 cM apart; A and G are 2.5 cM apart; and H and G are 20.4 cM apart. In which order are your genes located? a. H, B, A, G b. B, H, G, A c. A, B, H, G d. B, H, A, G

b (If crossing over occurs, half of the gametes formed are parental and the other half are recombinant.)

\Which of the following statements about gamete formation during meiosis is false? a. Parental gametes contain the same combinations of linked genes as found in the parent cell. b. Parental gametes can be formed only if there is no crossing over during meiosis. c. Recombinant gametes contain combinations of alleles not found in the parent cell. d. Complete linkage results in the formation of only parental gametes.

Okazaki fragments

_____are the short sections of DNA that are synthesized on the lagging strand of the replicating DNA.

helicase

breaks H-bonds between bases; binds at the replication fork

topoisomerase

breaks covalent bonds in DNA backbone; binds ahead of the replication fork

Heterochromatin

condensed chromatin few expressed genes

Euchromatin

less chromatin condensation many expressed genes

single-strand binding protein

prevents H-bonds between bases; binds after the replication fork