Genetics Final Exam Homework Review

Assuming that Hardy-Weinberg conditions apply, determine the TT frequency in population C. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.49

It is very unlikely that the offspring will have fully horse-like genetic characteristics. Calculate the chance that a gamete will contain all 32 horse chromosomes. Give the answer assuming that recombination does not occur. Express your answer as a fraction (example: (3/16)^2).

2.33×10^−10

How many chromosomes will be in each potato gamete? Enter your answer as a whole number.

24 chromosomes

Where is the enhancer sequence located? 1 2 3 4 5 6 7 8 9

6

How many chromosomes will the polyploid have after chromosome doubling? Enter your answer as a whole number.

60 chromosomes

Human late prophase karyotypes have about 2000 visible G bands. The human genome contains approximately 22,000 genes. Consider the region 4q32.3 through the end of the long arm of chromosome 4 that is identified on the late prophase chromosome in the following figure, and assume the entire region is deleted. Approximately how many genes will be lost as a result of the deletion? (Hint: Consider both how many interbands are lost as a result of the deletion and the gene density per interband.) Express your answer using two significant figures.

77

Where is the silencer sequence located? 1 2 3 4 5 6 7 8 9

9

Which of the following events could result in a frameshift mutation? -Incorporation of base analogs -Base deletion -Point mutation -Tautomeric shift

Base deletion A base deletion would shorten the DNA sequence and change the reading frame of the mRNA.

How is the Gal4 protein inactivated? See Section 13.1 (Page 479) . -By binding to galactose -By binding to the UASG -By binding to glucose-1-phosphate -By binding to the Gal80 protein

By binding to the Gal80 protein When Gal4 and Gal80 are bound to each other, the DNA-binding domain of Gal4 is inactive and blocks its ability to activate transcription. See Figure 13.10 for more information.

How is the trp repressor protein activated? -By binding the trp operator -By binding a small molecule inducer -By releasing bound tryptophan -By binding tryptophan

By binding tryptophan

Identify any family member who is alkaptonuric. M F C1 C2

C2

Identify the most likely wild-type codon(s) for position 243. GAA/G or AGT/C CAA CAA/G GAA or AGT/C

CAA/G

What are the examples of chemical mutagens? Check all that apply. -water -EMS -nitrogen -hydroxylamine

EMS AND hydroxylamine

Inversion loops do NOT form during meiosis in paracentric inversion heterozygotes. True False

False Loop formation allows pairing within inverted regions of homologous chromosomes regardless of whether the centromere is included in the inversion.

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

Fragile X syndrome - deletion 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 amino acid in histone tails is the most frequent target for acetylation? See Section 13.2 (Page 485) . -Threonine -Arginine -Serine -Lysine

Lysine In their unacetylated form, positively charged amino acids such as lysine promote nucleosome adherence to negatively charged DNA. Acetylation neutralizes the positive charge and relaxes the tight hold the nucleosomes have on DNA. See Figures 13.19 and 13.20 for more information.

What type of mutation results in a single amino acid substitution? -A single nucleotide insertion. -Silent. -Missense. -Nonsense.

Missense.

Heterozygosity for which of the following chromosome rearrangements will produce a dicentric bridge as a result of crossover in meiosis? -Pericentric inversion -Reciprocal translocation -Paracentric inversion -Large duplication

Paracentric inversion

Which type of chromosome rearrangement results in the fusion of the long arms of two nonhomologous chromosomes to produce a larger chromosome and a net reduction in chromosome number? -Unbalanced translocation -Reciprocal balanced translocation -Pericentric inversion -Robertsonian translocation

Robertsonian translocation

In the lac operon, what are the likely effects on operon gene transcription of the mutations identified below? Mutation of consensus sequence in the lac promoter. -Transcription is blocked. -Transcription is constitutive.

Transcription is blocked.

Thymine dimers can be repaired by Photoreactivation Repair or Nucleotide Excision Repair. True False

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

Assuming there are no other mutations in the genome, will this double-mutant yeast strain be able to grow on minimal medium? yes no

Yes

Which of the following proteins activates and maintains the lysogenic cycle of bacteriophage lambda? cro RecA Xis cI (λ repressor)

cI (λ repressor)

Which of the following spontaneous changes in DNA structure/sequence generally results from strand slippage? -expansion of trinucleotide repeat sequences -depurination -deamination of cytosine -tautomeric shift that changes the structure of a base

expansion of trinucleotide repeat sequences

In Drosophila, seven partial deletions (1 to 7) shown as gaps in the diagram below have been mapped on a chromosome. This region of the chromosome contains genes that express seven recessive mutant phenotypes, identified in the following table as a through g. A researcher wants to determine the location and order of genes on the chromosome, so he sets up a series of crosses in which flies homozygous for a mutant allele are crossed with flies that are homozygous for a partial deletion. The progeny are scored to determine whether they have the mutant phenotype ("m" in the table), or the wild-type phenotype ("+" in the table). Use the partial deletion map and the table of progeny phenotypes to determine the order of genes on the chromosome. Order the genes on the chromosome from left to right according to the diagram and data provided.

(Left) Gene E Gene B Gene D Gene G Gene A Gene F Gene C (Right)

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

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

What is meant by the term chromatin remodeling? Select the two correct answers. -Chromatin remodeling includes altering chromatin structure by altering the composition of histones within nucleosomes. -Chromatin remodeling includes altering chromatin structure by altering the positioning of nucleosomes with respect to specific DNA sequences. -Chromatin remodeling includes altering chromatin structure by genomic imprinting. -Chromatin remodeling includes altering chromatin structure by lncRNAs acting as scaffolds linking chromatin regulatory proteins. -Chromatin remodeling includes altering chromatin structure by altering the extent of DNA supercoiling.

-Chromatin remodeling includes altering chromatin structure by altering the composition of histones within nucleosomes. AND -Chromatin remodeling includes altering chromatin structure by altering the positioning of nucleosomes with respect to specific DNA sequences.

Calculate the frequency of the dominant allele at brachydactyly locus. Express your answer using four decimal places.

0.0001

Calculate the frequency of heterozygotes for brachydactyly. Express your answer using four decimal places.

0.0002

What is the frequency of the mutant (βSS ) allele in this population? Express your answer using two decimal places.

0.50

Transcriptional regulation of operon gene expression involves the interaction of molecules with one another and of regulatory molecules with segments of DNA. In this context, define and give an example of each of the following: Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

1. Promoter - a DNA sequence that binds RNA polymerase and regulates transcription. 2. Corepressor - a compound that interacts with another protein or compound to form an active repressor. 3. Repressor - a regulatory protein that may bind DNA to inhibit transcription. 4. Inducer - a compound that induces or activates transcription, such as lactose. 5. Operator - a DNA sequence that binds a regulatory protein.

Justify your answer in each case Drag codons on the left to the appropriate blanks on the right to complete the sentences.

1. The possible codons for Gly at position 211 are GGN. The possible codons for Arg are AGA/G, and those for Glu are GAA/G. For single-nucleotide substitutions to change the same Gly codon to Arg and Glu, the Gly codon must have been GGA/G. The Gly-to-Arg mutation would be GGA/G to AGA/G, and the Gly-to-Glu mutation would be GGA/G to GAA/G. 2. The possible codons for Ser 235 are TCN. The possible codons for Leu are TTA/G. For a single-nucleotide substitution to change Ser to Leu, the Ser codon must have been TCA/G. The Ser-to-Leu mutation would be TCA/G to TTA/G. 3. The possible codons for Gln 243 are CAA/G. The possible stop codons are TAA/G. For a single-nucleotide substitution to change Gln to Stop, the Gln codons could have been CAA/G and the stop codons could have been TAA/G. The Gln-to-Stop mutation would be CAA/G to TAA/G

Approximately what range of DNA fragment sizes do you expect to see in the stained electrophoresis gel? Express your answer as an integer.

145 bp

The figure (Figure 1) shows original chromosomes at the top and paired chromosomes undergoing crossing over at the bottom. You will be determining how the chromosomes connect after the crossover event has finished. Be sure to begin with the left end of each chromosome. Select the correct combination describing the left end and right end of the following chromosome found in the gametes. (?-a-d-c-b-e-?) 3-4 3-2 3-3' 3'-4' 3-2'

3-3' Chromatid 3 does not undergo genetic exchange and enters the gamete unchanged.

How many chromosomes will the hybrid have before chromosome doubling? Enter your answer as a whole number.

30 chromosomes

How many different genes are coordinately regulated by the Gal4 transcriptional activator? See Section 13.1 (Page 479) . 3 4 10 2

4 GAL1, GAL2, GAL7, and GAL10 are all regulated by the Gal4 protein. Each of the four genes has its own promoter, but transcription of the genes is regulated by another gene, GAL4, which encodes Gal4, a regulatory protein. See Figures 13.9 and 13.10 for more information.

Assume that the population described in Part A is in Hardy-Weinberg equilibrium. What is the frequency of females who have hemophilia A in this population?

4.00×10^−4 The problem states that the frequency of X 1 (the normal allele) is 0.98. The frequency of X 2 (the mutant allele that causes hemophilia A) is 0.02.Let p represent X 1 , or 0.98, and let q represent X 2, or 0.02. Recall that in order for a female be affected with an X-linked recessive disorder, the female needs two copies of the mutated allele. Thus q 2 will represent the frequency of females that are affected with the disorder:q 2 = (0.02) x (0.02) = 0.0004

Below is a form of chromatin often referred to as "beads on a string." Use this image to answer the questions. How many total polypeptide chains would be present in a single nucleosome core particle?

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

Now imagine that you have analyzed an additional VNTR. From the blot below, what is the expected frequency of occurrence? What is the predicted frequency? Enter your answer as an integer.

9.96×10^7 The frequency of occurrence for the five alleles is 1 in (6 × 31 × 51 x 15 x 700) = 1 in 99,603,000.

Red-green color blindness is an X-linked recessive disorder. In a population (n = 1000) that contains an equal number of males and females, the frequency of the normal allele, X 1, is 0.90 and the frequency of the mutant allele that causes color blindness, X 2, is 0.10. How many females in this population are carriers of the red-green colorblindness trait? Assume that the population is in Hardy-Weinberg equilibrium. 5 10 90 180 More information is needed to answer this question.

90 The Hardy-Weinberg equilibrium formula predicts that two alleles will be distributed in genotypes of frequencies as follows:f(A 1 A 1) = p 2f(A 1 A 2) = 2pqf(A 2 A 2) = q 2Recall that the sum of genotype frequencies is p 2 + 2pq + q 2 = 1.0. To determine the frequency of heterozygotes, you can use 2pq. Thus, the frequency of females who are carriers is 2(0.90)(0.10) = 0.18.The population size is 1000, with equal numbers of males and females. Thus, there are 500 females in the population. To determine the number of females who are carriers, multiply the total number of females by 2pq: 500 x 0.18 = 90.

What term best describes this kind of chromosome abnormality? -Unpaired loop. -Uniparental disomy. -A fusion of two acrocentric chromosomes (Robertsonian translocation). -Partial chromosome deletion.

A fusion of two acrocentric chromosomes (Robertsonian translocation).

Which of the following mutations would result in constitutive expression of the lac operon? -A mutation in lacY that prevents transport of lactose into the cell -A mutation in lacI that increases the affinity of repressor binding to the operator -A mutation in lacI that prevents the repressor from binding lactose -A mutation in lacI that prevents the repressor from binding to the operator

A mutation in lacI that prevents the repressor from binding to the operator

Which of the following mutations could lead to constitutive expression of the genes of the lac operon? -A mutation in the lac-Y gene -A super repressor mutation -A mutation in the operator sequence -A mutation in the lac-Z gene

A mutation in the operator sequence Such a mutation could prevent binding of the repressor, allowing expression under all conditions.

Correlate the following structures or complexes with their effects on eukaryotic gene expression. Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

A promoter is a D N A sequence where R N A polymerase binds and begins the process of transcription. An enhancer is a D N A sequence that binds regulatory proteins that interact with promoter-bound proteins to activate transcription. A silencer is a D N A sequence that binds regulatory proteins that inhibit transcription. RISC is the protein complex that is part of the R N A interference (R N A i) mechanism. It denatures short double-stranded R N As to single strands that carry out R N A i. Dicer is the enzyme complex that is active in R N Ai, where it cuts double-stranded regulatory R N As into 21-b p to 26-b p segments that are subsequently denatured by RISC.

What is the sequence of the outer base pairs of the operator that are key recognition sequences for repressor binding? A/T-A/T-A/T-A/T C/G-A/T-C/G-C/G A/T-C/G-A/T-A/T C/G-A/T-C/G-A/T

A/T-C/G-A/T-A/T The nucleotide sequences of operator DNA to which repressor protein binds consist of 14 base pairs. The outer 4 base pairs (half-site) at the ends of the operator (A/T-C/G-A/T-A/T) serve as key recognition sequences for repressor binding.

For which codon(s) of isoleucine could a single base change account for an amino acid change to methionine? Select all that apply. AUU AUG AUA AUC none of these codons

AUU AND AUA AND AUC There are three codons for isoleucine (AUA, AUC, AUU) and one codon for methionine (AUG). Since positions #1 and #2 are identical in the two sets of codons, then a single base change in the #3 position of any of the isoleucine codons to a G would result in a methionine codon.

Which of the following are examples of heterochromatin? Select all that apply. -Barr body -telomeric DNA -repetitive DNA -centromeric DNA

All of the above. Repetitive DNA is often associated with heterochromatin. At the centromere, these heterochromatic repetitive sequences facilitate binding of spindle fibers during segregation of homologous chromosomes and sister chromatids. Telomeric repetitive sequences are involved in the maintenance of the chromosome's structural integrity. In other cases, entire chromosomes can be heterochromatic as occurs during X-inactivation in mammalian females where the inactive chromosome forms a highly condensed Barr body.

Geographic separation or isolation is a primary contributing factor to which of the following? See Section 20.7 (Page 749) . -Temporal reproductive isolation. -Sympatric speciation. -Allopatric speciation. -Hybrid inviability.

Allopatric speciation. In allopatric speciation, populations are separated by a physical barrier. New species can develop in separate geographic locations as a consequence of their reproductive isolation.

Which of the following may be located tens of kilobases away from the transcriptional start site? See Section 13.1 (Page 479) . -An enhancer sequence -The proximal promoter elements -A TATA box -The core promoter

An enhancer sequence An enhancer sequence is a eukaryotic cis-acting DNA regulatory sequence to which transacting factors bind and stimulate transcription.

What do we call a cluster of genes whose transcription is controlled by a single promoter and associated regulatory sequences? -An operon -A transcription unit -An operator -A polyprotein sequence

An operon

List examples illustrating the phenomenon. Select all that apply. -Klinefelter syndrome -Angelman syndrome -Down syndrome -Turner syndrome -Prader-Willi syndrome

Angelman syndrome AND Prader-Willi syndrome

In order for MutS to recognize a base-pair mismatch, specific MutS amino acid residues must interact with the mismatched base of DNA.Which of the following is not a specific interaction between MutS and mismatched G-T bases? -Asp35 forms a hydrogen bond with the mismatched thymine base. -Phe36 stacks with the mismatched thymine base. -Asp35 forms a hydrogen bond with the mismatched guanine base. -Glu38 forms hydrogen bonds with the mismatched thymine base.

Asp35 forms a hydrogen bond with the mismatched thymine base. You are correct that the asp35 does not form a hydrogen bond with the mismatched thymine. It bonds to the mismatched guanine base. There are three specific interactions between MutS and the mismatched DNA G-T pair: The asp35 residue forms a hydrogen bond with the mismatched guanine base. The glu38 residue forms hydrogen bonds with the mismatched thymine base. The phe36 residue stacks with the mismatched thymine base.

Region 4 is deleted. -Attenuation cannot occur. -Attenuation is likely to be affected to some extent. -There is no effect on attenuation.

Attenuation cannot occur.

The eight uracil nucleotides immediately following region 4 are deleted. -Attenuation cannot occur. -Attenuation is likely to be affected to some extent. -There is no effect on attenuation.

Attenuation cannot occur.

The entire polypeptide coding sequence of trpL is deleted. -Attenuation is likely to be affected to some extent. -There is no effect on attenuation. -Attenuation cannot occur.

Attenuation cannot occur.

The entire trpL region is deleted. -Attenuation cannot occur. -Attenuation is likely to be affected to some extent. -There is no effect on attenuation.

Attenuation cannot occur.

The start (AUG) codon of the trpL polypeptide is deleted. -Attenuation is likely to be affected to some extent. -Attenuation cannot occur. -There is no effect on attenuation.

Attenuation cannot occur.

Two nucleotides are inserted immediately following the polypeptide start codon. -There is no effect on attenuation. -Attenuation cannot occur. -Attenuation is likely to be affected to some extent.

Attenuation cannot occur.

What is the likely effect of each of the following mutations of the trpL region on attenuation control of trp operon gene transcription Region 3 is deleted. -Attenuation is likely to be affected to some extent. -Attenuation cannot occur. -There is no effect on attenuation.

Attenuation cannot occur.

Twenty nucleotides are inserted into the trpL region immediately after the polypeptide stop codon. -Attenuation is likely to be affected to some extent. -There is no effect on attenuation. -Attenuation cannot occur.

Attenuation is likely to be affected to some extent.

Which of the following statements regarding mutation rates is correct? -Mutation rates for all genes of an organism are approximately the same. -Mutation rates are independent of exposure to external agents such as UV radiation. -Mutation rates are independent of gene size. -Average mutation rates vary among taxonomic groups.

Average mutation rates vary among taxonomic groups.

Why? -Because the force driving change is reproductive isolation, which has a larger impact on smaller populations. -Because the force driving change is random inbreeding, which has a larger impact on smaller populations. -Because the force driving change is founder effect, which has a larger impact on smaller populations. -Because the force driving change is random genetic drift, which has a larger impact on smaller populations.

Because the force driving change is random genetic drift, which has a larger impact on smaller populations.

This image shows the PAZ domain of Argonaute. It contains an OB beta barrel (OB stands for oligonucleotide-oligosaccharide binding) with two alpha helices at the end of the barrel. This OB subdomain is flanked by two additional helices. A prominent cleft between the two subdomains is observed. What is the role of this cleft in RNA binding? -Binds the 3' end of the guide strand -Binds the 5' end of the target mRNA strand -Binds the 3' end of the target mRNA strand -Binds the 5' end of the guide RNA

Binds the 3' end of the guide strand The 3', two-nucleotide overhang of the guide strand (pink) is buried in the deep cleft of the PAZ domain (red). The 5' end of the other strand, representing target mRNA (light blue), is sandwiched between the PAZ and N-terminal domains (blue).

This table shows the frequency of given genotypes before and after selection. Identify the genotype(s) with a relative fitness of 1.0. Select all that apply. CC Cc cc

Cc AND cc Relative fitness is a value that quantifies the reproductive success of other genotypes relative to the most favored genotype. It is possible for more than one genotype to have a relative fitness of 1.0.In this example, two genotypes (Cc and cc) have equal success in comparison to a third genotype (CC), which is being selected against. (Note that there are no survivors with the CC genotype after selection; thus that genotype has a relative fitness of 0).

Which of the following best describes the role of chaperone proteins in the regulation of gene expression by steroid hormones? -Chaperone proteins directly enhance transcription. -Chaperone proteins enter the cell and bind receptor molecules. -Chaperone proteins activate receptor proteins. -Chaperone proteins maintain functionality of the receptor.

Chaperone proteins maintain functionality of the receptor. Chaperone proteins maintain the functionality of the receptor prior to binding of the steroid hormone to the receptor.

Which structures can be involved in recombination? -Chromatids of nonhomologous chromosomes -Chromatids of homologous chromosomes -Chromosomes in different cells -Any two chromosomes

Chromatids of homologous chromosomes Chromatids of homologous chromosomes can recombine during meiosis.

Which process does not occur during recombination? -DNA polymerization -Ligation -Nicking of the sugar‑phosphate backbone -Strand displacement

DNA polymerization Recombination does not include the synthesis of new DNA.

Which enzyme cuts double-stranded RNAi precursors to 21 -25 base-pair fragments? See Section 13.3 (Page 499) . -Drosha -Argonaute -RNA-induced silencing complex (RISC) -Dicer

Dicer An enzyme known as Dicer cuts the double-stranded RNA into 21- to 25-bp fragments. These fragments are then bound by a protein complex called the RNA-induced silencing complex (RISC) that denatures the dsRNAs into single strands of 21 to 25 nucleotides. See Figure 13.26 for more information.

What is the process by which changes in allele frequencies occur in response to natural selection? See Section 20.2 (Page 734) . -Differential reproduction. -Convergent evolution. -Divergent evolution. -Genetic hitchhiking.

Differential reproduction. Natural selection favors the survival of individuals with a favored phenotype, which can result in an increased transmission of alleles associated with that phenotype.

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

Diploid number of 5

Explain the origin of DNADNA fragments seen in the gel. -Each band would represent the non-nucleosomal DNA. -Each band would represent the DNA bound by histones in the nucleosome core particle. -Each band would represent the linker DNADNA between nucleosomes.

Each band would represent the DNA bound by histones in the nucleosome core particle.

Show how it could be mediated by a repressor. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all.

Environmental induction of transcription through a transcriptional repressor could be accomplished by inhibiting the function of the repressor. For example, an environmental signal could inhibit repressor binding to a(n) activator, as is the case for yeast Gal80. In the absence of the repressor, the activator can recruit additional factors that convert closed chromatin to open chromatin and promote transcription.

Choose the correct definition of epigenetics. -Epigenetic refers to heritable states of chromatin structure. -Epigenetic refers to heritable traits encoded by extrachromosomal genes (such as those of mitochondrial genes). -Epigenetic refers to heritable traits, the expression of which is strongly dependent on environmental factors. -Epigenetic refers to heritable patterns of a gene's transcription activity.

Epigenetic refers to heritable states of chromatin structure.

Which hormone is the end product of a multistep biochemical pathway that begins with cholesterol and has several intermediates? See Section 20.8 (Page 750) . -Estrogen. -Testosterone. -Progesterone. -Corticosteroids.

Estrogen See Figure 20.15 for more information.

Sort each characteristic into the appropriate bin. Bin 1: Euchromatin Bin 2: Heterochromatin -less chromatin condensation -many expressed genes -few expressed genes -condensed chromatin

Euchromatin: -less chromatin condensation -many expressed genes Heterochromatin: -few expressed genes -condensed chromatin Chromatin exists in different states of compaction and relaxation that regulate access to DNA by regulatory proteins. Heterochromatic regions of chromosomes are relatively inaccessible to transcriptional proteins either transiently (in the case of facultative heterochromatin) or nearly always (in the case of constitutive heterochromatin). In contrast, in euchromatic chromosome regions, transcriptional proteins and enzymes are more easily able to gain access to DNA.

An earthquake destroys the bridge between the island and the mainland, making migration impossible for the deer. What do you expect will happen to allele frequencies in the two populations over the following 10 generations? -The allele frequencies in the two populations will be expected to remain the same as they were before the earthquake. -Eventually, the allele frequencies in the two populations will be expected to differ. -The allele frequencies in the two populations will be expected to differ from what they were before the earthquake but still will not differ from one another.

Eventually, the allele frequencies in the two populations will be expected to differ.

Bacteria can distinguish between a newly replicated DNA strand and the original template strand because the newly replicated strand is methylated, whereas the original template strand is not. True False

False Methylation occurs shortly after replication, so the original template strand is methylated and the newly replicated DNA strand is not.

A cross between a tetraploid and a diploid member of the same species will produce offspring that can undergo sexual reproduction. True False

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

Attenuator systems such as the one described for regulation of tryptophan synthesis would be just as likely to occur in eukaryotes as in prokaryotes. True False

False Regulation by attenuation requires that translation of a given transcript can begin before transcription is completed. This is not possible in eukaryotes, as the two processes are spatially separated by the nuclear membrane.

All compounds that have been found to be mutagenic in the Ames test are also carcinogenic. True False

False The Ames test is used as a preliminary screening tool. Not all compounds that give a positive Ames test are carcinogenic.

Galactosemia is an autosomal recessive disorder. If untreated, the disease can cause mental retardation in children. It is characterized by an inability to metabolize galactose (which comprises part of lactose found in milk), due to a deficient enzyme. In one population, the frequency of the normal allele, A1 , is 0.92. The frequency of the galactosemia allele, A2 , is 0.08.Assuming that this population is in Hardy-Weinberg equilibrium, calculate the frequency of homozygous dominant individuals, heterozygous individuals, and homozygous recessive individuals. Enter your numerical answers in the boxes below. Express each answer to four decimal places. Frequency of homozygous dominant individuals= Frequency of heterozygous individuals= Frequency of homozygous recessive individuals=

Frequency of homozygous dominant individuals= 0.8464 Frequency of heterozygous individuals= 0.1472 Frequency of homozygous recessive individuals= 0.0064 The frequencies p 2, 2pq, and q 2 predict genotype frequencies under assumptions of the Hardy-Weinberg equilibrium. The two equations needed for problems like this are:p + q = 1 (In this case, 0.92 + 0.08 = 1.)andp 2 + 2pq + q 2 = 1 (In this case, (0.92)2 + 2(0.92)(0.08) + (0.08)2 = 1, or 0.8464 + 0.1472 + 0.0064 = 1.)

Which of the following is NOT true of gene flow? See Section 20.4 (Page 738) . -Gene flow generates new alleles. -Gene flow can alter allele frequencies in an admixed population. -Gene flow slows the genetic divergence of geographically separated populations. -Gene flow tends to block speciation.

Gene flow generates new alleles. New alleles are generated by mutations, and NOT through gene flow. Gene flow alters allele frequencies.

Which of the following processes results in changes in allele frequencies in a population strictly due to chance? See Section 20.5 (Page 740) . -Genetic drift. -Natural selection. -Differential reproductive fitness. -Heterozygous advantage.

Genetic drift. Genetic drift refers to changes in allele frequencies in small populations that occur simply by chance (sampling error).

Under which of the following conditions will the lac operon be expressed at the highest level? -Glucose and lactose present -Glucose absent, lactose present -Glucose present, lactose absent -Glucose and lactose absent

Glucose absent, lactose present

Which nucleotide will base‑pair with the enol form of 5‑bromouracil? -Guanine -Adenine -Thymine -Cytosine

Guanine The enol form of 5‑bromouracil forms a base pair with guanine.

Which histone helps stabilize the solenoid structure? -H1 -H2A -H2B -H3 -H4

H1 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.

What general role does acetylation of histone protein amino acids play in the transcription of eukaryotic genes? -Histone acetylation events are most often associated with transcription proofreading. -Histone acetylation events are most often associated with transcription repression. -Histone acetylation events are most often associated with transcription activation. -Histone acetylation events are most often associated with transcription termination.

Histone acetylation events are most often associated with transcription activation.

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? -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. -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. -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. -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

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. 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.

How does the pairing diagrammed in part B differ from the pairing of chromosomes in an inversion heterozygote? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all. -partial duplication -mitosis -inversion -meiosis

In a(n) inversion heterozygote, both homologs must loop out with one loop twisted around to accommodate synapsis of all homologous regions. During meiosis for a partial duplication heterozygote, all regions of the normal homolog can pair without looping out; therefore, only the chromosome containing the partial duplication needs to loop out.

Why is inbreeding depression a serious concern for animal biologists involved in species-conservation breeding programs? -Inbreeding depression is counterproductive for restoring species. -Inbreeding depression limits species diversity. -Inbreeding depression is the main reason for extinction of rare species. -None of the above.

Inbreeding depression is counterproductive for restoring species.

Which of the following is a common consequence of allopolyploidy? -Increased fruit and flower size -Increased fertility -Increased recessive homozygosity -Decreased heterozygosity

Increased fruit and flower size

What is the primary function of an operator in the regulation of transcription in bacteria? -It is a binding site for a repressor protein. -It is a binding site for an activator protein. -It is a binding site for a corepressor. -It is a binding site for an inducer.

It is a binding site for a repressor protein.

Why are liver extracts used in the Ames test? -Liver enzymes may activate some innocuous compounds, making them mutagenic. -Liver enzymes activate the bacterial enzymes. -The bacteria require the nutrients present in the liver extract for growth. -A liver extract is necessary for the bacteria to produce histidine revertants.

Liver enzymes may activate some innocuous compounds, making them mutagenic. Some compounds are innocuous until they are activated metabolically by liver enzymes.

Alkaptonuria is a human autosomal recessive disorder caused by mutation of the HAOHAO gene that encodes the enzyme homogentisic acid oxidase. A map of the HAOHAO gene region reveals four BamHI restriction sites (B1 to B4) in the wild-type allele and three BamHI restriction sites in the mutant allele. BamHI utilizes the restriction sequence 5′5′-GGATCCGGATCC-3′3′ . The BamHI restriction sequence identified as B3 is altered to 5′5′-GGAACCGGAACC-3′3′ in the mutant allele. The mutation results in a Ser-to-Thr missense mutation. Restriction maps of the two alleles are shown below, and the binding sites of two molecular probes (probe A and probe B) are identified. DNADNA samples taken from a mother (M), father (F), and two children (C1 and C2) are analyzed by Southern blotting of BamHI-digested DNADNA. The gel electrophoresis results are illustrated. Using AA to represent the wild-type allele and aa for the mutant allele, identify the genotype of each family member. Enter your answers separated by commas.

M,F,C1,C2 Aa,Aa,AA,aa

Which of the following statements about the relationship between histone modification and chromatin packaging is true? -Methylation of histone tails generally results in tighter packaging of chromatin. -Acetylation of histone tails generally results in tighter packaging of chromatin. -Acetylation of histone tails generally has no effect on chromatin packaging. -Methylation of histone tails generally results in looser packaging of chromatin.

Methylation of histone tails generally results in tighter packaging of chromatin.

Histone protein H4 isolated from pea plants and cow thymus glands contains 102 amino acids in both cases. A total of 100 of the amino acids are identical between the two species. Give an evolutionary explanation for this strong amino acid sequence identity based on what you know about the functions of histones and nucleosomes. -Histone proteins were among the first to appear in evolutionary history; thus they were conserved among all eukaryotes. -The interaction between H4 and DNA strongly depends on the specific amino acid sequence of the histone protein which is under stabilizing natural selection. -Most of the amino acid residues of H4 interact either with DNA or other histone proteins; therefore, most of the H4 amino acid sequence is under functional constraint.

Most of the amino acid residues of H4 interact either with DNA or other histone proteins; therefore, most of the H4 amino acid sequence is under functional constraint.

Can Dr. Dopsis be sure the polyploid will have the characteristics he wants? -Yes, combining tomato and potato genes will give you characteristics of both tomatoes and potatoes. -No, the genetic interactions of tomato and potato genes in the same plant cannot be predicted.

No, the genetic interactions of tomato and potato genes in the same plant cannot be predicted.

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

Normal, because they have a normal amount of genetic material 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.

The frequency of carriers for a rare autosomal recessive genetic condition is 0.04 in a population. Assuming this population is in Hardy-Weinberg equilibrium, what is the allele frequency of the recessive allele? 0.2 0.4 0.64 0.8 Not enough information is provided.

Not enough information is provided. The Hardy-Weinberg equilibrium predicts that two alleles will be distributed in genotypes of frequencies as follows:f(A 1 A 1) = p 2f(A 1 A 2) = 2pqf(A 2 A 2) = q 2Recall that the sum of genotype frequencies is p 2 + 2pq + q 2 = 1.0. Given that the question only gives you the value for 2pq, it is impossible to solve for either p or q, and you would need more information to solve this problem.

The helicase motors of RecD and RecB pull the double-stranded DNA (dsDNA) into the pin of RecC, thus separating the duplex into 5'- and 3'-tailed single strands. The 3' tail follows a channel through the complex, emerging at the nuclease active site. The 5' tail is threaded through a different channel, also emerging near the nuclease active site. Before encountering a chi site, the RecB nuclease digests the 3' tail at a faster rate than the digestion of the 5' tail is digested, possibly because the 3' tail is more favorably located near the nuclease active site. Which of the following is true about nuclease digestion by RecB after the trimer encounters a chi site in the DNA? -Nuclease digestion of both the 3' and 5' tails ceases. -Nuclease digestion of the 5' tail ceases while the digestion rate for the 3' tail increases. -The nuclease digestion rates for both the 3' and 5' tails increases. -Nuclease digestion of the 3' tail ceases while the digestion rate for the 5' tail increases.

Nuclease digestion of the 3' tail ceases while the digestion rate for the 5' tail increases. There are two channels through which the 3' tail can exit. The 3'-ending strand can pass along the 3' channel through the protein, exiting between subdomain 1A of RecC and the nuclease domain of RecB (purple arrow). A second exit (yellow arrow) would bring the 3'-ending strand past the active site of the nuclease (green), where digestion would occur. Once the RecBCD trimer encounters a chi site, digestion of the 3' tail ceases. The presence of a flexible loop (a short alpha helix flanked by long, flexible linkers; light blue) appears to block the channel past the nuclease active site. With the 3' tail no longer being digested, the 5' tail is free from competition for access to the nuclease active site and thus its rate of digestion increases.

To experimental variety 2? Express your answer as an integer.

Number of chromosomes = 36

A boy with Down syndrome (trisomy 21) has 46 chromosomes. His parents and his two older sisters have a normal phenotype, but each sister has 45 chromosomes. How many chromosomes do you expect to see in karyotypes of the parents? -Both of parents are expected to have 46 chromosomes. -One parent is expected to have 46 chromosomes and the other is expected to have 45 chromosomes. -Both of parents are expected to have 45 chromosomes. -One parent is expected to have 46 chromosomes and the other is expected to have 47 chromosomes.

One parent is expected to have 46 chromosomes and the other is expected to have 45 chromosomes.

Use the information from Part C to determine how the operon is regulated for this genotype (I- P+ O+ Z+ Y-). -Operon is repressible; lacZ is only expressed in the absence of lactose. -Operon is constitutive; lacZ is transcribed even when lactose is absent. -Operon is noninducible; lacZ is never transcribed, even in the presence of lactose. -Operon is inducible; lacZ is transcribed only in the presence of lactose.

Operon is constitutive; lacZ is transcribed even when lactose is absent. Of the two structural genes, only lacZ is a wild-type sequence, meaning that only beta-galactosidase can be produced from this operon.The promoter sequence is normal, so it's possible for RNA polymerase to transcribe the operon, and produce beta-galactosidase.In terms of regulation, the operator region is wild type (O+), but no repressor is made (I- ). Without repressor protein, it's not possible to prevent transcription. As a result, beta-galactosidase is produced under all conditions; it is "constitutively" expressed.

Use the information from Part A to determine how the operon is regulated for this genotype (I+ P+ Oc Z− Y+ ). -Operon is repressible; permease is only produced in the absence of lactose. -Operon is inducible; permease is produced only in the presence of lactose. -Operon is constitutive; permease is produced even when lactose is absent. -Operon is noninducible; permease is never produced, even when lactose is present.

Operon is constitutive; permease is produced even when lactose is absent. For the I+ P+ Oc Z− Y+ genotype, of the two structural genes, only lacY is wild-type sequence, meaning only permease is produced. The gene for beta-galactosidase (lacZ) is mutated, meaning that no functional enzyme is produced. The promoter sequence is normal, so its possible for RNA polymerase to transcribe the operon, and produce permease.In terms of regulation, functional repressor (I+) is made, however the operator region is mutated (Oc). Because the operator sequence is altered, the repressor cannot bind to DNA and prevent transcription. As a result, permease is produced under all conditions; it is "constitutively" expressed.

Which of the following summarizes expression of the lac genes in this partical diploid: I+ P+ Oc Z− Y+ / I− P+ O+ Z+ Y− ? -Beta-galactosidase is made both in the presence of and in the absence of lactose; permease is only made when lactose is present. -Permease is made both in the presence of and in the absence of lactose; beta-galactosidase is only made when lactose is present. -Both permease and beta-galactosidase are made only when lactose is present. -Both permease and beta-galactosidase are made both in the presence of and in the absence of lactose.

Permease is made both in the presence of and in the absence of lactose; beta-galactosidase is only made when lactose is present. Repressor protein is made from the F' plasmid and diffuses in the cell. It can bind to the operator region on the bacterial chromosome, but cannot bind to the operator on the region on the F' plasmid. As a result, the wild-type lacY gene on the F' plasmid will be expressed constitutively, while the wild-type lacZ gene on the bacterial chromosome will be inducible. If there is no lactose present, beta-galactosidase will be produced. When lactose is present, both beta-galactosidase and permease will be prodcued.

Which Argonaute domain contains the active site? -N-terminal domain -Piwi domain -Middle domain -PAZ domain

Piwi domain The Piwi domain lies between the Middle and N-terminal domains. These three domains form the crescent-shaped base of Argonaute (see image), which contains a positively charged central groove for binding and positioning the negatively charged RNA backbone for cleavage. The Piwi domain contains a central, five-stranded beta sheet that is surrounded by alpha helices. This core structure identifies Piwi as an RNase H domain. The close structural similarity to RNase HI and RNase II indicates that the Piwi domain is responsible for cleavage of target mRNA in RNAi.

What are the roles of the Polycomb and Trithorax complexes in eukaryotic gene regulation? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all.

Polycomb (PcG) and Trithorax (Trx) are chromatin-modifier proteins that have opposite effects on chromatin structure. PcG complexes maintain chromatin in a repressed state through its H3K27me3 and HDAC activities; that is, they tri-methylate histone H3 on lysine 27 and remove acetyl groups from histones. Trx complexes maintain chromatin in an active state through its HAC and H3K27 HDMT activities; that is, they acetylate histones and demethylate histone H3 lysine 27.

What is the probability the next child of this couple will have a normal phenotype and have 46 chromosomes? Express your answer as fraction (example 1/5)

Probability 1/3

Where are CpG islands clustered in mammalian genomes? See Section 13.2 (Page 485) . -Introns -Centromeres -Promoters -Heterochromatin

Promoters CpG islands typically occur at or near the transcription start site of genes, particularly housekeeping genes, in vertebrates. A C (cytosine) base followed immediately by a G (guanine) base (a CpG) is rare in vertebrate DNA because the cytosines in such an arrangement tend to be methylated.

Suppose that 40% of all meioses in mainland-island hybrids involve recombination somewhere in the chromosome region between q2.1 and p2. What percentage of the gametes of hybrid deer are viable? Enter your answer as an integer.

Proportion = 80 %

To find insertions into a specific region on chromosome 3, the researcher separately crosses individual F2 male flies with eye pigment to females heterozygous for a deletion on chromosome 3. Since this deletion is homozygous lethal (as most large deletions are), she uses flies that have one copy of a normal chromosome 3 as well as one copy with the deletion. The normal (non-deleted) chromosome 3 also carries a dominant mutation called Serrate, which produces notched wings. (This non-deleted chromosome also does not allow for recombination with the chromosome with the deletion).The females are also homozygous for the white mutation, meaning that any progeny with eye pigment will have a P-element insertion with its miniwhite + marker. In this way, the researcher hopes to find P-element insertions in her screen that do not survive when paired with the deletion -- indicating that the insertions are in the region spanned by the deletion, and have caused a mutation in a gene essential for survival.When the researcher examines the progeny with eye pigment that result from these crosses, she finds two different categories of flies. Some crosses produce flies with eye pigment and notched wings, as well as flies with eye pigment and normal wings. Other crosses produce flies with eye pigment and notched wings, but no flies with eye pigment and normal wings. What might explain these results? Select all statements that are true. -Progeny with eye pigment have a P {miniwhite +} insertion they inherited from their father. -Progeny with notched wings inherited the non-deleted chromosome 3 from their mother. -Progeny with normal wings inherited the chromosome 3 with the deletion from their mother. -Progeny with notched wings and eye pigment have both a P {miniwhite +} insertion and the chromosome 3 with the deletion. -Progeny with normal wings and eye pigment have both a P {miniwhite +} insertion and the chromosome 3 with the deletion. -A cross that produces progeny with normal wings and eye pigment indicates that the P-element insertion has disrupted an essential gene in the region spanned by the deletion. -A cross that does not produce progeny with normal wings and eye pigment indicates that the P-element insertion has disrupted an essential gene in the region spanned by the deletion.

Progeny with eye pigment have a P {miniwhite +} insertion they inherited from their father. AND Progeny with notched wings inherited the non-deleted chromosome 3 from their mother. AND Progeny with normal wings inherited the chromosome 3 with the deletion from their mother. AND Progeny with normal wings and eye pigment have both a P {miniwhite +} insertion and the chromosome 3 with the deletion. AND A cross that does not produce progeny with normal wings and eye pigment indicates that the P-element insertion has disrupted an essential gene in the region spanned by the deletion. In this experiment, crosses that do not produce progeny with normal wings and eye pigment indicate that flies simultaneously homozygous for a P-element insertion (marked by the miniwhite + gene) and the deleted chromosome 3 (indicated by a lack of the dominant Serrate mutation associated with the non-deleted chromosome 3) do not survive. This result indicates that the new P-element insertion has disrupted an essential gene in the region spanned by the deletion. Screens like the one described here are commonly used by Drosophila researchers to identify new transposon insertions. By using an X-linked insertion as a starting point and screening for male-to-male transmission, researchers can be sure that they are testing only new insertions for specific effects -- such as removing the function of an essential gene in a specific area uncovered by a deletion.

Which of the following is a basic assumption underlying Hardy-Weinberg equilibrium? See Section 20.1 (Page 727) . -Migration alters allele frequencies in the population. - The population is subject to natural selection. -Population size is finite. -Random mating occurs in the population.

Random mating occurs in the population.

The repair of double-strand breaks in DNA is essential for maintaining the integrity of the genetic material and for cell viability. The homologous recombination pathway serves as a primary means of double-strand break repair in eubacteria. An initial step in this pathway involves the RecBCD trimeric protein recognizing blunt-ended DNA at the site of a break, followed by complex processing of DNA inward from the break, finally resulting in a 3' overhang of single-stranded DNA (ssDNA) coated by RecA protein. This serves as the substrate for homologous recombination via the RecA pathway. Launch the molecular model of the RecBCD protein to explore its structure. Then answer the questions. Which subunits of the RecBCD trimer show helicase structure and function? -RecB, RecC, and RecD -RecB and RecD -RecB and RecC -RecB only

RecB and RecD Subunits RecB and RecD are functional helicases, as both use ATP to move unidirectionally along a DNA duplex and to help separate the DNA into single stranded form. RecB also has a functional nuclease domain that enables the RecBCD protein to cleave single-stranded DNA and to help load the RecA protein onto the DNA strand containing the chi site after RecBCD completes its main function. RecC also has structural similarities to helicases suggesting it evolved from a helicase subunit. However, RecC no longer exhibits helicase function.

Which of the following is a common consequence of a viable trisomy? -Polyploidy -Increased cell size -Reduced fertility -Increased rate of cell division

Reduced fertility Trivalent synaptic structures form in meiosis I. Segregation of homologs results in the production of aneuploid gametes.

Parts of an operon that control the expression of both enzyme genes are called "regulatory regions." The genes that produce the enzymes are called "structural genes." Using the data in the table, determine which regions are regulatory regions and which are structural genes. Sort each region into the correct category.

Regulatory Region(s) Region A Region B Region D Structural Gene(s) Region C Region E In the operon, both structural genes (the genes that produce enzyme 1 and enzyme 2) are regulated by the same regulatory protein, the same binding site for the regulatory protein, and the same promoter sequence. Therefore, mutations in these regulatory regions would affect both enzyme 1 and enzyme 2. This is true of mutations in regions A, B, and D, indicating these are regulatory regions.In contrast, mutations in regions C and E affect only one of the two enzymes, indicating that these are the structural genes that encode the individual enzymes.

Bypass (or translesion) DNA polymerases in E. coli are unique in their ability to catalyze which of the following reactions? See Section 11.5 (Page 421) . -Synthesize long stretches (hundreds of nucleotides) of DNA -Replicate across DNA damage that stalls DNA pol III -Remove nucleotides using a 3′-to-5′ exonuclease activity -Synthesize DNA in the 3′-to-5′ direction

Replicate across DNA damage that stalls DNA pol III Translesion DNA polymerases only function in the synthesis of short DNA segments, and they lack the DNA proofreading capability of DNA pol I and DNA pol III.

The term heterochromatin refers to heavily condensed regions of chromosomes that are largely devoid of genes. Since few genes exist in those regions, they almost never decondense for transcription. At what point during the cell cycle would you expect to observe the decondensation of heterochromatic regions? -G0 phase -G1 phase -S phase -G2 phase -M phase

S phase

Which repair system uses the RecA and LexA proteins? -SOS repair -Photoreactivation repair -Proofreading repair -Mismatch repair

SOS repair RecA and LexA are active during the SOS response.

The Ames test is used to determine the potential mutagenicity of test compounds by screening for new mutations in which of the following organisms? See Section 11.3 (Page 411) . -Mouse -Escherichia coli -Yeast -Salmonella typhimurium

Salmonella typhimurium See Figure 11.15 for more information.

The diagram below shows a segment of DNA containing an imaginary gene (Z) and the primary RNA transcript that results from the transcription of gene Z. Exons are represented in green and introns are represented in blue. Consider the predominant types of alternative splicing events that occur in mammals. Which of the following choices represent mRNA molecules that could be produced from the primary RNA transcript by alternative RNA splicing? (In each choice, the yellow part on the left represents the 5' cap, and the yellow part on the right represents the poly-A tail.) Select all that apply.

See image. Alternative RNA splicing produces different mRNA molecules from the same primary RNA transcript. During alternative RNA splicing, introns are removed, and some exons may also be removed. The removal of different exons produces different mRNA molecules, which are then translated into different proteins. While intron retention is the most common type of alternative splicing event in plants, fungi, and protozoa, recall that it is a rare event in mammals. Alternative RNA splicing can greatly expand the number of proteins produced from the same gene.

Which of the following best describes the mechanism by which steroid hormones control gene expression? -Steroid hormones enter a cell, bind directly to hormone response elements (HREs), and enhance transcription. -Chaperone proteins transport steroid hormones into the cell and guide them to their target genes. -Steroid hormones transport mRNA from the nucleus into the cytoplasm, where it is translated into protein. -Steroid hormones that enter the cell activate receptors. These hormone-receptor complexes then bind HREs and influence gene expression.

Steroid hormones that enter the cell activate receptors. These hormone-receptor complexes then bind HREs and influence gene expression.

Why mules are generally sterile? -Synapsis in the mule germ cells is abnormal, which results in the production of unbalanced gametes that do not form viable zygotes. -Extra chromosome prevents equal division in meiosis that leads to the segregation errors. -All hybrid animals are sterile due to the unlike pairs inherited from the two species of parents. -Mating between a male donkey and a female horse gives only male generation.

Synapsis in the mule germ cells is abnormal, which results in the production of unbalanced gametes that do not form viable zygotes.

Spontaneous mutations can arise following DNA replication when DNA bases undergo a tautomeric shift. Tautomers have the same chemical composition, but the hydrogen atom is in a different position. This change can result in different hydrogen bonding configurations. Stable common tautomeric forms make up the standard base pairing rules (A-T and G-C). Rare unstable tautomers have altered base pairing rules, as shown in this figure. Use this partial DNA sequence (the original sequence with no mutations) to answer the following questions. 5' - C A A - 3' 3' - G T T - 5' Suppose that a transient tautomeric shift occurred in the guanine base to produce a rare tautomer in the partial DNA sequence just prior to a round of DNA replication. Which base would be added opposite this rare tautomer during DNA replication? G T U C A

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

Identify the most likely wild-type codon(s) for position 235. TTA/G TCA/G AGT/C TCC/T

TCA/G

The beads-on-a-string form of chromatin is known as which of the following? -The 300-nm extended chromatin -The 700-nm coiled chromosome -The 10-nm fiber -The 30-nm solenoid

The 10-nm fiber

Part C: You continue your experiments with the black mutation on chromosome II. You cross the reciprocal translocation strain to the black pure line to generate F1 flies that are both translocation heterozygotes and Bb heterozygotes. As before, you testcross F1 males and females in two separate experiments. CrossF1 males x pure black females (Note: There is no crossing over in the F1 males.) F2 Progeny: 1/2 black, fully fertile 1/2 wild-type, semi-sterile F1 females x pure black males (Note: Crossing over may occur in the F1 females.) F2 Progeny: 424 black, fully fertile 426 wild-type, semi-sterile 73 black, semi-sterile 77 wild-type, fully fertile What can you conclude from these results? Select the three correct statements. -The F2 progeny in both experiments contain recombinants. -The F2 progeny contain recombinants only when F1 females are testcrossed. -None of the F2 progeny classes require crossing over. -Some of the F2 progeny classes require crossing over; others do not. -Chromosome II is one of the chromosomes involved in the reciprocal balanced translocation.

The F2 progeny contain recombinants only when F1 females are testcrossed. AND Some of the F2 progeny classes require crossing over; others do not. AND Chromosome II is one of the chromosomes involved in the reciprocal balanced translocation. These two crosses show that crossing over is required to recombine the b allele onto a translocated chromosome (or, conversely, recombine the B allele onto a normal chromosome). This indicates that the black locus is on one of the chromosomes involved in the reciprocal translocation.

Part B: You continue your experiments with the sepia mutation on chromosome III. You cross the reciprocal translocation strain to the sepia pure line to generate F1 flies that are both translocation heterozygotes and Ss heterozygotes. As before, you testcross F1 males and females in two separate experiments. CrossF1 males x pure sepia females (Note: There is no crossing over in the F1 males.) F2 Progeny: 1/2 sepia, fully fertile 1/2 wild-type, semi-sterile F1 females x pure sepia males (Note: Crossing over may occur in the F1 females.) F2 Progeny: 497 sepia, fully fertile 493 wild-type, semi-sterile 4 sepia, semi-sterile 6 wild-type, fully fertile What can you conclude from these results? Select the three correct statements. -The F2 progeny in both experiments contain recombinants. -The F2 progeny contain recombinants only when F1 females are testcrossed. -None of the F2 progeny classes require crossing over. -Some of the F2 progeny classes require crossing over; others do not. -Chromosome III is one of the chromosomes involved in the reciprocal balanced translocation.

The F2 progeny contain recombinants only when F1 females are testcrossed. AND Some of the F2 progeny classes require crossing over; others do not. AND Chromosome III is one of the chromosomes involved in the reciprocal balanced translocation. These two crosses show that crossing over is required to recombine the s allele onto a translocated chromosome (or, conversely, recombine the S allele onto normal chromosome). This indicates that the sepia locus is on one of the chromosomes involved in the reciprocal translocation.

Which of the following best describes the interaction of the two clamp domains with the DNA grooves? Both clamp domains contact only the phosphates nearest the minor groove. -The clamp domain of the non-mismatch-binding monomer contacts the phosphates nearest the minor groove. The clamp domain of the mismatch-binding monomer contacts the phosphates nearest the major groove. -Both clamp domains contact only the phosphates nearest the major groove. -The clamp domain of the mismatch-binding (recognition) monomer contacts the phosphates nearest the minor groove. The clamp domain of the non-mismatch-binding monomer contacts phosphates nearest the major groove.

The clamp domain of the mismatch-binding (recognition) monomer contacts the phosphates nearest the minor groove. The clamp domain of the non-mismatch-binding monomer contacts phosphates nearest the major groove.

What is the cause of the decreased proportion of viable gametes in hybrids relative to the parental populations? -The decrease in viability is due to increased rate of crossing over. -The decrease in viability is due to random chromosome assortment during anaphase II. -The decrease in viability is due to the formation of duplications and deletions by crossing over within the inverted chromosome segment. -The decrease in viability is due to the different number of genes that come from each parent.

The decrease in viability is due to the formation of duplications and deletions by crossing over within the inverted chromosome segment.

Consider synapsis in prophase II of meiosis for two plant species that each carry 36 chromosomes. Species AA is diploid and species BB is triploid. What characteristics of homologous chromosome synapsis can be used to distinguish these two species? -The diploid will have a combination of trivalents, bivalents, and univalents whereas the triploid will have 18 bivalents. -The diploid will have a combination of trivalents, bivalents, and univalents whereas the triploid will have univalents. -The diploid will have 18 bivalents whereas the triploid will have 18 trivalents. -The diploid will have 18 bivalents whereas the triploid will have a combination of trivalents, bivalents, and univalents.

The diploid will have 18 bivalents whereas the triploid will have a combination of trivalents, bivalents, and univalents.

Which of the following statements most accurately describes the transposable genetic elements found in corn? -The earlier in kernel development a reversion event occurs, the larger the spot of tissue exhibiting the reversion phenotype. -Transposable elements found in maize are strictly autonomous. -Transposable elements are found only in corn. -Transposons in corn lack inverted repeats.

The earlier in kernel development a reversion event occurs, the larger the spot of tissue exhibiting the reversion phenotype. The earlier in development the reversion occurs, the greater the number of divisions the reverted cell type undergoes before kernel development is complete.

Compare and contrast the transcriptional regulation of GALGAL genes in yeast with that of the laclac genes in bacteria. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all.

The enzymes catalyzing lactose and galactose are produced from genes whose transcription is activated only when these sugars are present glucose has been depleted. Gene transcription from the bacterial l a c operon is induced by the binding of the lactose metabolite, allolactose, to the repressor protein. This process opens the laclac operator and promoter region for RNA polymerase and activates transcription. Induction of transcription of GAL genes by galactose is more indirect. The yeast GAL genes have their transcription activated by binding of Gal4 protein to UAS elements affiliated with each gene. Gal4 binding is induced when it is released from Gal80, which is caused by the binding of Gal3 to Gal80, which occurs only when Gal3 is bound to galactose and cellular glucose has been depleted.

How do the expected results support the 10-nmnm-fiber model of chromatin? -The only protection against digestion by DNase I indicates that the chromatin is folded into higher-order structures. -The only protection against digestion by DNase I indicates that no other proteins occur in this region. -The only protection against digestion by DNase I indicates that no other proteins and no higher-order packing of the chromatin at this region occurred. -Results won't support the 10-nmnm-fiber model of chromatin.

The only protection against digestion by DNase I indicates that no other proteins and no higher-order packing of the chromatin at this region occurred.

The two DNA and polypeptide sequences shown are for alleles at a hypothetical locus that produce different polypeptides, both five amino acids long. In each case, the lower DNA strand is the template strand: allele A1A1:5' . . . ATGCATGTAAGTGCATGA . . . 3' 3' . . . TACGTACATTCACGTACT . . . 5'A1A1 polypeptide N - Met-His-Val-Ser-Ala - C allele A2A2:5' . . . ATGCAAGTAAGTGCATGA . . . 3'3' . . . TACGTTCATTCACGTACT . . . 5'A2A2 polypeptide N - Met-Gln-Val-Ser-Ala - C Based on DNA and polypeptide sequences alone, determine which allele is dominant and which is recessive? --allele A1A1 is dominant, allele A2A2 is recessive -allele A2A2 is dominant, allele A2A2 is recessive -The information is not enough to determine it.

The information is not enough to determine it.

What would be the effect of a mutation in the lacI gene that prevented the repressor from binding to lactose? -The lac Z, Y, and A genes would be induced by lactose. -The lac Z, Y, and A genes would not be expressed. -The lac Z, Y, and A genes would be repressed by lactose. -The lac Z, Y, and A genes would be expressed constitutively.

The lac Z, Y, and A genes would not be expressed. If lactose could not bind to the repressor, the repressor would stay bound to the operator and repress the transcription of the lac Z, Y, and A genes.

A 1-mLmL sample of the bacterium E. coli is exposed to ultraviolet light. The sample is used to inoculate a 500-mLmL flask of complete medium that allows growth of all bacterial cells. The 500-mLmL culture is grown on the benchtop, and two equal-size samples are removed and plated on identical complete-medium growth plates. Plate 1 is immediately wrapped in a dark cloth, but plate 2 is not covered. Both plates are left at room temperature for 36 hours and then examined. Plate 2 is seen to contain many more growing colonies than plate 1. Thinking about DNA repair processes, how do you explain this observation? -Visible light stimulates formation of thymine dimers that activate nucleotide excision repair systems to repair DNA damage and prevent cell death. -The level of DNA damage inflicted by UV light was high enough to require the photoreactive repair pathway to repair the DNA damage sufficiently to prevent cell death. -The results are puzzling, as the plate kept in the dark would be expected to have more colonies than the plate kept in the light, as protection from light facilitates DNA repair. -DNA double-stranded breaks inflicted by UV light were sufficiently repaired by the photoreactive repair pathway to prevent cell death.

The level of DNA damage inflicted by UV light was high enough to require the photoreactive repair pathway to repair the DNA damage sufficiently to prevent cell death.

Ten nucleotides are inserted between regions 2 and 3 of trpL . -Attenuation cannot occur. -There is no effect on attenuation. -Attenuation is likely to be affected to some extent.

There is no effect on attenuation.

Two nucleotides are inserted into the trpL region immediately after the polypeptide stop codon. -Attenuation cannot occur. -There is no effect on attenuation. -Attenuation is likely to be affected to some extent.

There is no effect on attenuation.

Tautomers of nucleotide bases are isomers that differ from each other in the location of one hydrogen atom in the molecule. True False

True Nucleotide tautomers differ only in the bonding location of one hydrogen atom.

Below is a partial DNA sequence (the original sequence with no mutations); only the coding strand is shown. Assume the sequence is transcribed and translated from left to right with the reading frame as indicated.5' - | GGC | GTG | GTA | TTA | GCG | - 3' This sequence represents a single mutation. Mutation #1: 5' - | GGC | GCG | GTA | TTA | GCG | - 3' Complete these sentences about this mutated DNA sequence.

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

Which type of DNA damage is repaired by the enzyme photolyase? See Section 11.4 (Page 416) . -Apurinic sites -Single-strand breaks in the DNA backbone -Thyminedimers -Methylated or alkylated bases

Thyminedimers See Table 11.4 for more information.

Mutation of the CAP binding site of the lac promoter. -Transcription is blocked. -Transcription is constitutive.

Transcription is blocked.

Mutation of the lac I gene affecting the allosteric site of the protein. -Transcription is blocked. -Transcription is constitutive.

Transcription is blocked.

Mutation of the lac I gene affecting the DNA-binding site of the protein. -Transcription is blocked. -Transcription is constitutive.

Transcription is constitutive.

Mutation of the repressor binding site on the operator sequence. -Transcription is blocked. -Transcription is constitutive.

Transcription is constitutive.

The process that determines the length of heteroduplex DNA on the chromatids is called branch migration. True False

True The crossbridge DNA structure formed after the initial nick is sealed can migrate along the chromatid. This process is called branch migration, and it increases the length of heteroduplex DNA.

The placement of the operator sequence between the promotor and the structural genes is critical to the proper function of the lac operon. True False

True When the repressor binds to the operator, RNA polymerase cannot transcribe the structural genes.

In the presence of DNA lesions in the cell, the stability of p53 protein increases. True False

True p53 blocks progression of the cell cycle—a response that is appropriate in the presence of unrepaired DNA lesions.

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

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

Thymine dimers are most commonly caused by exposure to See Section 11.3 (Page 411) . -UV irradiation. -DNA intercalating agents. -alkylating agents. -X-rays.

UV irradiation

Will this hybrid be infertile? no yes

yes

In terms of its involvement in mutagenesis, 5BU is best described as _______. -a base analog that can cause either A-T > G-C or G-C > A-T transitions -a chemical that alters the structure of nitrogenous bases -a rare form of adenine that can base pair with cytosine -a base analog that, if incorporated into a DNA molecule during replication, remains permanently in its rare form

a base analog that can cause either A-T > G-C or G-C > A-T transitions In its common form, 5BU can pair with adenine and in its rare form it can pair with guanine.

All of the following are characteristics of insertion sequences elements except _______. -a copy of the insertion sequence becomes integrated at a new location -there can be more than one copy of an IS element in a bacterial genome -they encode protein -they are flanked by inverted repeats

a copy of the insertion sequence becomes integrated at a new location The insertion sequence itself, not a copy, is excised from its original location and transferred to a new one.

In eukaryotes, homologous recombination is initiated by See Section 11.6 (Page 422) . -single-strand breaks in both homologs. -a single-strand break in one homolog. -a double-strand break in one homolog. -a double-strand breaks in both homologs.

a double-strand break in one homolog

How is the transcription of heat shock genes activated in E. coli? -by expression of an alternative sigma factor that forms an alternate RNA polymerase holoenzyme that specifically transcribes heat shock genes -by activation of a heat shock repressor protein that prevents transcription of non-heat shock genes -by antitermination, which allows transcription to extend into heat shock genes -by inactivation of a heat shock repressor that induces transcription of heat shock genes

by expression of an alternative sigma factor that forms an alternate RNA polymerase holoenzyme that specifically transcribes heat shock genes

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 _______. -chromosome with some deleted regions -chromosome with some duplicated regions -chromosome with two centromeres -completely normal chromosome

chromosome with two centromeres When the centromere is contained within the inversion, it is neither duplicated nor deleted.

The MutS dimer clasps the DNA mismatch between which two domains? -mismatch and core domains -clamp and core domains -core and connector domains -clamp and mismatch domains

clamp and mismatch domains This image shows the MutS dimer clasping the DNA at the location of the mismatched base pair. Only the mismatch binding (blue) and clamp (green) domains clasp the DNA.

Which bacteria grow on the agar plate if the Ames test is positive? his− prototrophs his + auxotrophs his + prototrophs his − auxotrophs

his + prototrophs The bacteria used in the Ames test to evaluate mutagenicity are his− auxotrophs. If the Ames test is positive, these bacteria have reverted back to wild type and are his + prototrophs

What process is responsible for position effect variegation? -inactivation of a gene due to deletion -gene duplication -inactivation of a gene due to its incorporation in a heterochromatic region of the chromosome -activation of a gene by an enhancer

inactivation of a gene due to its incorporation in a heterochromatic region of the chromosome

What growth information is used to make these determinations? -information about environment temperature -information about growth conditions -information about medium content -None of the above.

information about medium content

For the following lac operon haploid genotypes identify the mode of transcription of operon genes and indicate determine whether the strain is lac + or lac- . I+P+O+I+P+O+Z+Y−Z+Y− -is lac- and the genes are inducibly transcribed. -is lac + and the genes are constitutively transcribed. -is lac- and the genes are not transcribed. -is lac- and the genes are constitutively transcribed. -is lac+ and the genes are inducibly transcribed. -is lac+ and the genes are not transcribed.

is lac- and the genes are not transcribed.

For the cross in Part A, in which parent and during what meiotic division might nondisjunction have occurred? Select all that apply. -maternal meiosis I -maternal meiosis II -paternal meiosis I -paternal meiosis II

maternal meiosis II

What type of aneuploidy is responsible for Turner syndrome in humans? -monosomy YO -trisomy 21 -monosomy XO -trisomy 13 -trisomy 18

monosomy XO

Which repair process(es) use(s) a DNA polymerase? Select all that apply. translesion DNA synthesis -nucleotide excision repair -base excision repair -nonhomologous end joining -photoreactivation repair -homologous recombination (synthesis-dependent strand annealing)

nucleotide excision repair AND base excision repair AND nonhomologous end joining AND homologous recombination (synthesis-dependent strand annealing) Many DNA repair processes (base excision, nucleotide excision, translesion DNA synthesis) use a DNA polymerase when a template strand is available to direct synthesis/repair of the other strand. Even in cases where there is no such template strand (e.g., a double-strand break), homologous recombination can use the identical sister chromatid like a template for a DNA polymerase to fill in the double-strand break gaps.

Classify the nature of the mutation in colony 1. -adenine auxotrophic mutation -temperature sensitive mutation -temperature resistive mutation

temperature sensitive mutation

Classify the nature of the mutation in colony 2. -adenine auxotrophic mutation -temperature resistive mutation -temperature sensitive mutation

temperature sensitive mutation

How will the mutation affect the determination of the lytic or lysogenic life cycle in mutant 2 phage strains? -the mutation will inhibit lysogeny establishment -the mutation will make it difficult to reverse lysogeny -the mutation will have no effect

the mutation will make it difficult to reverse lysogeny

What type(s) of mutations are likely produced by chemical #3? Select all that apply. transitions frameshifts transversions

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

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

two haploid gametes fuse to form a diploid cell Fusion of haploid gametes occurs after meiosis.

What is the frequency of mating between heterozygotes for albinism? Express your answer using five decimal places.

0.00097

Experiments by Charles Yanofsky in the 1950s and 1960s helped characterize the nature of tryptophan synthesis in E.coli. In one of Yanofsky's experiments, he identified glycine (Gly) as the wild-type amino acid in position 211 of tryptophan synthetase, the product of the trpA gene. He identified two independent missense mutants with defective tryptophan synthetase at these positions that resulted from base-pair substitutions. One mutant encoded arginine (Arg) and another encoded glutamic acid (Glu). At position 235, wild-type tryptophan synthetase contains serine (Ser), but a base-pair substitution mutant encodes leucine (Leu). At position 243, the wild-type polypeptide contains glutamine, and a base-pair substitution mutant encodes a stop codon. Identify the most likely wild-type codon(s) for position 211. GGC GGA/G GAA or AGT/C CAA

GGA/G

Which of the following describes the molecular mechanism of heterochromatin protein-1 (HP-1)? See Section 13.2 (Page 485) . -HP-1 acetylates histones, resulting in the formation of transcriptionally active euchromatin. -HP-1 demethylates histones, resulting in the formation of transcriptionally repressed heterochromatin. -HP-1 methylates guanine when it appears in the sequence GATC, resulting in the formation of transcriptionally silent heterochromatin. -HP-1 binds to methylated histones and maintains heterochromatin in a transcriptionally inactive state.

HP-1 binds to methylated histones and maintains heterochromatin in a transcriptionally inactive state. HP-1 binding helps condense chromatin structure to silence gene expression. See Figure 13.14 for more information.

Under what conditions do the different stem-loop structures occur, and what effect do they have on transcription of the trp genes? Sort items into the correct bins to associate the specific events with the conditions in which they are seen.

High tryptophan Region 1 - Region 2 stem loop Region 3 - Region 4 stem loop trp genes not transcribed Low tryptophan Region 2 - Region 3 stem loop trp genes transcribed If tryptophan is present, coupled transcription/translation proceeds quickly because there are plenty of tRNAs charged with tryptophan to bind to the UGG codons in the leader sequence. This pulls region 1 and region 2 into the ribosome complex quickly, leaving region 3 to bind to region 4, forming a termination stem-loop. Transcription is terminated and the trp genes are not expressed because tryptophan levels are high, and tryptophan is the final product of the enzymes encoded by the trp operon. If tryptophan is low/absent, coupled transcription/translation stalls because tRNAs charged with tryptophan are scarce. Slow transcription leaves regions 2, 3 and 4 outside the ribosome complex. This allows region 2 to bind to region 3, preventing the formation of the 3-4 termination stem loop. Transcription continues and the trp genes are expressed.

I+P+OCI+P+OCZ−Y+Z−Y+ -is lac- and the genes are constitutively transcribed.is lac + and the genes are inducibly transcribed. -is lac + and the genes are not transcribed. -is lac + and the genes are constitutively transcribed. -is lac- and the genes are inducibly transcribed. -is lac- and the genes are not transcribed.

is lac- and the genes are constitutively transcribed.is lac + and the genes are inducibly transcribed.

How do UV-induced DNADNA lesions lead to mutation? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

See image.

Match the following mutation descriptions with the type(s) of mutations they refer to. Drag the appropriate items to their respective bins.

See image.

Assuming that Hardy-Weinberg conditions apply, determine the Tt frequency in population B. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.50

In its rare form, 5BU pairs with guanine. True False

True In its more common form, 5BU will pair with adenine. In its rare form, it pairs with guanine.

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? ~11 ~146 ~200 ~54 ~34

~54 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."

Calculate the frequency of the recessive (t) allele for nontasting in population A. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.56

Calculate the frequency of the recessive allele at brachydactyly locus. Express your answer using four decimal places.

0.9999

Now that you have identified the number of genotypes predicted for a gene that has three alleles, can you calculate phenotype frequencies when just allele frequencies are known? For the ABO blood group in humans, there are three alleles (IA , IB , and i), six possible genotypes (IAIA , IBIB , IAIB , IAi, IBi, and ii), and four possible phenotypes (A, B, AB, and O). Recall that IA is dominant to i, IB is dominant to i, and IA and IB are codominant. In a given population, the allele frequencies are as follows: IA = 0.15IB = 0.25i = 0.60 Assuming this population is in Hardy-Weinberg equilibrium, determine the expected phenotype frequencies. Enter your answers to four decimal places. Frequency of the A phenotype= Frequency of the B phenotype= Frequency of the AB phenotype= Frequency of the O phenotype=

Frequency of the A phenotype= 0.2025 Frequency of the B phenotype= 0.3625 Frequency of the AB phenotype= 0.075 Frequency of the O phenotype= 0.36 If a gene has three alleles, then six genotypes are predicted by applying the Hardy-Weinberg equilibrium equation. In the case of ABO blood groups, four phenotypes are possible: A, B, AB, and O. To determine the frequency of each phenotype, let p designate the frequency of allele IA , let q designate the frequency of allele IB , and let r designate the frequency of allele i. The A phenotype is composed of genotypes IAIA and IAi. Thus, the frequency of individuals with phenotype A will be p 2 + 2pr. The B phenotype is composed of genotypes IBIB and IBi. Thus, the frequency of individuals with phenotype B will be q 2 + 2qr. The AB phenotype is determined by one genotype: IAIB . Thus, the frequency of individuals with phenotype AB will be 2pq. The O phenotype is determined by one genotype: ii. Thus, the frequency of individuals with phenotype O will be r 2.

Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A 2 A 2) is 0.0004.Assuming that this population is under Hardy-Weinberg equilibrium, calculate all allele frequencies and genotype frequencies. Enter your numerical answers in the boxes below. Express each answer to four decimal places. Frequency of the A1 allele= Frequency of the A2 allele= Frequency of homozygous dominant individuals= Frequency of heterozygous individuals= Frequency of homozygous recessive individuals=

Frequency of the A1 allele= 0.98 Frequency of the A2 allele= 0.02 Frequency of homozygous dominant individuals= 0.9604 Frequency of heterozygous individuals= 0.0392 Frequency of homozygous recessive individuals= 0.0004 This problem requires you to use the square root method to solve for allele frequencies. Note that this method can only be used when a population is in Hardy-Weinberg equilibrium and the two alleles in question exhibit a dominant and recessive relationship. For this problem, calculate q by taking the square root of q 2. Then calculate p using the equation p + q = 1. Knowing both p and q, you can calculate the remaining genotype frequencies given that f(A1A1) = q 2 and f(A1A2) = 2pq. Thus, q = 0.02 and p = 0.98. For the genotype frequencies, f(A1A1) = 0.9604, f(A1A2) = 0.0392, and f(A2A2) =0.0004.

In the mid-1960s, George Streisinger proposed that strand slippage during DNA replication (shown in the figure) altered the number of DNA repeats found in trinucleotide repeat disorders. Formation of a hairpin structure in the region of repeated DNA during strand slippage could facilitate this process. Which trinucleotide repeat disorder would not be consistent with the hairpin formation part of this model? -Friedreich ataxia (GAA repeat) -Huntington disease (CAG repeat) -Jacobsen syndrome (CGG repeat) -Fragile X syndrome (CGG repeat)

Friedreich ataxia (GAA repeat) To form a hairpin, there must be complementary bases within the repeat. Friedreich ataxia's repeat (GAA repeat) does not fit this model since guanine can not pair with adenine. All of the other repeats contain a guanine and a cytosine that can base pair to stabilize the hairpin structure.

Rank the following levels of chromatin compaction in eukaryotes from the least compact to the most compact.

(Least) naked DNA nucleosome solenoid loop domains chromatid metaphase chromosome (Most) Chromatin compaction is required for the nucleus to accommodate genomes that are often more than 1,000 times longer than the nuclear diameter. As the 2nm wide DNA is organized into chromatin, each level of compaction has a chromatin structure with a signature diameter (2nm DNA, 11nm nucleosome, 30nm solenoid, 300nm loop domains, 700nm chromatid, and 1400nm metaphase chromosome).

Experimental evidence demonstrates that the nucleosomes present in a cell after the completion of S phase are composed of some "old" histone dimers and some newly synthesized histone dimers. Describe the general design for an experiment that uses a protein label such as 35S to show that nucleosomes are often a mixture of old and new histone dimers following DNA replication. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all.

(See image.) unlabeled methionine G1 35S-methionine S unlabeled methionine S S electron microscopy are a mixture of old and new most or all S 1/2 1/4 contain either old or new 1/2 S 1/4 1/8

Down syndrome is caused by trisomy 21, the presence of three copies of chromosome 21. The extra copy usually results from nondisjunction during meiosis. In some cases, however, the extra copy results from a translocation of most of chromosome 21 onto chromosome 14. A person who has had such a translocation in his or her gamete-producing cells is a carrier of familial Down syndrome. The carrier is normal because he or she still has two copies of all the essential genes on chromosome 21, despite the translocation. However, the same may not be true for the carrier's offspring. The diagram shows the six possible gametes that a carrier of familial Down syndrome could produce. Suppose that a carrier of familial Down syndrome mated with a person with a normal karyotype. Which gamete from the carrier parent could fuse with a gamete from the normal parent to produce a trisomy-21 zygote? Drag the appropriate labels to their respective targets. Drag one of the gametes to the target of Group 1 in the diagram. Drag one of the zygotes to the target of Group 2.

(See picture) A carrier of familial Down syndrome has two copies of chromosome 21 and a normal phenotype. However, one of those copies has been translocated to another chromosome, often chromosome 14. Some of the carrier's gametes will contain both the normal and the translocated chromosome 21. If one of those gametes fuses with a gamete from a person with a normal karyotype, a zygote with trisomy 21 will result.

The diagram below shows two normal chromosomes in a cell. Letters represent major segments of the chromosomes. The following table illustrates some structural mutations that involve one or both of these chromosomes. Identify the type of mutation that has led to each result shown. Drag one label into the space to the right of each chromosome or pair of chromosomes. You can use a label once, more than once, or not at all.

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

Having assembled a map of the five deletions, you now decide to map several loci known to be in this region relative to the deletions. You select alleles of these loci that are recessive to their corresponding wild-type alleles. For each locus, you cross a homozygous recessive line to the deletion heterozygote stocks. For example, for locus A and DF1, you perform the following cross: aa ×× DF1/+In this cross, you notice that ½ of the F1 progeny have the recessive (aa) phenotype. You conclude that the A locus is deleted by the DF1 deletion. You continue this type of analysis for all five loci (A - E), crossing recessive homozygotes to each deletion heterozygote. The results are shown in the chart. A "+" indicates that all of the F1 progeny are wild-type. A "-" indicates that ½ of the progeny have the recessive phenotype. Use the data in the chart to place the five loci on the map below. Drag the label for each locus to the correct target to indicate its location on the chromosome. Two targets should be left blank because none of the loci are located there.

(See picture) Deletions typically act as recessive alleles at the loci they delete. When a deletion heterozygote is crossed with an individual homozygous for a recessive allele at a locus that falls within the deleted region, half of the progeny unexpectedly show the homozygous recessive phenotype. This effect was historically known as pseudodominance, or "false dominance." The effect is not due to the recessive allele suddenly becoming dominant to wild-type, but rather because the deleted region acts as a recessive allele at the deleted locus. Mapping to deletion breakpoints is a useful way for geneticists to quickly narrow down the map location of a locus.

How would chromosomes align during synapsis in an individual who is heterozygous for the translocation?

(See picture) Homologs that are heterozygous for a reciprocal translocation undergo unorthodox synapsis during meiosis and form a cruciform, or cross-like, configuration, in which regions of homologous chromosomes pair.

Suppose a diploid cell with three pairs of homologous chromosomes (2n = 6) enters meiosis. How many chromosomes will the resulting gametes have in each of the following cases? Drag one label into each space at the right of the table. Labels can be used once, more than once, or not at all.

(See picture) If one chromosome pair undergoes nondisjunction in meiosis I, half the gametes will have an extra chromosome (n +1), and half will be missing a chromosome (n - 1). If all chromosome pairs undergo nondisjunction in meiosis I, half the gametes will have twice the normal haploid number of chromosomes (2n), and half will have no chromosomes. If one chromosome undergoes nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have an extra chromosome (n +1), and one-quarter will be missing a chromosome (n - 1). If all chromosomes undergo nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have twice the haploid number (2n), and one-quarter will have no chromosomes.

You cross stocks heterozygous for these deletions to each other, and record whether the deletions are lethal or viable as double heterozygotes. For example, flies that are simultaneously heterozygous for deletion DF1 and deletion DF3 die (indicated with an X in the chart). In contrast, flies simultaneously heterozygous for deletions DF1 and DF2 survive (indicated with a check mark).Use the data in the chart to identify the location of each deletion on the map below. The top line on the map represents the undeleted chromosome; on the five other lines, the extent of each deletion is represented by the brackets. Drag the label for each deletion next to the appropriate chromosome. The location of one deletion (DF1) has been identified for you.

(See picture) Large deletions are typically homozygous lethal. In addition, when two different deletions overlap each other, they are also lethal as double heterozygotes because the same chromosomal material is missing in both deletions. This fact can be exploited to map deletions to each other; deletions that overlap will not survive as double heterozygotes, whereas deletions with no overlap will survive.

You decide to cross the reciprocal translocation strain to a pure (black, sepia) line to generate female F1 flies that are both translocation heterozygotes and BbSs dihybrids. You then backcross these F1 females to males from the pure (black, sepia) line. This diagram shows synapsis in the F1 females. (In the diagram, NII = normal chromosome II; TII = translocated chromosome with the chromosome II centromere; NIII = normal chromosome III; TIII = translocated chromosome with the chromosome III centromere.)You wish to recover several categories of F2 flies. What event must occur to produce each category of F2 flies listed in the table below? Answers to the first scenario have been provided as an example. Complete the table by dragging the labels to the correct locations. Blue labels are not reusable. Pink labels can be used once, more than once, or not at all. (Note: CO = crossover; NCO = no crossover.)

(See picture) One way to detect a reciprocal balanced translocation is to realize that crossing over is required to recombine loci on the affected chromosomes in a translocation heterozygote. This differs from the expected situation, where loci known to be on separate chromosomes recombine through independent assortment. Apparent linkage between loci known to be on separate chromosomes is thus a diagnostic clue for a reciprocal balanced translocation between the two chromosomes.

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

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

When this trisomic male fly undergoes meiosis to form sperm, what genetic combinations are possible, and in what overall proportions will they be produced? In the chart below, the (n) and (n+1) reciprical products of a given meiosis are organized in rows. (Note that crossing over does not occur in male Drosophila. Also, assume that chromosome segregation is random and that all sperm survive.) Complete the chart by dragging the chromosome labels to the correct locations. Then use the pink labels to indicate the expected proportion of all gametes that would have each chromosome combination. Genotype labels are not reusable. Proportion labels in pink can be used once, more than once, or not at all.

(See picture) When a 2n+1 organism undergoes meiosis, the resulting gametes will have one of six possible chromosome combinations: three that are (n), and three that are (n+1), each with a frequency of 1/6. Organizing the gametes into the reciprocal products of a given meiosis is a useful way to ensure you have identified all six possible combinations.

he F' plasmid appears to "rescue" mutant strain 3 by restoring wild-type inducible expression. The F' plasmid does not change the phenotype of mutant strain 4; there is still lower than normal expression in the absence of glucose. There are two possible mutations that could lead to low expression of the lac operon even in the absence of glucose -- a mutation in the gene for the CAP protein (known as the crp gene) or a mutation in the CAP-cAMP binding site. Recall that binding of the CAP-cAMP complex to DNA facilitates transcription of the lac operon when glucose levels are low or absent.To determine the effect of introducing a wild type F' plasmid into a bacterial cell with each of these mutations, you draw out the two possible scenarios. For each of the partial diploids, determine whether expression of the lac operon will be inducible (wild type) or low expression in the absence of glucose (mutant). Then use this information to identify the mutations in mutant strain 3 and mutation strain 4. Select the two correct answers. -F' cap+ / crp+ cap- shows low expression in the lactose only condition. Mutation 4 is a CAP-cAMP binding site mutation. -F' cap+ / crp+ cap- shows high expression in the lactose only condition. Mutation 3 is a CAP-cAMP binding site mutation. -F' cap+ / crp- cap+ shows high expression in the lactose only condition. Mutation 3 is a mutation in the crp gene that produces CAP. -F' cap+ / crp- cap+ shows low expression in the lactose only condition. Mutation 4 is a mutation in the crp gene that produces CAP.

-F' cap+ / crp+ cap- shows high expression in the lactose only condition. Mutation 3 is a CAP-cAMP binding site mutation. AND -F' cap+ / crp- cap+ shows low expression in the lactose only condition. Mutation 4 is a mutation in the crp gene that produces CAP. In a strain with a mutation in the gene that produces the CAP protein, there is no CAP protein produced. The F' plasmid has a copy of the lac operon, but not a copy of the wild-type crp gene (which is location elsewhere in the genome). Without CAP protein, neither the lac operon on the plasmid nor the lac operon on the bacterial chromosome can be transcribed most efficiently. Even though both lac operons have a normal CAP-cAMP binding site, without CAP protein, RNA polymerase will have a lower affinity for the promoter sequence, resulting in lower than normal expression of the operon when glucose is absent. In a strain with a mutation in the CAP-cAMP binding site, the CAP protein is made, but is unable to bind to the mutant binding site. As a result, expression of the lac operon is low in the absence of glucose. Introducing a wild-type copy of the lac operon (which has a normal CAP-cAMP binding site) would restore normal high levels of expression of the lac operon when glucose is absent. This expression would, of course, be from the plasmid, not the bacterial chromosome.

In all organisms, certain genes are expressed at any given time while other genes are not. Both prokaryotes and eukaryotes regulate gene expression at the transcription stage. However, the greater complexity of eukaryotic cells makes it possible for gene expression to be regulated at many other stages as well. The diagram below shows different stages at which gene expression may be regulated in eukaryotes. Which statements about the modification of chromatin structure in eukaryotes are true? Select all that apply. -Methylation of histone tails in chromatin can promote condensation of the chromatin. -Acetylation of histone tails is a reversible process. -Acetylation of histone tails in chromatin allows access to DNA for transcription. -DNA is not transcribed when chromatin is packaged tightly in a condensed form. -Some forms of chromatin modification can be passed on to future generations of cells. -Deacetylation of histone tails in chromatin loosens the association between nucleosomes and DNA.

-Methylation of histone tails in chromatin can promote condensation of the chromatin. AND -Acetylation of histone tails is a reversible process. AND -Acetylation of histone tails in chromatin allows access to DNA for transcription. AND -DNA is not transcribed when chromatin is packaged tightly in a condensed form. AND -Some forms of chromatin modification can be passed on to future generations of cells. One of the mechanisms by which eukaryotes regulate gene expression is through modifications to chromatin structure. When chromatin is condensed, DNA is not accessible for transcription. Acetylation of histone tails reduces the attraction between neighboring nucleosomes, causing chromatin to assume a looser structure and allowing access to the DNA for transcription. If the histone tails undergo deacetylation, chromatin can recondense, once again making DNA inaccessible for transcription. Recent evidence suggests that methylation of histone tails can promote either the condensation or the decondensation of chromatin, depending on where the methyl groups are located on the histones. Thus, methylation can either inactivate or activate transcription, and demethylation can reverse the effect of methylation. Changes in chromatin structure may be passed on to future generations of cells in a type of inheritance called epigenetic inheritance.

The UG4UG4 gene is expressed in stem tissue and leaf tissue of the plant Arabidopsis thaliana. To study mechanisms regulating UG4UG4 expression, six small deletions of DNADNA sequence upstream of the gene-coding sequence are made. The locations of deletions and their effect on UG4UG4 expression are shown here. Explain the differential effects of deletions B and F on expression in the two tissues. Select all that apply. -The deletion in mutant F has no effect on UG4UG4 expression in either tissue, indicating that this region contains sequences which downregulate transcription in those cells. -Mutant B only mildly affects UG4UG4 transcription in leaves, but knocks it out almost entirely in stems. This indicates the use of different promoter sequences in the transcription of UG4UG4 in these tissues. -Mutant B only mildly affects UG4UG4 transcription in leaves, but knocks it out almost entirely in stems. This indicates the use of different homologous chromosomes in these tissues. -The deletion in mutant F has no effect on UG4UG4 expression in either tissue, indicating that this region contains no sequences required for promoter action in those cells. -Mutant B only mildly affects UG4UG4 transcription in leaves, but knocks it out almost entirely in stems. This indicates the use of different transcription factors in these tissues. -The deletion in mutant F has no effect on UG4UG4 expression in either tissue, indicating that this region contains no sequences important for transcription whatsoever.

-Mutant B only mildly affects UG4UG4 transcription in leaves, but knocks it out almost entirely in stems. This indicates the use of different promoter sequences in the transcription of UG4UG4 in these tissues. AND -The deletion in mutant F has no effect on UG4UG4 expression in either tissue, indicating that this region contains no sequences required for promoter action in those cells.

You further investigate the regulatory proteins that bind to Regions 1, 2 and 3 using gel shift assays. A gel shift assay uses labeled DNA as a target for regulatory protein binding. A labeled DNA fragment that fails to bind a protein will migrate quickly on an electrophoretic gel. A labeled DNA fragment that does bind a protein will migrate more slowly because the DNA + protein complex is larger than the DNA alone. Thus, the label associated with the DNA will "shift" to a higher location on the gel, as shown in this diagram. To investigate regulatory protein binding to Regions 1, 2, and 3, you will use labeled DNA fragments with some of the same point mutations as in Part B, incubate them with proteins extracted from heart or lung tissue, and conduct a gel shift assay. How would gel shift assay results be affected by these mutations? Select the two correct answers. -Regions 1 and 3 bind protein from heart tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue. -Region 2 binds protein from heart tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue. -Regions 1 and 3 bind protein from lung tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue. -Region 2 binds protein from lung tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue.

-Regions 1 and 3 bind protein from heart tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue. AND -Region 2 binds protein from lung tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue. Regions 1 and 3 bind regulatory proteins found in heart tissue and Region 2 binds a regulatory protein in lung tissue. Point mutations that abolish reporter gene expression also prevent binding of these regulatory proteins to their target sequences.

Part A: You start your experiments with the eyeless mutation on chromosome IV. You cross the reciprocal translocation strain to the eyeless pure line to generate F1 flies that are both translocation heterozygotes and Ee heterozygotes. You decide to testcross F1 males and females in two separate experiments to take advantage of the fact that crossing over does not occur in male Drosophila. Cross F1 males x pure eyeless females (Note: There is no crossing over in the F1 males.): F2 Progeny: ¼ eyeless, fully fertile ¼ wild-type, semi-sterile ¼ eyeless, semi-sterile ¼ wild-type, fully fertile F1 females x pure eyeless males (Note: Crossing over may occur in the F1 females.): F2 Progeny: ¼ eyeless, fully fertile ¼ wild-type, semi-sterile ¼ eyeless, semi-sterile ¼ wild-type, fully fertile What can you conclude from these results? Select the two correct statements. -The F2 progeny in both experiments contain recombinants. -The F2 progeny contain recombinants only when F1 females are testcrossed. -None of the F2 progeny classes require crossing over. -Chromosome IV is one of the chromosomes involved in the reciprocal balanced translocation.

-The F2 progeny in both experiments contain recombinants. AND -None of the F2 progeny classes require crossing over. The two crosses show it is possible to recombine the e allele into a gamete with translocated chromosomes (or, conversely, recombine the E allele into a gamete with normal chromosomes) even when crossing over is not possible. This shows that the recombination is based on chromosome assortment. Since it is possible to recombine the eyeless locus with the translocated chromosomes without crossing over, this indicates that eyeless is not on one of the chromosomes involved in the reciprocal translocation.

The diagram below shows two stretches of DNA in the genome of an imaginary eukaryotic cell. The top stretch of DNA includes the fantasin gene, along with its promoter and one of its enhancers. The bottom stretch of DNA includes the imaginin gene, its promoter, and one of its enhancers. The slash marks (//) indicate that more than 1,000 nucleotides separate the promoter and enhancer of each gene. Which statements about the regulation of transcription initiation in these genes are true? Select all that apply. -Control elements A, B, and C are proximal control elements for the fantasin gene. -The fantasin gene will be transcribed at a high level when activators specific for control elements A, B, and C are present in the cell. -Both the fantasin gene and the imaginin gene will be transcribed at high levels when activators specific for control elements A, B, C, D, and E are present in the cell. -Both the fantasin gene and the imaginin gene will be transcribed at high levels whenever general transcription factors are present in the cell. -The fantasin gene and the imaginin gene have identical enhancers. -The imaginin gene will be transcribed at a high level when repressors specific for the imaginin gene are present in the cell. -Control elements C, D, and E are distal control elements for the imaginin gene.

-The fantasin gene will be transcribed at a high level when activators specific for control elements A, B, and C are present in the cell. AND -Both the fantasin gene and the imaginin gene will be transcribed at high levels when activators specific for control elements A, B, C, D, and E are present in the cell. AND -Control elements C, D, and E are distal control elements for the imaginin gene. Only certain genes are transcribed in a eukaryotic cell at any particular time. The regulation of transcription initiation depends on the interaction of specific transcription factors with specific control elements in enhancers. In the imaginary eukaryotic cell used as an example here, the enhancers for the fantasin gene and imaginin gene are unique because they contain different sets of control elements (A, B, and C for the fantasin gene; C, D, and E for the imaginin gene). Each gene will be transcribed at a high level when activators specific for all of the control elements in its enhancer are present in the cell.

Now, for each region of the lac operon on the bacterial chromosome, I− P+ O+ Z+ Y−, determine whether the region is wild type (that is, it produces a functional protein or it's a correct protein binding sequence) or whether the region is mutated. Select all that apply. -The promoter sequence is correct/functional. -The repressor protein is produced. -The operator sequence is correct/functional. -Beta-galactosidase is produced from the lacZ gene. -Permease is produced from the lacY gene.

-The promoter sequence is correct/functional. AND -The operator sequence is correct/functional. AND -Beta-galactosidase is produced from the lacZ gene. This lac operon, I- P+ O+ Z+ Y-, has a normal promoter sequence. Of the two structural genes (lacZ and lacY), only lacZ has a wild-type sequence. In terms of regulation in response to lactose, the operon does not produce any repressor protein (I-), however the operator sequence is wild type (O+).

In this problem you will explore how to solve problems involving partial diploid lac operon bacterial strains.Bacterial strains that are "partially diploid" have two copies of the lac operon because they aquired a plasmid carrying just the lac operon region. One copy of the lac operon region is on the recipient's bacterial chromosome, and the other copy is on the F' plasmid that was introduced into the cell by conjugation. Partial diploid genotypes are written with the F' segment first and the recipient chromosome next. You create a lac operon partial diploid with this genotype:F' I+ P+ Oc Z− Y+ / I− P+ O+ Z+ Y−To determine which genes are transcribed and under what conditions, you need to first consider each genotype separately, and then together. For each region of the lac operon on the F' plasmid, I+ P+ Oc Z− Y+ , determine whether the region is wild type (that is, it produces a functional protein or it's a correct protein binding sequence) or whether the region is mutated. Select all that apply. -The promoter sequence is correct/functional. -The repressor protein is produced. -The operator sequence is correct/functional. -Beta-galactosidase is produced from the lacZ gene. -Permease is produced from the lacY gene.

-The promoter sequence is correct/functional. AND -The repressor protein is produced. AND -Permease is produced from the lacY gene. This lac operon I+ P+ Oc Z− Y+ has a normal promoter sequence. Of the two structural genes (lacZ and lacY) only lacY has a wild-type sequence.In terms of regulation in response to lactose, the operon makes functional repressor (I+ ), however the operator sequence is mutated (Oc ).

You are investigating an autosomal recessive condition in an extended family. The gene responsible for the condition is known, but the specific allele in the family under study is uncharacterized. As part of your research, you examine mRNA expression for the parents, their two unaffected children, and their affected child with a Northern blot using a full-length cDNA probe.Your results, shown here, are as follows:Both parents have low, but detectable, mRNA expression.The unaffected children have either normal or low mRNA expression.The affected child has no detectable mRNA expression.,What possible molecular defect could explain these results? Select every legitimate possibility from the choices below. -The recessive allele could be a mutation in the transcribed region that makes the mRNA unstable and causes it to degrade. -The recessive allele could be a mutation in regulatory DNA that prevents chromatin from being remodeled from a closed to an open state. -The recessive allele could be a mutation in a splice donor or acceptor sequence that prevents proper mRNA splicing, leaving an intron in the spliced transcript. -The recessive allele could be a premature stop codon that prevents the translation of a full-length protein. -The recessive allele could be a mutation in the promoter region that prevents RNA polymerase II from binding.

-The recessive allele could be a mutation in the transcribed region that makes the mRNA unstable and causes it to degrade. AND -The recessive allele could be a mutation in regulatory DNA that prevents chromatin from being remodeled from a closed to an open state. AND -The recessive allele could be a mutation in the promoter region that prevents RNA polymerase II from binding. Your observation that the affected child has no detectable mRNA has several possible explanations. You decide that further experiments are needed to distinguish between them.

Describe the importance of this process to transcription. Select the four correct answers. -These alterations can be used to activate transcription of a gene by opening its enhancer sequences. -These alterations can be used to repress transcription of a gene by opening its enhancer sequences. -These alterations can be used to repress transcription of a gene by hiding its promoter sequences. -These alterations can be used to activate transcription of a gene by hiding its enhancer sequences. -These alterations can be used to repress transcription of a gene by opening its promoter sequences. -These alterations can be used to activate transcription of a gene by opening its promoter sequences. -These alterations can be used to activate transcription of a gene by hiding its promoter sequences. -These alterations can be used to repress transcription of a gene by hiding its enhancer sequences.

-These alterations can be used to activate transcription of a gene by opening its enhancer sequences. AND -These alterations can be used to repress transcription of a gene by hiding its promoter sequences. AND -These alterations can be used to activate transcription of a gene by opening its promoter sequences. AND -These alterations can be used to repress transcription of a gene by hiding its enhancer sequences.

What are the sources of dsRNA? Select all that apply. -infection and transcription of a DNA virus -replication of an ssRNA -infection of a cell by an RNA virus -transcription of both strands of a repetitive DNA sequence -transcription of genes that encode microRNA

-infection of a cell by an RNA virus AND -transcription of both strands of a repetitive DNA sequence AND -transcription of genes that encode microRNA

Compare and contrast promoters and enhancers with respect to their location (upstream versus downstream), orientation, and distance (in base pairs) relative to a gene they regulate. promoter sequence is likely to be: Check all that apply. -orientation independent -located downstream of the gene it controls -located upstream of the gene it controls -orientation dependent -located either nearby or at great distance from the gene it controls -located within a few dozens nucleotides of the gene it controls

-located upstream of the gene it controls AND -orientation dependent AND -located within a few dozens nucleotides of the gene it controls

Identify the evolutionary forces that can cause allele frequencies to change from one generation to the next. Check all that apply. -extinction -inbreeding -migration -random genetic drift -natural selection -mutation

-migration AND -random genetic drift AND -natural selection AND -mutation

Which of the following mutations would result in higher-than-normal expression of the trp genes in presence of tryptophan? Select all that apply. -mutations in Region 3 that prevent 3-4 stem loop formation -mutations in Region 2 that prevents 2-3 stem loop formation -mutations in Region 1 that prevent 1-2 stem loop formation

-mutations in Region 3 that prevent 3-4 stem loop formation AND -mutations in Region 1 that prevent 1-2 stem loop formation The 3-4 stem loop terminates transcription (termination stem loop). Mutations that reduce the percentage of complementary base pairs binding these two regions destablize the termination stem loop and increase the probability of transcription of the trp operon. The 2-3 stem loop prevents termination of transcription. Mutations that reduce the percentage of complementary base pairs binding region 2 and region 3 destabilize the 2-3 stem loop. If region 3 cannot bind to region 2, it is more likely to bind to region 4, forming a termination stem loop. As a result, mutations in region 2 that prevent the 2-3 stem loop formation increase the probability of transcription termination. The 1-2 stem loop causes a minor pause in coupled transcription/translation. Mutations that reduce the percentage of complementary base pairs binding region 1 and region 2 destabilize the 1-2 stem loop. If region 2 cannot bind to region 1, its more likely to bind to region 3, forming an anti-termination loop. This increases the probability that transcription will continue.

What process is the researcher intending to detect with the use of these restriction enzymes? Select all that apply. -mutations that create new SmaI recognition sites -mutations that disrupt SmaI recognition sites -mutations that create new PvuII recognition sites -mutations that disrupt PvuII recognition sites

-mutations that create new SmaI recognition sites AND -mutations that disrupt SmaI recognition sites AND -mutations that create new PvuII recognition sites AND -mutations that disrupt PvuII recognition sites

enhancer sequence is likely to be: Check all that apply. -orientation independent -located downstream of the gene it controls -located within a few dozens nucleotides of the gene it controls -orientation dependent -located upstream of the gene it controls -located either nearby or at great distance from the gene it controls

-orientation independent AND -located downstream of the gene it controls AND -located upstream of the gene it controls AND -located either nearby or at great distance from the gene it controls

Calculate the frequency of heterozygotes for albinism. Express your answer using four decimal places.

0.0311

Given only the data in the table, for which part(s) of the operon is it possible to identify the exact region it corresponds to? Select all that apply. -the gene for enzyme 2 -the promoter region -the operator region -the gene for the repressor protein -the gene for enzyme 1

-the gene for enzyme 2 AND -the promoter region AND -the gene for enzyme 1 Of the five mutant strains, three have distinctive effects. A mutation in region A eliminates expression of both enzymes. This suggests that region A is the promoter sequence, where RNA polymerase binds to transcribe both enzyme genes. A mutation in region C eliminates expression of enzyme 2. This suggests that region C is the gene for enzyme 2. A mutation in region E eliminates the expression of enzyme 1. This suggests that region E is the gene for enzyme 1. The other two mutations, in region B and in region D, both result in constitutive expression. This suggests that these two regions are involved in negative control -- a repressor binding to an operator region. With just this information, however, you cannot determine which region is the gene for the repressor protein and which region is the operator.

Trinucleotide repeat disorders are hereditary diseases caused by mutant genes containing an increased number of repeats of a DNA trinucleotide sequence. Which sequence(s) contain a trinucleotide repeat? Select all that apply. ...CACGGCGGCGGCGGCGGCATCGC... ...TTCACTGTCACTGTCACTGTCACTGTCC... ...CACGGAAGAAGAAGAAGAAATAGAC... ...AGCGACAGCAGCAGCAGCAGCAAGT... ...GGCAGGCAGGCAGGCAGGCAGGCTG...

...CACGGCGGCGGCGGCGGCATCGC... AND ...CACGGAAGAAGAAGAAGAAATAGAC... AND ...AGCGACAGCAGCAGCAGCAGCAAGT... A trinucleotide repeat must contain the same three nucleotides adjacent to each other over and over again....CAC GGC GGC GGC GGC GGC ATCGC... contains a GGC repeat....CACG GAA GAA GAA GAA GAA ATAGAC... contains a GAA repeat....AGCGA CAG CAG CAG CAG CAG CAAGT... contains a CAG repeat.

Albinism, an autosomal recessive trait characterized by an absence of skin pigmentation, is found in 1 in 4000 people in populations at equilibrium. Brachydactyly, an autosomal dominant trait producing shortened fingers and toes, is found in 1 in 6000 people in populations at equilibrium. Calculate the frequency of the recessive allele at albinism locus. Express your answer using four decimal places.

0.0158

Cystic fibrosis (CF) is the most common autosomal recessive disorder in certain Caucasian populations. In some populations, approximately 1 in 2000 children have CF. Determine the frequency of CFCF carriers in this population. Assume this population is in Hardy-Weinberg equilibrium for the purposes of this calculation. Enter your answer to three decimal places (example: 0.254 or 10.000).

0.043

If a recessive disease is found in 50 out of 100,000 individuals, what is the frequency of the heterozygote carriers for this disease? 0.0005 0.022 0.043 0.956

0.043 If q 2 = 0.0005, then q = 0.022 and p = 1 − q = 0.978. The heterozygote frequency is 2pq, or 2 (0.978) (0.022) = 0.043.

Assuming that Hardy-Weinberg conditions apply, determine the tt frequency in population C. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.09

f this trisomic male is crossed with a (ey−ey− ey−ey−, gw−gw− gw−gw−) female, what proportion of the progeny will be phenotypically wild-type? Express your answer as a fraction (example: 3/5).

0.167 For the progeny of this cross to be wild-type, they need to receive both the (ey−gw+ey−gw+) and (ey+gw−ey+gw−) chromosomes from the trisomic male. This combination is present in 1/6 of the trisomic male's sperm.

Assuming that Hardy-Weinberg conditions apply, determine the TT frequency in population A. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.20

Assuming that Hardy-Weinberg conditions apply, determine the TT frequency in population B. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.25

Assuming that Hardy-Weinberg conditions apply, determine the tt frequency in population B. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.250

Calculate the frequency of the recessive (t) allele for nontasting in population C. Enter your answer to one decimal place (example: 0.2 or 10.0).

0.3

Assuming that Hardy-Weinberg conditions apply, determine the tt frequency in population A. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.31

Assuming that Hardy-Weinberg conditions apply, determine the Tt frequency in population C. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.42

The frequency of tasters and nontasters of PTC varies among populations. In population A, 69 percent of people are tasters (an autosomal dominant trait) and 31 percent are nontasters. In population B, tasters are 75 percent and nontasters 25 percent. In population C, tasters are 91 percent and nontasters are 9 percent. Calculate the frequency of the dominant (T) allele for PTC tasting in population A. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.44

In a Hardy-Weinberg population, 16% of individuals display the homozygous recessive phenotype. What proportion of the population will be heterozygous carriers of the recessive allele? See Section 20.1 (Page 727) . 0.24 0.6 0.48 0.36

0.48 16% = q2 = 0.16. Therefore, q = 0.4 and p = 0.6. The frequency of heterozygous carriers will be 2pq (2 × 0.4 × 0.6), or 0.48

Assuming that Hardy-Weinberg conditions apply, determine the Tt frequency in population A. Enter your answer to two decimal places (example: 0.25 or 10.00).

0.49

Calculate the frequency of the dominant (T) allele for PTC tasting in population B. Enter your answer to one decimal place (example: 0.2 or 10.0).

0.5

Calculate the frequency of the recessive (t) allele for nontasting in population B. Enter your answer to one decimal place (example: 0.2 or 10.0).

0.5

Sickle cell disease (SCD) is found in numerous populations whose ancestral homes are in the malaria belt of Africa and Asia. SCD is an autosomal recessive disorder that results from homozygosity for a mutant β-globin gene allele. Data on one affected population indicates that approximately 25 in 100 newborn infants have SCD. What is the frequency of the wild-type (βAA ) allele in this population? Express your answer using two decimal places.

0.50

What is the frequency of carriers of SCD in the population? Express your answer using two decimal places.

0.50

If this trisomic male is crossed with a (ey−ey− ey−ey−, gw−gw− gw−gw−) female, what proportion of the progeny will be phenotypically eyeless? Express your answer as a fraction (example: 3/5).

0.500 For the six categories of gamete produced by the trisomic male, three do not have an ey+ey+ allele present and would thus form eyeless progeny when fused with an (ey−gw−ey−gw−) gamete from the female.

If this trisomic male is crossed with a (ey−ey− ey−ey−, gw−gw− gw−gw−) female, what proportion of the progeny will be phenotypically gawky? Express your answer as a fraction (example: 3/5).

0.500 For the six categories of gamete produced by the trisomic male, three do not have an gw+gw+ allele present and would thus form gawky progeny when fused with an (ey−gw−ey−gw−) gamete from the female.

Calculate the frequency of the dominant (T) allele for PTC tasting in population C. Enter your answer to one decimal place (example: 0.2 or 10.0).

0.7

Calculate the frequency of the dominant allele at albinism locus. Express your answer using four decimal places.

0.9842

Genomic DNADNA from the nematode worm Caenorhabditis elegans is organized by nucleosomes in the manner typical of eukaryotic genomes, with 145 bpbp encircling each nucleosome and approximately 55 bp in linker DNADNA. When C. elegans chromatin is carefully isolated, stripped of nonhistone proteins, and placed in an appropriate buffer, the chromatin decondenses to the 10-nmnm fiber structure. Suppose researchers mix a sample of 10-nmnm-fiber chromatin with a large amount of the enzyme DNaseI that randomly cleaves DNA in regions that are not protected by bound protein. Next, they remove the nucleosomes, separate the DNA fragments by gel electrophoresis, and stain all the DNA fragments in the gel. How many bands will be visible on the gel? 1 2 3 4

1

Imagine that you have been called as an expert witness in forensic genetics in a criminal case. You have compared DNA fingerprints from a bloodstain collected from the scene of the crime with a blood sample from the defendant. Some of your results are shown in the figure below. In your expert opinion, what is the predicted frequency for occurrence of this pattern in the general population? 1 in 6 1 in 15 1 in 31 1 in 52 1 in 2790

1 in 2790 The frequency of occurrence for the three alleles is 1 in (6 × 31 × 15) = 1 in 2790.

Use the data from Part B to calculate the distance (in map units) from the sepia locus to the translocation breakpoint. -0.5 map units -1 map unit -2 map units -4 map units

1 map unit For the sepia cross using F1 females, two of the output products of meiosis are different from either input gamete that made up the F1 females: the s allele recombined onto a translocated chromosome, and the S allele recombined onto a normal chromosome. Since these two recombinant products have a frequency of 1% in the F2 (calculated as follows: (4 + 6)/1000 = 0.01), you can infer that there is a distance of 1 map unit between the sepia locus and the translocation breakpoint on chromosome III.

Binding of the guide strand by the PAZ domain cleft involves several conserved amino acid residues. Answer the following two questions: 1) Which PAZ residue stacks with the terminal base of the guide strand? 2) Which PAZ residue hydrogen-bonds with the 2' OH group on the second-to-last ribose in the guide strand? 1) Tyrosine190; 2) Tryptophan213 1) Tryptophan213; 2) Tyrosine190 1) Tryptophan213; 2) Histidine217 1) Histidine217; 2) Tyrosine190

1) Tryptophan213; 2) Tyrosine190 The image shows the interaction between the conserved Tryptophan213 residue (blue) and terminal base of the guide strand RNA (pink) within the PAZ domain cleft. Tryptophan is a neutral-nonpolar amino acid with a planar aromatic ring. The proximity of the Tryptophan213 residue allows for it to stack on top of the last nitrogenous base in the guide strand (which is also an aromatic planar surface). This creates an attractive noncovalent interaction between aromatic rings, thereby helping to stabilize the 3' end of the guide strand. The image shows the conserved Tyrosine190 residue (blue) hydrogen-bonded (red line) to the 2' OH on the second-to-last ribose in the guide strand (pink). Notice that this binding interaction takes place between the guide strand RNA and the OB beta barrel of the PAZ domain.

The figure (Figure 1) shows original chromosomes at the top and paired chromosomes undergoing crossing over at the bottom. You will be determining how the chromosomes connect after the crossover event has finished. Be sure to begin with the left end of each chromosome. Select the correct combination describing the left end (unprimed number) and right end (primed number) of the following chromosome found in the gametes. (?-A-B-C-D-E-?) 1-3' 1-2' 1'-1 1-3 1-1'

1-1' Chromatid 1 does not undergo genetic exchange and enters the gamete unchanged.

The operon model describes how bacteria control the production of groups of enzymes. In this model, synthesis of the messenger RNA coding for these enzymes is switched on or off by regulatory proteins. Can you match terms related to operons to their definitions? Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

1. A(n) operon is a stretch of DNA consisting of an operator, a promoter, and genes for a related set of proteins, usually making up an entire metabolic pathway. 2. The genes of an operon is/are arranged sequentially after the promoter. 3. A(n) promoter is a specific nucleotide sequence in DNA that binds RNA polymerase, positioning it to start transcribing RNA at the appropriate place. 4. A(n) regulatory gene codes for a protein, such as a repressor, that controls the transcription of another gene or group of genes. 5. Regulatory proteins bind to the operator to control expression of the operon. 6. A(n) repressor is a protein that inhibits gene transcription. In prokaryotes, this protein binds to the DNA in or near the promoter. 7. A(n) inducer is a specific small molecule that binds to a bacterial regulatory protein and changes its shape so that it cannot bind to an operator, thus switching an operon on. An operon is made up of a promoter and the genes of the operon. The promoter, which includes an operator, is the stretch of DNA where RNA polymerase binds. Regulatory proteins bind to the operator. The genes of the operon code for a related set of proteins. A regulatory gene located away from the operon codes for a protein that controls the operon.

The ancient remains of several human bodies have been discovered and legally need to be returned to the tribe that contains the closest descendents. You have been called as a witness in the resulting legal case to give your opinion. The jury is confused about the analysis you have done. Could you please explain it to them again? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

1. After obtaining the samples (preserved bone and muscle fragments), I conducted a DNA extraction. 2. Because I suspected the ancient DNA would be degraded, I pretreated it with random primers and the Klenow fragment (a procedure found to help repair damage and aid amplification) and amplified the DNA using polymerase chain reaction. 3. To make sure that the DNA I analyzed was indeed from the ancient sample, during these procedures, I used both positive and negative controls. 4. Following amplification, I used restriction enzymes to cut out the DNA fragments with the genes I wanted to study. 5. Then I used gel electrophoresis to separate the DNA fragments. 6. I performed a Southern blot, in which I transferred the DNA onto a filter and then incubated the filter with a radiolabeled DNA sequence complementary to the gene I was interested in. 7. This final step with the radiolabeled DNA sequence is called hybridization to a probe. I can repeat this step for multiple genes. 8. I used several highly variable loci to create a specific DNA fingerprint. 9. Finally, I compared the sample to modern DNA samples in order to determine the group most closely related to the individual who the sample was taken from. First, you need to extract the DNA from the sample. This (and all subsequent steps) must be done very carefully to avoid contamination. The polymerase chain reaction will allow you to amplify the DNA (random priming will allow you to amplify all the DNA rather than a specific fragment).A positive control (like a modern DNA sample) would verify that the technique is working. A negative control (lack of DNA) would show a lack of contamination. Restriction enzymes can be used to isolate the genes of interest, including VNTRs. During gel electrophoresis, smaller fragments will migrate faster than larger fragments. Southern blotting is used to prepare DNA for probing. The probe hybridizes to the gene of interest and can be visualized (by autoradiography for a radioactive probe).

Define and give an example of each of the following: Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

1. Attenuation - a mechanism of transcriptional regulation in which transcription level is modified (attenuated) to meet environmental requirements. 2. Negative regulation - a process of transcriptional regulation through which binding of regulatory proteins to DNA blocks transcription. 3. Allostery- a process by which the stereochemistry of a protein is altered to change its interaction capabilities. 4. Positive regulation- a process of transcription regulation through which the binding of regulatory proteins to DNA activates transcription.

The second pattern you wish to obtain expresses the UASG-GFP construct only in the cells in row 2, as shown here.How will you drive UASG-GFP expression specifically in row 2? For each regulatory DNA construct, decide if it should express Gal4, Gal80, or not be used in order to produce the desired pattern. (Note: You realize there are two ways to obtain this pattern, so you decide not to use regulatory DNA construct 3 in this experiment.) Drag the labels on the left to the appropriate blanks on the right to complete the sentences. Labels may be used once, more than once, or not at all.

1. Construct 1 should express Gal 4. 2. Construct 2 should express Gal 80.Construct 3 is not used. 3. Construct 4 should express Gal 4. This pattern requires the action of three constructs. Regulatory DNA construct 4 can be used to express Gal4 in cells B1/B2/B3, C1/C2/C3, and D1/D2/D3. Cell A2 can be added to this pattern by using regulatory DNA construct 1 to express Gal4. The pattern can then be refined to be specific to row 2 by expressing Gal80 with regulatory DNA construct 2, which will block the action of Gal4 in rows 1 and 3.

The final pattern you wish to obtain expresses the UASG-GFP construct only in cell C2.How will you drive UASG-GFP expression specifically in cell C2? For each regulatory DNA construct, decide if it should express Gal4, Gal80, or not be used in order to produce the desired pattern. Drag the labels on the left to the appropriate blanks on the right to complete the sentences. Labels may be used once, more than once, or not at all.

1. Construct 1 should express Gal 80. 2. Construct 2 should express Gal 80. 3. Construct 3 should express Gal 4. 4. Construct 4 should not be used. The desired pattern in this case also requires three constructs. Only regulatory constructs 3 and 4 drive expression in cell C2, but regulatory construct 4 also drives expression in cell D2, a cell unique to that construct. Since Gal80 cannot be expressed in cell D2 to block UASG-GFP expression, regulatory construct 3 must be used to express Gal4. Then, to inhibit Gal4 in the undesired cells, Gal80 can be expressed with regulatory constructs 1 and 2.

The Gal80 / Gal4 / UASG system is used by yeast to regulate expression of genes required for galactose metabolism. Gal4 is a transcription factor that acts as a homodimer to bind regulatory DNA called the Upstream Activating Sequence (Galactose), or UASG. Binding of Gal4 to UASG induces transcription of galactose-metabolizing genes. Gal80 acts as an inhibitor of Gal4 by binding to it and preventing it from binding to UASG. This system has been used to drive gene expression in other organisms, including the fruit fly Drosophila melanogaster. You study nervous system development in Drosophila embryos. You have identified a set of 12 cells that differentiate into various neurons and glia later in development.As part of studying these cells, you identify regulatory DNA from 4 genes expressed in a subset of these cells. Connecting the regulatory DNA from these 4 genes directly to a GFP reporter gives you the patterns shown below.While these expression patterns are useful, you decide that you also want to obtain GFP expression patterns that are specific to fewer cells. You decide to use the Gal80 / Gal4 / UASG system to accomplish this. You already have a UASG-GFP construct, so you focus on using combinations of your four characterized regulatory DNA constructs to drive the expression of Gal4 or Gal80 to produce the expression patterns you want. The first pattern you wish to obtain expresses the UASG-GFP construct in cells B1/B2/B3 and D1/D2/D3, as shown here.How will you drive UASG-GFP expression specifically in cells B1/B2/B3 and D1/D2/D3? For each regulatory DNA construct, decide if it should express Gal4, Gal80, or not be used in order to produce the desired pattern. Drag the labels on the left to the appropriate blanks on the right to complete the sentences. Labels may be used once, more than once, or not at all.

1. Construct 1 should not be used. 2. Construct 2 should not be used. 3. Construct 3 should express Gal 80. 4. Construct 4 should express Gal 4. The desired pattern can be obtained by using two constructs. Regulatory DNA construct 4 can be used to express Gal4 in cells B1/B2/B3, C1/C2/C3, and D1/D2/D3. Using regulatory DNA construct 3 to express Gal80 will block the action of Gal4 in cells C1/C2/C3 (and have no effect in cells A1/A2/A3), giving the desired expression pattern for UASG-GFP.

You decide to perform a similar analysis using a GFP reporter gene in place of Gene 2. The full length construct has the proper expression pattern for Gene 2 (expression in wing and leg discs). As in Part A, you make a number of deletions of the full length construct and score them for expression of GFP. The six constructs and data table with expression results are shown below. What can you conclude from these data? Answer each question below by dragging the terms on the left to the appropriate blanks on the right. Not all terms will be used.

1. For which gene or genes does the insulator region act as an insulator? both Gene 1 and Gene 2 2. How many enhancers control the expression of Gene 2 in wing and leg discs? two enhancers, one for each disc type This experiment refutes your previous hypothesis from Part A--that there is single enhancer (EWL); instead, you now conclude that there are two separate enhancers -- one for wing discs (EW) and one for leg discs (EL).In addition, this experiment confirms that the insulator sequence separates EA from the two enhancers for Gene 2.

You are studying the regulatory DNA of a mouse gene expressed in developing heart, liver, and lung tissue. Your preliminary work has shown that heart and lung expression of this gene is controlled by a short fragment of DNA just upstream of the promoter. Based on this result, you decide to investigate this region further to understand its function. You decide to compare this sequence to the regulatory DNA of the same gene found in rats and humans. Using genome databases, you identify and align the three sequences, shown here. What can you conclude from these results? Drag the terms on the left to the appropriate blanks on the right. Not all terms will be used.

1. How many regions in the DNA sequences have been conserved among all three species? three 2. In general, DNA regions that are conserved between distantly-related species provide evidence of conserved function. 3. Regulatory DNA that is conserved between distantly-related species may function as regions that bind regulatory proteins. From these experiments, you observe three regions that are conserved between the mouse, rat, and human orthologs of this gene. (Ortholog is the term for a gene that is structurally and functionally conserved between related organisms). From this observation, you hypothesize that this region of regulatory DNA binds three regulatory proteins - one at each conserved region.

Prior to knocking each gene down with RNAi, you use antibodies to determine the wild-type expression pattern of the two novel proteins (NP1 and NP2) in relation to αα-spectrin. You note that NP2 is localized to the entire cell cortex in a pattern similar to αα-spectrin, but that NP1 is localized to only a subset of the cell cortex. Having determined the wild-type pattern for αα-spectrin, NP1, and NP2, you then assay the expression of these three proteins when NP1 is knocked down with RNAi. In the 12 cells you are interested in, you drive expression of UAS-NP1 RNAi in two stripes (cells B1/B2/B3 and D1/D2/D3).As before, you use antibodies to determine the expression pattern of NP1, NP2 and αα-spectrin. You obtain these results. What can you conclude from these results? Answer the questions by dragging the terms on the left to the appropriate blanks on the right. Terms may be used once, more than once, or not at all.

1. In cells expressing the UAS-NP1 RNAi construct, is NP1 protein present or absent? absent 2. In cells expressing the UAS-NP1 RNAi construct, is NP2 protein present or absent? presentAlso, is the location of NP2 in cells expressing UAS-NP1 RNAi normal or abnormal? normal 3. In cells expressing the UAS-NP1 RNAi construct, is αα-spectrin protein present or absent? present In cells expressing the UAS-NP1 RNAi construct, is α-spectrin protein present or absent? presentAlso, is the location of α-spectrin in cells expressing UAS-NP1 RNAi normal or abnormal? present Also, is the location of αα-spectrin in cells expressing UAS-NP1 RNAi normal or abnormal? normal In your first experiment, you determine that expressing the UAS-NP1 RNAi construct in cells B1/B2/B3 and D1/D2/D3 eliminates the expression of NP1, but has no effect on the expression or localization of NP2 or αα-spectrin.

You then repeat the experiment, but with a UAS-NP2 RNAi construct. You obtain these results. What can you conclude from these results? Answer the questions by dragging the terms on the left to the appropriate blanks on the right. Terms may be used once, more than once, or not at all.

1. In cells expressing the UAS-NP2 RNAi construct, is NP1 protein present or absent? presentAlso, is the location of NP1 in cells expressing UAS-NP2 RNAi normal or abnormal? abnormal 2. In cells expressing the UAS-NP2 RNAi construct, is NP2 protein present or absent? absent 3. In cells expressing the UAS-NP2 RNAi construct, is αα-spectrin protein present or absent? present In cells expressing the UAS-NP2 RNAi construct, is α-spectrin protein present or absent? presentAlso, is the location of α-spectrin in cells expressing UAS-NP2 RNAi normal or abnormal? present Also, is the location of αα-spectrin in cells expressing UAS-NP2 RNAi normal or abnormal? normal In your second experiment, you determine that expressing the UAS-NP2 RNAi construct in cells B1/B2/B3 and D1/D2/D3 eliminates the expression of NP2 in these cells. Additionally, removing NP2 causes the mislocalization of NP1; in wild-type cells, NP1 is restricted to a defined portion of the cell cortex.In the absence of NP2, however, NP1 is not restricted to the cortex, but is rather mislocalized throughout the cell. This indicates that NP2 is necessary for the proper localization of NP1.

Now use all the data to match the unknown mutations to specific mutation types. Drag each specific mutation into the correct sentence.

1. Mutation 1 is lacZ- or lacP-. 2. Mutation 2 is lacIs. 3. Mutation 3 is cap-cAMP binding site mutation. 4. Mutation 4 is CAP gene mutation. 5. Mutation 5 is lacOc. 6. Mutation 6 is lacI-. The F' lac+ / Mutation 1 strain shows wild-type inducible expression, while the F' lac+ / Mutation 2 strain still shows no expression of the operon. This indicates that Mutation 1 is either a mutation in the lacZ gene or a mutation in the promoter sequence. It's not possible to tell the difference between these two possibilities based on the current data. Mutation 2 is a super-repressor (Is) mutation.The F' lac+ / Mutation 3 strain shows high expression without glucose (wild type), while the F' lac+ / Mutation 4 strain shows low expression in the absence of glucose. This indicates that Mutation 3 is a mutation in the CAP-cAMP binding site. Mutation 4 is a mutation in the gene that produces the CAP protein (crp).The F' lac+ / Mutation 5 strain shows constitutive expression, while the F' lac+ / Mutation 6 strain is inducible. This indicates that Mutation 5 is an Oc mutation. Mutation 6 is an I- mutation.

Autonomous elements can transpose by themselves even though they cannot produce transposase. True False

False Transposase is required for transposition of both autonomous and non-autonomous transposons, but autonomous elements produce their own transposase.

To test the hypothesis that the three conserved regions function as regulatory DNA, you make several transgenic constructs using a GFP reporter gene and you evaluate their expression in developing heart and lung tissue. Construct 1 is the wild-type construct. Notice that this construct has normal expression in both tissues. For Constructs 2-8, you introduced point mutations into each of the three conserved regions (indicated by ***). Expression results for each of these constructs are indicated in the chart. To test the hypothesis that the three conserved regions function as regulatory DNA, you make several transgenic constructs using a GFP reporter gene and you evaluate their expression in developing heart and lung tissue. Construct 1 is the wild-type construct. Notice that this construct has normal expression in both tissues. For Constructs 2-8, you introduced point mutations into each of the three conserved regions (indicated by ***). Expression results for each of these constructs are indicated in the chart.

1. Region 1 is necessary but not sufficient for gene expression in the heart because expression is absent when this region is mutated.Which other region, if any, is also required for gene expression in the heart? Region 3 2. Region 2 is necessary and sufficient for gene expression in the lung because expression is absent when this region is mutated.Which other region, if any, is also required for gene expression in the lungs? no region 3. Region 3 is necessary but not sufficient for gene expression in the heart because expression is absent when this region is mutated.Which other region, if any, is also required for gene expression in the heart? Region 1 From these experiments, you conclude that Regions 1 and 3 are necessary, but not individually sufficient, for expression in heart tissue. Conversely, you conclude that expression in lung tissue is mediated solely through Region 2.

In a partial diploid strain, both copies of the lac operon are together in the same cell. To determine the overall regulation of lac operon genes in a partial diploid strain, here are the most important things to understand: Proteins can diffuse around the cell and potentially interact with either copy of the lac operon (known as trans-acting). DNA binding sites can only regulate genes on the same copy (known as cis-acting). For the partial diploid you've created, F' I+ P+ Oc Z− Y+ / I− P+ O+ Z+ Y− , you already know that the promoter sequences for copies are functional, so you can focus on the repressor proteins and operator regions that control expression. Indicate the source of repressor protein and the operator region(s) to which it binds. Then indicate how this affects expression of lac genes from the F' plasmid and from the bacterial chromosome.

1. Repressor protein is made from the F' plasmid. 2. Repressor protein is able to bind to the operator region on the bacterial chromosome. 3. Functional beta-galactosidase protein could be made from the bacterial chromosome . 4. Functional permease protein could be made from the F' plasmid. Repressor protein is only produced from the F' plasmid (I+); the bacterial chromosome has a mutated sequence (I-). In terms of operator sequences, only the operator sequence on the bacterial chromosome (O+) can bind the repressor; the F' plasmid has a mutated operator region (Oc). LacZ gene is wild type (Z+) only on the bacterial chromosome; lacY is wild type (Y+) only on the F' plasmid.

DNAse I is an enzyme that cleaves DNA accessible to it.When chromatin is remodeled during active gene transcription, regulatory DNA becomes accessible to cleavage by DNAse I.In contrast, the regulatory DNA of genes not being actively transcribed is resistant to DNAse I - induced cleavage.You obtain DNA from the affected child and her older brother and treat both samples with DNAse I. With the treated DNA, you separate DNA fragments with gel electrophoresis, blot the DNA to a membrane, and probe the membrane with a DNA probe that spans the entire gene along with some flanking DNA.Your results are shown in this gel. What can you conclude from these results? Complete the sentences by dragging the terms on the left to the appropriate blanks on the right. Not all terms will be used.

1. This region of DNA from the older brother is completely sensitive to DNAse I digestion. 2. This region of DNA from the affected child is partially sensitive to DNAse I digestion. 3. These results support the hypothesis that this DNA region in the affected child cannot remodel chromatin to allow for proper gene expression. Taken together, the results of the expression construct analysis and the DNAse I sensitivity assay support the hypothesis that the defect in the affected child is due to regulatory DNA that will not remodel, and thus not allow for transcription factors to bind to initiate gene expression.

Populations evolve when the frequencies of different genetic alleles in the population change over time. Three major mechanisms cause allele frequencies to change--natural selection, genetic drift, and gene flow. Before beginning this tutorial, watch the Mechanisms of Evolution BioFlix animation. Pay particular attention to: conditions or events that result in natural selection, genetic drift, or gene flow; how each of these mechanisms affects a population In the beetles described in the animation, there were two alleles for color, brown and green. Suppose that you discover a very small population of these beetles, consisting of the individuals shown below. How can you calculate the frequency of each allele in this population? Drag the terms or numbers on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

1. To calculate the frequency of the brown allele, count the number of brown alleles and divide by the total number of alleles in this population. 2. In this beetle population, the number of brown alleles is 8. 3. In this beetle population, the total number of alleles for color is 20. 4. The frequency of the brown allele in this beetle population is 0.4. 5. The frequency of the green allele in this beetle population is 0.6. Allele frequencies measure the relative proportions of different alleles in a population. By looking at how allele frequencies change over time, you can see how populations are evolving.

Use the data from Part C to calculate the distance (in map units) from the black locus to the translocation breakpoint. -7.5 map units -10 map units -15 map units -20 map units

15 map units For the black cross using F1 females, two of the output products of meiosis are different from either input gamete that made up the F1 females: the b allele recombined onto a translocated chromosome, and the B allele recombined onto a normal chromosome. Since these two recombinant products have a frequency of 15% in the F2 (calculated as follows: (73 + 77)/1000 = 0.15), you can infer that there is a distance of 15 map units between the black locus and the translocation breakpoint on chromosome II.

Which two domains/subdomains of the RecB subunit display typical helicase motor domain activity? -1A and 2A -1A and 1B -2A and 3 -2A and 2B

1A and 2A Subdomains 1A and 2A show motor function typically found in helicases, as they use the energy from ATP to help drive the RecBCD protein unidirectionally along duplexed DNA. As a consequence of their motor function, these subdomains help pull the duplexed DNA into the RecC pin causing the DNA to split into single strands. Subdomain 1B contains an insertion of residues that form an extended arm that runs adjacent to and contacts the duplex DNA. Subdomain 2B is the major interface between RecB and RecC. The third carboxy-terminal nuclease domain, responsible for cleaving both the 3'-tailed strand (darker purple) and the 5'-tailed strand (bright purple), is connected to the remainder of the RecB subunit by a long 70-residue linker.

The figure (Figure 1) shows original chromosomes at the top and paired chromosomes undergoing crossing over at the bottom. You will be determining how the chromosomes connect after the crossover event has finished. Be sure to begin with the left end of each chromosome. Select the correct combination describing the left end and right end of the following chromosome found in the gametes. (?-A-B-C-d-a-?) 2-3' 2'-2 2-4 2'-1' 2-2'

2-4 Chromatid 2 undergoes genetic exchange with chromatid 4, which results in a dicentric chromosome with a duplication and a deletion.

Hemophilia A is an X-linked recessive disorder characterized by lack of a functional Factor VIII, a protein involved in blood clotting. In one population, the frequency of the normal allele, X 1, is 0.98 and the frequency of the mutant allele that causes hemophilia A, X 2, is 0.02. What percentage of males in this population have hemophilia A?

2.00 % The problem states that the frequency of X 1 (the normal allele) is 0.98. The frequency of X 2 (the mutant allele that causes hemophilia A) is 0.02.Recall that human males have only one X chromosome. Thus, the probability that a male will receive any given allele is the frequency of that allele (0.02, in this case). This means that the percentage of males who have hemophilia is 2%.

Referring to the image, there appears to be a span of 13 "beads" present. Ignoring the DNA before the first bead and after the 13th bead, how much DNA is present in the photo?

2546 base pairs Since there are 13 nucleosome core particles connected by 12 linker regions: (13 x 146) + (12 x 54) = 2546 bp

Given the frequency you reported in Part E, how many individuals out of 280 million Americans would you expect to have this pattern? Round to the nearest integer. 1 3 30 100

3 The frequency of occurrence for the five alleles is 1 in 99,603,000. Therefore, the number of people in the population expected to have this pattern is 280,000,000 ×× 1/99,603,000 = 3.

Select the two possible sequences of the wild-type DNA template strand. Notation: A/G if the nucleotide could be either purine, T/C if it could be either pyrimidine, N if any nucleotide could occur at a site. 3'-N-C-T-T-G-T-C-T-C-G-C-A-T-A-T-G-T-A-A-G-A-5' 3'-C-G-N-A-T-G-T-A-T-A-C-G-C-T-C-T-G-T-T-C-N-5' 3'-T-C-T/C-T-A-C-A-T-A-T-G-C-G-A-G-A-C-A-A-G-N-5' 3'-G-C-N-T-A-C-A-T-A-T-G-C-G-A-G-A-C-A-A-G-N-5' 3'-A-G-A-A-T-G-T-A-T-A--C-G-C-T-C-T-G-T-T-C-N-5'

3'-T-C-T/C-T-A-C-A-T-A-T-G-C-G-A-G-A-C-A-A-G-N-5' AND 3'-G-C-N-T-A-C-A-T-A-T-G-C-G-A-G-A-C-A-A-G-N-5'

The length of linker DNA segments varies somewhat among organisms. The length of the linker DNA in sea urchin is 110 base pairs. Suppose it were possible to move the entire region of DNA (92,672 base pairs long) from Drosophila melanogaster into the genome of a sea urchin. Approximately how many nucleosomes would be required to organize this region into the 10nm fiber structure?

362 nucleosomes Each nucleosome repeat is composed of the DNA (146 bp) wrapped around the core nucleosome particle plus the linker DNA (110 bp in this case). Therefore each repeat accounts for 256 (146 + 110) base pairs of DNA. Since the region is 92,672 bp long, then approximately 362 nucleosomes (92,672/256) would be required to organize the region into the 10nm fiber structure

The figure (Figure 1) shows original chromosomes at the top and paired chromosomes undergoing crossing over at the bottom. You will be determining how the chromosomes connect after the crossover event has finished. Be sure to begin with the left end of each chromosome. Select the correct combination describing the left end and right end of the following chromosome found in the gametes. (?-e-b-c-D-E-?) 4'-2' 4'-1 4-3 4'-3 4-2'

4'-2' Chromatid 4 undergoes genetic exchange with chromatid 2, which results in an acentric chromosome with a duplication and a deletion.

Suppose that the transient rare guanine tautomer shifted back to the common guanine tautomer prior to a second round of replication. Which DNA sequence(s) would be present in the sister chromatids after this second round of replication? Select all that apply. 5' - U A A - 3' 3' - A U U - 5' 5' - C A A - 3' 3' - G T T - 5' 5' - T A A - 3' 3' - G T T - 5' 5' - T A A - 3' 3' - A T T - 5' 5' - C G G - 3' 3' - G C C - 5' none of these

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

If four different VNTR alleles have frequencies of 1 in 10, 1 in 20, 1 in 50, and 1 in 500 in a population of 250 million, how many people would have all four alleles? 5,000,000 1 50 5

50 The frequency of occurrence for the four alleles is 1 in (10 × 20 × 50 × 500) = 1 in 5,000,000, and the number of people in the population who have this pattern is 250,000,000/5,000,000 = 50.

Which of the following best describes the overall composition (by mass) of eukaryotic chromatin? -100% DNA, 0% protein -15% DNA, 85% protein -50% DNA, 50% protein -95% DNA, 5% protein

50% DNA, 50% protein

A large gene-containing region of DNA in the fruit fly Drosophila melanogaster‎ is 92,672 base pairs long. The length of linker DNA in Drosophila‎ is 35 base pairs. Approximately how many nucleosomes are required to organize this region of DNA into the 10nm fiber structure?

512 nucleosomes Each nucleosome repeat is composed of the DNA (146 bp) wrapped around the core nucleosome particle plus the linker DNA (35 bp in this case). Therefore each repeat accounts for 181 (146 + 35) base pairs of DNA. Since the region is 92,672 bp long, then approximately 512 nucleosomes (92,672/181) would be required to organize the region into the 10nm fiber structure.

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? -3 -6 -2 -4

6 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.

Applying Hardy-Weinberg equilibrium, how many genotypes are predicted for a gene that has three alleles? 3 4 6 9

6 If a gene has three alleles, then six genotypes are predicted by applying the Hardy-Weinberg equilibrium equation. If the alleles are represented by the variables p, q, and r, then the sum of the genotype frequencies is represented by p 2 + 2pq + q 2 + 2pr + r 2 + 2qr = 1. The distribution of alleles in the aforementioned genotypes is determined by the trinomial expansion (p + q + r)2.

Dr. Ara B. Dopsis has an idea he thinks will be a boon to agriculture. He wants to create the "pomato," a hybrid between a tomato (Lycopersicon esculentum) that has 12 chromosomes and a potato (Solanum tuberosum) that has 48 chromosomes. Dr. Dopsis is hoping that his new pomato will have tuber growth like a potato and the fruit production of a tomato. He joins a haploid gamete from each species to form a hybrid and then induces doubling of chromosome number. How many chromosomes will be in each tomato gamete? Enter your answer as a whole number.

6 chromosomes

Mating between a male donkey (2n=62)(2n=62) and a female horse (2n=64)(2n=64) produces sterile mules. Recently, however, a very rare event occurred−−a female mule gave birth to an offspring by mating with a horse. Determine how many chromosomes are in the mule karyotype. Express your answer as an integer.

63 chromosomes

How many chromosomes does the mule-horse offspring carry? -63 or 64 -62 -62 or 64 -62 or 63 -64 -63 -62, 63, and 64 are all possible.

63 or 64

In chickens, having feathers on legs is dominant to having featherless legs. In a population of 200 chickens, 128 have featherless legs. How many chickens are heterozygous for the "feathered leg" allele? Assume the population is in Hardy-Weinberg equilibrium. -32 -36 -64 -72 -Not enough information is provided.

64 This problem also requires you to use the square root method to solve for allele frequencies. (Note that this method can only be used when the population is in Hardy-Weinberg equilibrium and the two alleles in question exhibit a dominant and recessive relationship. Also, note that you need to be aware of what the question is asking -- is it asking for a frequency or for a total number of individuals? If it is asking for a total number of individuals that possess a given genotype or allele, be sure your answer is a whole number. So, to solve this problem, you first calculate the frequency of homozygous recessive individuals, f(A2A2) by dividing the total affected by the total population, or 128/200 = 0.64, which is equal to q 2. Then solve for q by taking the square root of 0.64; thus q = 0.8. Using the equation p + q = 1, solve for p, which equals 0.2. To solve for the frequency of heterozygotes, calculate 2pq = 0.32. Note that the problem doesn't end here. Since the question asks for the number of individuals (and not the frequency), you need to multiple the population number by the heterozygote frequency: 200 x 0.32 = 64 chickens.

Based on your analysis, what is the most likely molecular abnormality causing the disease allele? -A pre-mRNA spicing mutation causes a deletion of exon 1. -A pre-mRNA spicing mutation causes a deletion of exon 2. -A pre-mRNA spicing mutation causes a deletion of exon 3. -A pre-mRNA spicing mutation causes a deletion of exon 4.

A pre-mRNA spicing mutation causes a deletion of exon 3.

Describe the following: Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

A(n) upstream activator sequence is an enhancer-like D N A sequence that binds regulatory proteins that interact with promoter-bound proteins to activate transcription. A(n) insulator sequence shields genes from enhancer effects by promoting the formation of specific D N A loops that protect particular genes from enhancers. A(n) silencer sequence represses transcription by binding proteins that block enhancer mediated transcription. Enhanceosome is a protein complex that assemble at enhancers to facilitate transcription. RNARNA interference is the posttranscriptional regulation of m R N As by regulatory R N A molecules.

The amino acid changes noted in Parts A-C are the result of changes in DNA sequence.For each of these amino acid changes, evaluate whether a single base change in the given codons could account for the change in the amino acid. For which codon(s) could a single base change account for this amino acid change? Lysine to Asparagine Select all that apply. AAA AAG UUU none of these codons

AAA AND AAG There are two codons for lysine (AAA, AAG) and two codons for asparagine (AAC, AAU). Since positions #1 and #2 are identical in the two sets of codons, then a single base change in the #3 position of the lysine codons can result in either asparagine codon (AAA/G -> AAC/U).

The MutS protein of Escherichia coli, shown in a complex with a DNA substrate containing a G-T base mismatch, is a key component of the macromolecular assemblage responsible for repairing certain types of DNA mutations. MutS is responsible for recognizing and binding to base pair mismatches, and recruits other key proteins required for repair, MutH and MutL, to the mismatch site. Launch the molecular model of the MutS DNA mismatch repair protein to explore its structure. Then answer the questions. MutS is a large protein. Two MutS monomers combine to form a functional MutS homodimer. Each monomer is composed of six distinct domains.This image shows a single MutS monomer. Identify the domains labeled 1-4. -ATPase (1), connector (2), core (3), mismatch-binding (4) -ATPase (1), core (2), connector (3), mismatch-binding (4) -Clamp (1), ATPase (2), core (3), mismatch-binding (4)

ATPase (1), core (2), connector (3), mismatch-binding (4) The four domains labeled in the image are: The ATPase domain (purple) which binds and hydrolyzes ATP The core domain (red) which has two regions that form a helical bundle, with two additional alpha helices extending as levers toward the DNA The connector domain (light blue) which connects the mismatch domain to the core domain The mismatch-binding domain (dark blue) binds to DNA containing the mismatched bases Now that you have identified these four domains, you may want to go back to the tutorial and review the remaining two domains before proceeding.

Researchers collecting data on an island population of rabbits have found that the mutation rate of the recessive allele, d, is 8 x10-5. Also, the relative fitness of the phenotype encoded by dd, has been found to be 0.80.If the researchers let this rabbit population come to equilibrium, determine the allele frequency of the recessive allele. Enter your answer as a decimal expressed to two decimal places. (Example: 0.08) Allele frequency of d at equilibrium =

Allele frequency of d at equilibrium = 2.00×10^−2 In cases such as this one, where selection is acting against the recessive phenotype, the frequency of mutant alleles in the population is a balance of natural selection versus the frequency of mutation.To calculate the frequency of the recessive allele at equilibrium (q E), you first need to calculate the selection coefficient (s), which = 1 - w or 1 - 0.8 = 0.2.Then, you can calculate the frequency of the recessive allele using this formula:q E = μs−−√μswhere μ equals the rate of mutation.q E = 0.00008/0.2−−−−−−−−−√0.00008/0.2 = 0.0004−−−−−√0.0004 = 0.02

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

An inversion heterozygote for a large inversion A large heterozygous inversion will result in greater numbers of genes that are duplicated or deleted when crossing over occurs.

In what way does position effect variegation (PEV) of Drosophila eye color indicate that chromatin state can affect gene transcription? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all. -euchromatic -not all cells -all cells -centromeric -red -white

At its normal location, all cells of the eye expresses the w + allele, leading to a uniformly red pigmented eye. At a more centromeric location, not all cells of the eye express the w + gene, because it is packaged in heterochromatin. Since some cells are red, it is clear that the w + allele is functional, so it is not the mutational inactivation of the white gene that explains the white cells.

The active site of the Argonaute protein is structurally similar to the active sites of proteins in the RNase H family. A conserved motif of three amino acids coordinates metal cation cofactors (usually magnesium ions) in the enzyme's active site to trigger cleavage of the target mRNA. What are these three conserved residues? -Aspartate628, Lysine558, and Arginine745 -Aspartate628, Aspartate558, and Histidine745 -Aspartate628, Arginine558, and Histidine745 -Arginine628, Arginine558, and Histidine745

Aspartate628, Aspartate558, and Histidine745 The Piwi domain displays RNase H family structure. RNase H family enzymes typically carry a catalytic "DDE" motif aspartate (D), aspartate (D), glutamate (E) in their active sites. These three highly conserved acidic residues are required for catalysis. The Argonaute Piwi domain has a "DDH" motif (aspartate, aspartate, histidine) which is analogous to DDE. The key active site residues are Aspartate628, Aspartate558 and Histidine745. The close proximity of these residues to the target mRNA cleavage site supports their role in RNA cleavage (slicing). These residues coordinate metal ions (usually magnesium) required for the cleavage reaction (hydrolysis).

In a sample population of chickens, genotype BB has a relative fitness of 1.0. Genotype Bb has a selection coefficient of 0.3, and genotype bb has a selection coefficient of 0.4.What is the relative fitness of genotype Bb? What is the relative fitness of genotype bb? Enter your answer as two decimals (each expressed to one decimal place) separated by a comma. (Example: 0.3, 0.2) Bb, bb=

Bb, bb= 0.700,0.600 Both the Bb and bb genotypes are less fit than the BB genotype, and each has a different selection coefficient. To determine relative fitness, you need to subtract the selection coefficient from 1 for each genotype. For genotype Bb, s = 0.3. The relative fitness (w) of Bb = 1 - s or 1 - 0.3 = 0.7. For genotype bb, s = 0.4. The relative fitness (w) of bb = 1 - s or 1 - 0.4 = 0.6.

Why? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

Because the hexaploid line contributes 3 of each of its 8 different homologous chromosomes to the experimental varieties. These chromosomes will form abnormal structures during synapsis in meiosis I, thus preventing the formation of balanced gametes.

In a different population, suppose that 38.40% of people are blood type A, and 46.24% are blood type O. Assuming this population is in Hardy-Weinberg equilibrium, what percentage of the population is blood type B? What percentage is blood type AB? Enter your answers to two decimal places. Blood Type B= Blood Type AB=

Blood Type B= 11.52 % Blood Type AB= 3.84 % The key to solving the problem above is to calculate allele frequencies for IA , IB , and i, and to then use that information to calculate the percentage of people with the B phenotype (genotypes IBIB and IBi) and the percentage of people with the AB phenotype (genotype IAIB ).The question tells you that 46.24% of the population is blood type O. This is equal to r 2. The question also tells you that 38.40% of the population is blood type A. This is equal to p 2 + 2pr. Using this information, you can solve for both p and r, as shown below. Solve for r by taking the square root of r 2. The square root of 0.4624 is 0.68. Now that you know r = 0.68, you can solve for p using the equation p 2 + 2pr = 0.3840. Thus, p 2 + 1.36p = 0.3840. Rearrange the equation to be p 2 + 1.36p - 0.3840 = 0. Now you have a quadratic equation that you can solve. You may recall from high school math class that there is more than one way to solve a quadratic equation. One way is to use the quadratic formula: x=−b±b2−4ac−−−−−−√2ax=−b±b2−4ac2a Recall that the formula above is used to solve ax 2 + bx + c = 0. Thus, plugging in the information from your quadratic equation, p 2 + 1.36p - 0.3840 = 0, gives you: p=−1.36±1.362−4(1)(−0.3840)−−−−−−−−−−−−−−−−−√2(1)p=−1.36±1.362−4(1)(−0.3840)2(1) p = 0.24 Now that you have identified p (0.24) and r (0.68), you can solve for the frequency of the IB allele (q) using the equation p + q + r = 1.

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

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

Now consider a single base mutation in a codon for leucine that creates a codon for phenylalanine. A true reversion changes the phenylalanine codon back to a codon for leucine. Which of the following leucine codon(s) could be mutated once to form a phenylalanine codon, and then mutated at a second site to restore a leucine codon? Select all that apply. (Note that two different positions in the codon must be mutated.) CUG CUC CUA UUA UUG CUU none of these codons

CUC AND UUA AND UUG AND CUU There are six codons for leucine (UAA, UUG, CUA, CUC, CUG, CUU) and two codons for phenylalanine (UUC, UUU). Leucine codons (CUA, CUG) do not share two of the codon positions with phenylalanine. CUA vs. UUC or UUU CUG vs. UUC or UUU The remaining four leucine codons share two of the codon positions with phenylalanine, therefore they are capable of being mutated to a phenylalanine codon. A subsequent mutation at a second site restores a leucine codon in each case. For example, if the original leucine codon was UUA, then an A to C transversion mutation at position #3 would create a phenylalanine codon (UUC). A subsequent U to C transition mutation at position #1 would create a true reversion back to a leucine codon.

For which codon(s) could a single base change account for this amino acid change? Leucine to Glutamine Select all that apply. CUC CUG UUA CUA UUG CUU none of these codons

CUG AND CUA There are six codons for leucine (UUA, UUG, CUA, CUG, CUC, CUU) and two codons for glutamine (CAA, CAG). Only two leucine codons share two of the codon positions with glutamine. CUA shares two positions with CAA. CUG shares two positions with CAG. For both leucine codons, a base change from U to A will change them to glutamine codons.

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

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

Which enzyme is responsible for proofreading during replication? -DNA polymerase -DNA glycosylase -RNA polymerase -DNA ligase

DNA polymerase DNA polymerase performs proofreading functions during replication using its 3′ to 5′ exonuclease capability.

If a DNA replication error is detected by DNA polymerase, how is it corrected? -This is mostly not corrected, since the mutation is not significant. -DNA polymerase removes the last nucleotide added using its 5'-to-3' exonuclease activity and adds the correct nucleotide in its place. -DNA polymerase stops replication in the place of mismatch, the replication starts once again at the beginning. -DNA polymerase removes the last nucleotide added using its 3'-to-5' exonuclease activity and adds the correct nucleotide in its place.

DNA polymerase removes the last nucleotide added using its 3'-to-5' exonuclease activity and adds the correct nucleotide in its place.

For 5BU to cause a transition mutation, which of the following must occur? -DNA with incorporated 5BU must replicate. -DNA with incorporated 5BU must not replicate. -It must undergo at least two form changes. -It must be incorporated into DNA in its rare form.

DNA with incorporated 5BU must replicate. 5BU must undergo a form change, but that is not sufficient to cause transition. The form change must be followed by replication.

In a population of birds in Africa, it was observed that birds with small or large beaks could efficiently crack and eat small or large seeds, respectively. Birds with intermediate beaks had trouble with both types of seeds. What type of selection would be expected to occur in this population if small and large seeds were the only types of food available to these birds? -Directional -There would be no selection in this population. -Disruptive -Stabilizing

Disruptive Birds with an intermediate beak phenotype are at a disadvantage in this population and will be selected against because they are less fit.

RecB and RecC are structurally similar; both contain three domains (1A and 1B, 2A and 2B and 3). However, the function of these domains differs between the two proteins. Domains 1 and 2 in both RecB and RecC are structurally homologous to SF1 helicases, but only RecB still shows helicase function. Domains 1 and 2 of RecC evolved a new function, which is to form a channel for directing the 3' DNA strand down to the RecB nuclease. Domain 3 of RecB is a nuclease that digests both 3' and 5' ends. This domain in RecC has the novel function of splaying the DNA duplex into single strands. Given this information, identify the specific domains of RecC. A labeled image of RecB is provided for comparison. -Subdomain 2A (1), subdomain 1B (2), domain 3 (3), subdomain 2B (4), subdomain 1A (5) -Domain 3 (1), subdomain 2B (2), subdomain 1B (3), subdomain 2A (4), subdomain 1A (5) -Domain 3 (1), subdomain 2B (2), subdomain 2A (3), subdomain 1B (4), subdomain 1A (5)

Domain 3 (1), subdomain 2B (2), subdomain 1B (3), subdomain 2A (4), subdomain 1A (5) A comparison of the RecB and RecC models reveals that one subunit is inverted relative to the other. In RecC, domain 3 splits the duplex DNA into 3' and 5' strands; this means it is located at the top of the image near the DNA molecule (1). The homologous domain in RecB is the active nuclease (domain 3), which is found at the opposite end where the split DNA strands will be digested. Although domains 1 and 2 of RecB and RecC are homologous, only RecB still exhibits helicase function. Domains 1 and 2 of RecC evolved a new function, which is to form a channel for directing the 3' DNA strand down to the RecB nuclease (not shown, but located at the bottom of the image).

There are several possible mutations in the trp operon: trpP - is a mutation in the promoter sequence that prevents RNA polymerase from binding to the promoter and initiate transcription of the trp operon genes. trpOc is a mutation in the operator sequence that prevents the trp repressor protein from binding to the operator to block transcription of the trp operon genes. trpR- is a mutation in the repressor protein that either prevents repressor protein from being made or produces a mutant repressor that cannot bind to the operator region. trpRs is a mutation in the repressor protein that causes the repressor to be more sensitive to tryptophan levels. This mutant repressor binds the operator region even when tryptophan is low. Determine whether each mutation is a trans-acting factor or a cis-acting factor, and whether it is dominant or recessive to wild type. Drag each mutation into the correct bin.

Dominant/Trans trpRs Recessive/Trans trpR- Dominant/Cis trpOc Recessive/Cis trpP- The repressor protein is a trans-acting factor because it can diffuse around the cell and interact with other strands of DNA. Mutations in the repressor gene that prevent the repressor from binding to the operator (trpR-) are recessive to wild-type repressor. Partial diploid cells that are trpR+/trpR- are regulated normally (repressible) because the wild-type repressor is capable of binding to operator regions on both DNA sequences. Mutations in the repressor gene that cause the repressor to bind to the operator even when the cell is starved for tryptophan (trpRs ; super-repressor) are dominant to wild-type repressor (trpR+ ). Partial diploid cells that are trpR+/trpRs show little or no expression because even though wild-type repressor only binds to DNA in the presence of tryptophan, the mutant repressor binds to DNA under all conditions. DNA sequences such as the promoter and operator regions are cis-acting because they can only regulate the genes on the same strand of DNA. Mutations in the operator region that prevent repressor binding (trpOc ) are dominant to wild- type operator. Partial diploids with the genotype trpO +/ trpO c are constitutively expressed. The wild-type operator regions is only able to regulate genes on that piece of DNA; trp operon genes with the trpOc mutation continue to be constitutively expressed. Mutations in the promoter region that prevent RNA polymerase from binding (trpP -) are recessive to wild-type promoter sequences. Partial diploids that are trpP+/ trpP- show normal regulation (repressible).

Explain how the inducibility of a gene−−for instance in response to an environmental cue−−could be mediated by an activator. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

Environmental induction of transcription through a transcriptional activator could be accomplished by controlling activator binding to chromatin. For example, an environmental signal could be required to allow the activator into the nucleus, as is the case for yeast Pho4. The activator binds to cis-regulatory sites and recruits additional factors, including histone acetyl transferases and chromatin remodeling complexes, which convert closed chromatin to open chromatin. Such chromatin is then accessible to transcription factors, which promote RNA polymerase binding and transcription initiation and elongation.

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. -Even though eye color phenotypic variegation exists within the eye, all cells have the same genotype. -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. -When heterochromatin spreading encompasses the new location of w + allele, the gene is transcribed, producing white eye patches. -When heterochromatin spreading encompasses the new location of w + allele, the gene is not transcribed, producing white eye patches. -When heterochromatin spreading does not reach the new location of the w + allele, the gene will be transcribed, producing red eye patches.

Even though eye color phenotypic variegation exists within the eye, all cells have the same genotype. AND When heterochromatin spreading encompasses the new location of w + allele, the gene is not transcribed, producing white eye patches. AND When heterochromatin spreading does not reach the new location of the w + allele, the gene will be transcribed, producing red eye patches. 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.

Given the information available from deletion analysis, can you give a molecular explanation for the observation that ME1ME1 expression appears to turn on and turn off at various times during normal mouse development? -Expression of ME1ME1 is regulated by a promoter and an inhibitor, which can be regulated such that ME1ME1 expression is activated at some developmental stages and then repressed at others. -Expression of ME1ME1 is regulated by an enhancer and a silencer, which can be regulated such that ME1ME1 expression is activated at some developmental stages and then repressed at others. -Expression of ME1ME1 is a non-regulated process which is turned on and turned off randomly. -None of the above.

Expression of ME1ME1 is regulated by an enhancer and a silencer, which can be regulated such that ME1ME1 expression is activated at some developmental stages and then repressed at others.

Now predict the outcome of a partial diploid of your novel dominant negative mutation (trpR d-) and a trpR s mutation. trpRd- trpO+ / trpRs trpO+ How would trp genes be regulated in this partial diploid? -Expression of trp genes will be constitutive. trpRd- is recessive to trpRs . -There will be no expression of trp genes. trpRd- is recessive to trpRs . -There will be no expression of trp genes. trpRd- is dominant to trpRs . -Expression of trp genes will be constitutive. trpRd- is dominant to trpRs .

Expression of trp genes will be constitutive. trpRd- is dominant to trpRs . The repressor requires a trpR dimer to bind to the operator region in order to prevent transcription. The trpRd- mutation produces a repressor protein that cannot bind to the operator region, whereas the trpR s mutation produces a repressor protein that can bind the operator region when tryptophan is low. trpRd- /trpRd- dimers would be unable to bind to the operator. trpRd- /trpR s would also be unable to bind to the operatore due to the trpRd- subunit. Only trpR s /trpR s dimers would be able to bind to the operator and repress transcription. Transcription will occur when tryptophan is present because trpRd- /trpRd- and trpRd- /trpR s dimers will not be able to bind to the operator region. Unlike with wild-type strains, transcription will occur when tryptophan is present-- it will be constitutive. This means that trpRd- is dominant to trpR s .This reasoning is similar to the explanation for why trpRd- is dominant to trpR+ . Binding to trpRd- causes repressor proteins of other genotypes to be unable to bind the operator region.

You make a partial diploid of this new strain by introducing a wild type F' plasmid. trpRnovel trpO+ / trpR+ trpO + Given what you know about the "novel" mutation, how will expression of the trp operon be regulated in this partial diploid? -Expression of trp genes will be repressible. trpRnovel is dominant to trpR+ . -Expression of trp genes will be constitutive. trpRnovel is dominant to trpR+ . -Expression of trp genes will be repressible. trpRnovel is recessive to trpR+ . -Expression of trp genes will be constitutive. trpRnovel is recessive to trpR+ .

Expression of trp genes will be constitutive. trpRnovel is dominant to trpR+ . To bind to the operator region, the repressor must consist of a complex composed of two subunits of repressor polypeptide in the form of a dimer. The novel mutant repressor can form dimers either with other mutant repressor proteins or with wild type repressor proteins. Several configurations are possible - mutant/mutant dimer, mutant/wild-type dimer, and wild-type/wild-type dimer. Any complex that contains a mutant repressor protein will be unable to bind the oeprator and prevent transcription.This means that the trp genes will be transcribed even in the presence of tryptophan in this partial diploid strain; the "novel" mutation is dominant to wild-type repressor (trpR+ ). This type of mutation is called a dominant negative, often indicated by a "d-" superscript (trpR d-)

Introducing the F' plasmid into the cells with Mutation 5 does not appear to have any effect on the mutant phenotype (it exhibits constitutive expression). The F' plasmid appears to "rescue" Mutation 6 by returning that cell line to wild-type expression. There are two possible mutations that could lead to constitutive expression -- a loss of function mutation of the operator (Oc) or the repressor protein(lacI-). To determine the effect of introducing a wild-type F' plasmid into a bacterial cell with each of these mutations, you draw out the two possible scenarios. For each of the possible partial diploids, determine whether expression of the lac operon will be inducible (wild type) or constitutive (mutant). Then use this information to identify the mutations In Mutation 5 and Mutation 6. Select the two correct answers. F' I+ / I- is inducible. Mutation 6 is a lacI- mutation. F' I+ / I- is constitutive. Mutation 5 is a lacI- mutation F' O+ / Oc is inducible. Mutation 6 is a lacOc mutation. F' O+ / Oc is constitutive. Mutation 5 is a lacOc mutation.

F' I+ / I- is inducible. Mutation 6 is a lacI- mutation. AND F' O+ / Oc is constitutive. Mutation 5 is a lacOc mutation. Repressor proteins are trans-acting; proteins can diffuse around the cell and regulate genes on any piece of DNA. This means that the functional repressor made from the F' plasmid will be able to regulate both the lac genes on the plasmid and the lac genes on the bacterial chromosome. As a result, expression of the lac operon in the F' I+ / I- partial diploid will be inducible. Operator regions are cis-acting; they only regulate the genes on their DNA strand. If a bacterial strain has a mutation in the operator region, expression will be constitutive because the repressor cannot bind. The wild-type operator introduced on the F' plasmid only regulates genes on the plasmid, not genes on the bacterial chromosome. The lac genes on the F' plasmid will be inducible, but the bacterial chromosome will still be expressed constitutively. As a result, the F' O+ / Oc partial diploid strain will show constitutive expression of the operon.

In order to create the possibility of generating a trisomy, nondisjunction must occur during meiosis II. True False

False Nondisjunction during either meiosis I or meiosis II creates gametes that will generate trisomies if fertilized.

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

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

If the frequency of the M allele in the human MN blood group system is 0.65 in a population at equilibrium, then the frequency of the N allele must be 0.04. True False

False The sum of the allele frequencies must equal 1, so the frequency of the N allele must be 0.35.

There are an unlimited number of VNTR alleles for each locus, such that every individual has unique alleles at each VNTR locus. True False

False There are many VNTR alleles for each locus but not an unlimited number. People who are closely related are more likely to have similar DNA fingerprints.

Expression of the trp operon is regulated by the level of free tryptophan in the cell both through repressor action and attenuation. This chart shows the percent expression of the trp operon in the presence and absence of tryptophan for wild-type (trpR+ or trp+) and repressor mutants (trpR-).Use this information to determine the levels of expression for the following genotypes and conditions. Rank the genotypes and conditions from highest to lowest level of trp operon expression.

Highest trp+ / tryptophan absent trpL- / tryptophan present trpR- / tryptophan present trp+ / tryptophan present trpP- / tryptophan absent Lowest The trp operon is a negative repressible system; the presence of tryptophan prevents transcription of the trp operon. Tryptophan acts as a corepressor by binding to and activating the trp repressor protein (trpR). Tryptophan also plays a role in promoting attenuation, which leads to the termination of transcription, through the structure of the leader sequence (trpL). The highest level of trp operon expression (100%) would be in a wild-type strain (trp+) in the absence of tryptophan. Without tryptophan, the repressor cannot bind to the operator to stop transcription and the folding of the mRNA favors anti-termination, increasing the likelihood of transcribing the trp genes. To determine the effect of trpR- versus trpL- on expression in the absence of tryptophan, first consider the information in the chart. Full expression of the operon in the absence of tryptophan in a wild-type strain is 100%; when tryptophan is present, expression drops to a basal level of 8%. The chart indicates that when the repressor is mutated (trpR-), expression jumps from 8% to 33%. If the only two repression mechanisms are the repressor protein and attentuation, this suggests that the 67% of expression that is still inhibited must be mediated by attentuation. By this logic, strains mutant in trpL should have higher expression than strains mutant in trpR. A wild-type genotype would express a very low level of trp genes in the presence of tryptophan. Even though tryptophan would promote binding of the repressor to the operator and increase the likelihood of attenuation, neither process is permanent/irreversible. The repressor protein binds and unbinds, and as a result, a low level of trp genes will be transcribed (8%). The lowest level of trp expression (0%) would be in the genotype with a promoter mutation (trpP-). Without a wild-type promoter sequence (trpP), RNA polymerase will be unable to bind and transcribe the genes .

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

Homologs fail to separate during meiosis I. Ordinarily, homologs separate during meiosis I. Failure of this separation is one of the ways in which nondisjunction can occur.

What is the primary function of β-galactosidase? -Hydrolysis of the disaccharide lactose to produce glucose and galactose -Acetylation of lactose -Transport of lactose into the cell -Hydrolysis of galactose to produce glucose

Hydrolysis of the disaccharide lactose to produce glucose and galactose

Which of the following statements regarding familial Down syndrome is false? -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. -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. -Heterozygous carriers of the Robertsonian translocation that cause familial Down syndrome can have children who are also carriers of the translocation. -It accounts for approximately 5% of all Down syndrome cases.

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. 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 is a common consequence of inbreeding in a population? See Section 20.6 (Page 742) . -Increased reproductive fitness. -Increased mutation rate. -Increased frequency of homozygous genotypes. -Increased frequency of heterozygous genotypes.

Increased frequency of homozygous genotypes. The principal genetic consequences of inbreeding are an increase in the frequency of homozygous genotypes in a population and a decrease in the frequency of heterozygous genotypes relative to the frequencies expected from random matings.

The pattern of binding of repressor and Cro to these sites determines whether the phage grows lytically or lysogenically. Which of the following accurately describes the regulation of lytic and lysogenic cycles by bacteriophage repressor? -In the lysogenic cycle, repressor protein is produced, preventing the expression of Cro and genes for viral replication. In the lytic cycle, Cro protein is produced, preventing the expression of repressor protein; genes for viral replication are expressed. -In the lysogenic cycle, Cro protein is produced, preventing the expression of the repressor protein; genes for viral replication are expressed. In the lytic cycle, repressor protein is produced, preventing the expression of Cro; genes for viral replication are not expressed. -In the lysogenic cycle, there is no expression of either repressor or Cro protein; genes for viral replication are not expressed. In the lytic cycle, both repressor and Cro protein are produced; genes for viral replication are expressed. -In the lysogenic cycle, both repressor and Cro proteins are produced; genes for viral replication are expressed. In the lytic cycle, there is no expression of either repressor or Cro protein; genes for viral replication are not expressed.

In the lysogenic cycle, repressor protein is produced, preventing the expression of Cro and genes for viral replication. In the lytic cycle, Cro protein is produced, preventing the expression of repressor protein; genes for viral replication are expressed.

The researcher then crosses the F1 males (XP{miniwhite+}/ Y; transposase / +) to female flies that are homozygous for the white mutation. She selects F2 males that have eye pigment present (indicating that they have a P {miniwhite +} insertion somewhere in their genome). Why does the researcher select these F2 males? Select all statements that are true. -In this cross, most F2 males will have eye pigment because they inherit the XP{miniwhite+} chromosome with a P-element that did not move. -In this cross, most F2 males will have no eye pigment because they cannot inherit the X chromosome that had the original P {miniwhite +} insertion on it. -In this cross, F2 males with eye pigment result from the P-element moving to a new location on an autosome. -In this cross, F2 males with eye pigment result from the P-element remaining at its original location on the X chromosome.

In this cross, most F2 males will have no eye pigment because they cannot inherit the X chromosome that had the original P {miniwhite +} insertion on it. AND In this cross, F2 males with eye pigment result from the P-element moving to a new location on an autosome. By selecting F2 males with eye pigment, the researcher is selecting against flies that carry the original XP{miniwhite+}chromosome. Because F1 males cannot transmit an X chromosome to their male offspring, any F2 males with eye pigment must result from a P-element transposition event from the original location on the X chromosome to a new location on one of the autosomal chromosomes.

What is inbreeding depression? -Inbreeding depression refers to the increased fitness or viability of a species caused by inbreeding, which drives genetic loci to homozygosity. -Inbreeding depression refers to the increased fitness or viability of a species caused by inbreeding, which drives genetic loci to heterozygosity. -Inbreeding depression refers to the reduced fitness or viability of a species caused by inbreeding, which drives genetic loci to heterozygosity. -Inbreeding depression refers to the reduced fitness or viability of a species caused by inbreeding, which drives genetic loci to homozygosity.

Inbreeding depression refers to the reduced fitness or viability of a species caused by inbreeding, which drives genetic loci to homozygosity.

To determine the identity of each strain, you add a F' plasmid containing a wild-type version of the lac operon to each mutant strain, F' I+ cap+ P+ O+ Z+ Y+ . (Note: "cap+ " represents the CAP-cAMP binding site in the lac operon. The gene that produces the catabolite activator protein (CAP) is not part of the lac operon region and thus not on the F' plasmid.) F' I+ cap+ P+ O+ Z+ You measure beta-galactosidase levels for these partial diploid strains and obtain these results. You decide to focus first on Mutation 1 and Mutation 2. The F' plasmid appears to "rescue" mutant strain 1 by restoring wild-type inducible expression. The F' plasmid does not change the phenotype of mutant strain 2; there is still no expression of the lac operon. There are three possible mutations that could lead to lack of expression: a super-repressor (Is), a promoter mutation (P-) or a mutation in the lacZ gene (Z- ). To determine the effect of introducing a wild-type F' plasmid into a bacterial cell with each of these mutations, you draw out the three possible scenarios. Determine the expression of the lac operon for each possible partial diploid and for the "unknown" mutation partial diploids (Mutation 1 and Mutation 2).

Inducible (wild type): F' lac+ / P- F' lac+ / Z- F' lac+ / Mutation 1 No expression (mutant): F' lac+/ Is F' lac+ / Mutation 2 The super-repressor protein (Is) is trans-acting; it can diffuse around the cell and bind to any lac operator regions. Introduction of a wild type lacI gene (I+) will add normal repressor protein to the bacterial cell with the super-repressor. The normal repressor protein will bind when lactose is absent, and will not bind when lactose is present. The super-repressor will bind under both conditions. As a result, in this partial diploid, there would be no expression of the lac operon. The lacZ gene produces beta-galactosidase. Introduction of a wild-type lacZ gene (Z+) into a bacterial strain with a mutation in lacZ (Z-) will restore expression of the lac operon. The lacZ gene on the bacterial chromosome will not produce any beta-galactosidase, but the F' plasmid contains a wild-type copy of the lac operon. As a result, in this partial diploid, expression of the lac operon would be inducible (wild type). The lacP region is the promoter sequence for the lac operon. Introduction of a wild-type lac operon with a functional promoter into a bacterial strain with a mutation in lacP (P-) will restore expression of the lac operon. The promoter on the bacterial chromosome will not be functional, but the F' plasmid contains a wild-type copy of the lac operon. As a result, in this partial diploid, expression of the lac operon would be inducible (wild type).

In which population do you expect to see the greatest allele frequency change? Living in a mainland forest. Inhibiting the forest on an island.

Inhibiting the forest on an island.

Which of the following transposition events is most likely to result in a loss of function mutation? -Insertion of an IS element within the coding region of a gene -Transposition of an IS element that contains a stop codon within the inverted repeat sequence -Insertion near the promoter region of a gene

Insertion of an IS element within the coding region of a gene Insertions of DNA elements within coding regions have the greatest potential for altering amino acid sequence and/or causing truncation of a polypeptide due to frameshift.

What is the role of glucose in catabolite repression? -It stimulates transcription from the lac operon, causing an increase in cAMP levels in the cell. -It decreases the levels of cAMP in the cell, repressing transcription from the lac operon. -It increases the levels of cAMP in the cell, stimulating transcription from the lac operon. -It represses transcription from the lac operon, causing a decrease in cAMP levels in the cell.

It decreases the levels of cAMP in the cell, repressing transcription from the lac operon. Glucose decreases the levels of cAMP in the cell, preventing formation of the CAP-cAMP complexes necessary for the stimulation of transcription from the lac operon.

Which of the following best describes the biological role of the lac operon? -It ensures that a cell dedicates resources to the production of enzymes involved in lactose metabolism only when lactose is available in the environment. -It ensures that bacterial cells produce lactose only when no other food sources are available. -It prevents other sugars from being metabolized until all available lactose has been used. -It ensures that a cell produces enzymes involved in lactose metabolism in a constitutive manner.

It ensures that a cell dedicates resources to the production of enzymes involved in lactose metabolism only when lactose is available in the environment. The cell expends energy to produce the proteins necessary for lactose metabolism only when lactose is present.

Why is aneuploidy in animals generally detrimental? -It results in an imbalance of gene products from affected chromosomes, which alters normal development. -It results in DNA replication errors due to the aberrant number of chromosomes. -Synthesis of extra chromosomes creates a metabolic burden that decreases cell viability. -It results in chromosome segregation errors in mitosis due to aberrant chromosome numbers.

It results in an imbalance of gene products from affected chromosomes, which alters normal development.

Members of the Argonaute protein family play key roles in RNA interference pathways (RNAi). RNAi lowers gene expression in response to double-stranded RNA (dsRNA), whether endogenous (e.g. pre-miRNA) or exogenous (e.g. viral genomes). The RNAi process is initiated when Dicer, an RNase III family enzyme, cleaves dsRNA into short duplexes (~21 base pairs), each strand possessing a single-stranded, two-nucleotide overhang on its 3' end and a phosphate on its 5' terminus. These dsRNAs (siRNAs or miRNAs) are then loaded onto Argonaute (Ago). Once engaged by Ago, one of the siRNA or miRNA strands, the passenger strand, is cut and discarded. The remaining single-stranded RNA (ssRNA) is termed the guide strand. Ago acts as the scaffold for the recognition of complementary target sequences by the guide strand. Additionally, as part of the RNA-induced silencing complex (RISC), Ago can catalyze the cleavage (slicing) of target mRNAs. Launch the molecular model of Argonaute to explore its structure. Then, answer the questions. Identify the domains of Argonaute labeled 1-3. -Middle domain (1), Piwi domain (2), N-terminal domain (3) -Piwi domain (1), Middle domain (2), PAZ domain (3) -PAZ domain (1), Middle domain (2), N-terminal domain (3) -PAZ domain (1), Piwi domain (2), N-terminal domain (3)

Middle domain (1), Piwi domain (2), N-terminal domain (3) The Argonaute protein contains four domains: Middle, Piwi, N-terminal, and PAZ domains. The N-terminal domain includes a "stalk" that helps support the Paz domain. The Middle, Piwi, and N-terminal domains together form the "crescent" of the Argonaute protein. The yellow structure is an inter-domain connector that clasps all four domains.

Which of the following processes tends to increase genetic variation in a population? See Section 20.3 (Page 737) . -Genetic bottlenecks. -Natural selection. -Nonrandom mating. -Mutation.

Mutation. Mutation creates new alleles, which increases the genetic variation in a population.

Why does deletion E lower expression of UG4UG4 in leaf tissue but not in stem tissue? -Mutations in regions close to the transcription start site that prevent expression typically affect a promoter; therefore, mutation E deletes a required promoter sequence for UG4UG4 transcription in leaf tissue but not in stem tissue. -Mutations in regions at a distance from the transcription start site that prevent expression typically affect an enhancer; therefore, mutation E deletes a required enhancer sequence for UG4UG4 transcription in leaf tissue but not in stem tissue. -Mutations in regions close to the transcription start site that prevent expression typically affect the start codon; therefore, mutation E deletes a required start codon sequence for UG4UG4 transcription in leaf tissue but not in stem tissue. -Mutations in regions at a distance from the transcription start site that prevent expression typically affect a silencer; therefore, mutation E deletes a required silencer sequence for UG4UG4 transcription in leaf tissue but not in stem tissue.

Mutations in regions at a distance from the transcription start site that prevent expression typically affect an enhancer; therefore, mutation E deletes a required enhancer sequence for UG4UG4 transcription in leaf tissue but not in stem tissue.

Why does deletion D raise UG4UG4 expression in leaf tissue but not in stem tissue? -Mutations that increase expression typically remove methylated sites; therefore, mutant D lacks a methylated site that inhibits the transcription of UG4UG4 in stem tissue but not in leaf tissue. -Mutations that increase expression typically remove transcription termination points; therefore, mutant D lacks a transcription termination point that terminates the transcription of UG4UG4 in leaf tissue but not in stem tissue. -Mutations that increase expression typically remove silencers; therefore, mutant D lacks a silencer sequence that regulates the level of transcription of UG4UG4 in leaf tissue but not in stem tissue. -Mutations that increase expression typically remove parts of promoters; therefore, mutant D lacks a part of promoter sequence that regulates the level of transcription of UG4UG4 in leaf tissue but not in stem tissue.

Mutations that increase expression typically remove silencers; therefore, mutant D lacks a silencer sequence that regulates the level of transcription of UG4UG4 in leaf tissue but not in stem tissue.

The human genome contains 3.4×1093.4×109 base pairs. The length of linker DNADNA in mammals is 50 base pairs. Approximately how many nucleosomes are required to organize the 10-nm-fiber structure of the genome? Enter your answer using two significant figures (example: 1.8×1041.8×104).

N = 1.7×107 nucleosomes

Two experimental varieties of rice are produced by crossing a hexaploid line that contains 72 chromosomes and a tetraploid line that contains 48 chromosomes. Experimental variety 1 contains 60 chromosomes, and experimental variety 2 contains 84 chromosomes. Do you expect both experimental lines to be fertile? -Yes, both experimental line will be fertile. -No, only tetraploid line will be fertile. -No, only hexaploid line will be fertile. -Neither line is expected to be fertile.

Neither line is expected to be fertile.

Imagine you are carrying out research on the lac operon. You isolate six mutations in the lac operon by measuring the amount of beta-galactosidase made in mutant cell line under three different conditions: no lactose/no glucose; lactose only; and lactose/glucose. Your results are shown in the table. You notice that two mutations result in no expression of the lac operon (Mutations 1 and Mutation 2), two mutations result in low expression of the lac operon even in the presence of lactose and the absence of glucose (Mutation 3 and Mutation 4), and two mutations result in constitutive expression of the lac operon (Mutation 5 and Mutation 6). First, think about what types of mutations could cause the phenotypes you see. Sort each mutation into the bin that describes its expression pattern.

No expression: lacP- (promoter mutation) lacZ- (beta-galactosidase mutation) Is -(super-repressor) Low expression: mutation in gene for CAP CAP-cAMP binding site mutation Constitutive expression: lacI- (repressor mutation) Oc (operator mutation) Mutations in the promoter sequence prevent RNA polymerase from binding and transcribing the operon; this results in no expression. Binding of the CAP protein (bound to cAMP) to the CAP-cAMP binding site facilitates the binding of RNA polymerase to the promoter sequence. Mutations in the gene that encodes the CAP protein or in the CAP-cAMP binding site make binding of RNA polymerase to the promoter less likely. This results in a lower than normal expression of the operon. Loss of function mutations in the repressor protein or in the operator region to which it binds prevent the repressor from blocking transcription. This results in constitutive expression of the operon.Gain of function mutations in the gene for the repressor (super-repressor) cause the repressor to bind to the operator even in the presence of lactose; this results in no expression. Mutations in the lacZ gene which encodes beta-galactosidase result in no expression of beta-galactosidase.

If this trisomic fly is the progeny of a male fly of genotype (ey+ey+ ey−ey−, gw+gw+ gw−gw−) crossed to a female fly with genotype (ey−ey− ey−ey−, gw−gw− gw−gw−), what can you conclude about the events that led to its formation? -Nondisjunction occurred in the female parent at the meiosis I division to produce an (n+1) egg that fused with a normal (n) sperm. -Nondisjunction occurred in the male parent at the meiosis II division to produce an (n+1) sperm that fused with a normal (n) egg. -Nondisjunction occurred in the female parent at the meiosis II division to produce an (n+1) egg that fused with a normal (n) sperm. -Nondisjunction occurred in the male parent at the meiosis I division to produce an (n+1) sperm that fused with a normal (n) egg.

Nondisjunction occurred in the male parent at the meiosis I division to produce an (n+1) sperm that fused with a normal (n) egg. The (ey−gw−ey−gw−) chromosome in the trisomic male must have come from the female parent, and the other two chromosomes must have come from the male parent. Since these two chromosomes have distinct alleles of eyey and gwgw, they were not sister chromatids (i.e., replicated copies of the same chromosome). That means that they must have been paired as homologous chromosomes at synapsis, and remained in the same cell through a nondisjunction event at the meiosis I division.

How many chromosomes from the tetraploid lines are contributed to experimental variety 1? Express your answer as an integer.

Number of chromosomes = 24

How many chromosomes from the hexaploid line are contributed to experimental variety 1? Express your answer as an integer.

Number of chromosomes = 36

To experimental variety 2? Express your answer as an integer.

Number of chromosomes = 48

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

Partial duplication and deletion

Common baker's yeast (Saccharomyces cerevisiae) is normally grown at 37∘∘C, but it will grow actively at temperatures down to approximately 20∘∘C. A haploid culture of wild-type yeast is mutagenized with EMS. Cells from the mutagenized culture are spread on a complete-medium plate and grown at 25∘∘C. Six colonies (1 to 6) are selected from the original complete-medium plate and transferred to two fresh complete-medium plates. The new complete plates (shown below) are grown at 25∘∘C and 37∘∘C. Four replica plates are made onto minimal medium or minimal plus adenine from the 25∘∘C complete-medium plate. The new plates are grown at either 25∘∘C or 37∘∘C, as indicated in the diagram. Which colonies are prototrophic and which are auxotrophic? Drag the appropriate items to their respective bins.

Prototrophic: 1, 2, 3, 6 Auxotrophic: 4, 5

Which of the following is a consequence of inactivation of both p53 alleles in a cell? -Programmed cell death (apoptosis) of the cell -Relative instability of the non-functional protein -A reduction in kinase activity of the cyclin-Cdk complex -Reduction of p21 synthesis

Reduction of p21 synthesis Functional p53 is required for activation of the gene that encodes p21.

Describe photoreactivation and nucleotide excision. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

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Diagram the mechanisms by which dsRNA are produced and processed into small RNARNAs. Drag the appropriate labels to their respective targets.

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A hereditary disease is inherited as an autosomal recessive trait. The wild-type allele of the disease gene produces a mature mRNA that is 1250 nucleotides (nt) long. Molecular analysis shows that the mRNA consists of four exons: A mother and father with two healthy children and two children with the disease have northern blot analysis performed in a medical genetics laboratory. The results of the northern blot for each family member are shown. Identify the genotype of each family member using the sizes of mRNAs to indicate each allele. (For example, a person who is homozygous wild type is indicated as "1250/12501250/1250.") Drag the appropiate labels to their respective targets.

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Choose the gel electrophoresis band patterns for all the genotypes that could be found in children of this couple.

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Choose the synapsis of these homologs during prophase I in hybrids produced from the cross of mainland with island deer.

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Choose which of the following genotypes for lac operon haploids have the following phenotypic characteristics: Drag the appropriate items to their respective bins.

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Compare DNADNA double-strand breaks in bacteria versus the double-strand breaks that precede homologous recombination in eukaryotes. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

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Diagram the pairing of these homologous chromosomes in prophase I. Drag the appropriate labels to their respective targets.

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During mismatch repair, why is it necessary to distinguish between the template strand and the newly made daughter strand? Describe how this is accomplished. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all.

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Explain how the DNADNA sequence change results in a Ser-to-Thr missense mutation. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

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Explain why CpG dinucleotides are hotspots of mutation. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

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In a mainland-island hybrid deer, recombination takes place in band q1 of the homologous chromosomes. Complete the diagram of the gametes that result from this event. Drag the appropriate labels to their respective targets.

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Populations evolve for many reasons. Suppose there is a population of plants that have either purple flowers or white flowers, and the allele for purple flowers is dominant. This means that plants with two purple alleles have purple flowers. Plants with one purple allele and one white allele also have purple flowers. Only plants with two white alleles have white flowers. For each event or condition described below, answer the following questions. Which mechanism of evolution is at work? How does this event affect the population's gene pool? Do the frequencies of the two alleles change, and if so, how? Drag the labels to their appropriate locations in the table below. Drag the pink labels onto the pink targets in the table to indicate which mechanism of evolution is at work. Then drag the blue labels onto the blue targets to indicate the effect on allele frequencies. Labels can be used once, more than once, or not at all.

See image. Different events and conditions result in natural selection, genetic drift, and gene flow. In turn, these mechanisms cause allele frequencies to change--and populations to evolve--in different ways. Natural selection, which results from differential survival and reproduction, is the only mechanism of evolution that consistently causes a population to become better adapted to its environment. While genetic drift and gene flow may cause alleles to increase in frequency regardless of whether they are harmful, neutral, or beneficial, natural selection causes only beneficial alleles to become more common in a population. Think about the events and conditions described in this exercise. Can you see how only those associated with natural selection produce a better match between the plants and their environment?

During Drosophila development, adult structures are formed from clusters of cells called imaginal discs, which are recognizable during larval and pupal stages. You are interested in the genes expressed in imaginal discs, and you have identified two such genes that are closely linked together. Gene 1 is expressed in the imaginal discs that will become the adult antennae. Gene 2 is expressed in the discs that will become the adult wings and legs. A map of the two genes is shown here. The arrows indicate the start of transcription. You decide to make a construct that places the DNA between the genes upstream to a reporter gene (GFP) in place of Gene 1. The full-length construct has the correct expression pattern in antennal discs. You then create several constructs that delete sections of the upstream DNA and score them for expression. The seven constructs are shown below; the black bars indicate the region(s) deleted in each construct. The data table on the right shows the expression results for each construct. From these results, you hypothesize that the sequence you have examined contains a promoter for gene 1 (P1), an enhancer that controls antennal expression (EA), an enhancer that controls leg and wing expression (EWL), and an insulator region (I). Using the data, identify the function of each DNA segment by dragging the labels to their correct locations. Labels can be used once, more than once, or not at all.

See image. From these experiments, you can conclude that: The DNA adjacent to the start of transcription for Gene 1 is its promoter region (P1), since this sequence is necessary for expressing Gene 1 in any tissue. Adjacent to the promoter is a sequence required to express Gene 1 in antennal discs (EA), consistent with this region acting as an antennal enhancer sequence. Upstream of the EA enhancer is a sequence that, when deleted, allows another enhancer to drive expression of Gene 1 in a region it is not normally expressed - suggesting that this sequence is an insulator sequence (I) that normally shields P1 from an enhancer further upstream that directs expression in wing and leg discs. Further upstream we find a sequence that can induce Gene 1 expression in wing and leg discs if the insulator is missing (EWL) - suggesting that this sequence is regulatory DNA for Gene 2.

Use the expression data for constructs 8 - 13 from Part B (repeated below) to determine the locations of EW and EL, as well as the promoter for Gene 2 (P2). Using the data, identify the function of each DNA segment by dragging the labels to their correct locations.

See image. From this experiment, you conclude that EW is adjacent to the insulator (I), and that EL lies between EW and P2. The expression patterns of Gene 1 and Gene 2 are thus controlled by four cis-acting regulatory sequences.

A helix-turn-helix motif is found in many prokaryotic transcription factors, and in a modified form (the homeo domain) in some eukaryotic transcription factors. Choose the image that properly indicates the helix-turn-helix motif in this repressor protein.

See image. Helices two (yellow) and three (purple) of each amino-terminal domain form a helix-turn-helix motif found in many prokaryotic transcription factors, and in a modified form (the homeo-domain) in some eukaryotic transcription factors. This motif comprises ~20 residues. Helix three inserts into the major groove of operator DNA.

You hypothesize that αα-spectrin may be responsible for the correct localization of NP2. If this hypothesis is correct, what would you expect for the localization of NP1 and NP2 in cells expressing a UAS-αα-spectrin RNAi construct that eliminates αα-spectrin protein expression? Complete this table by dragging labels to the correct locations. Labels may be used once, more than once, or not at all.

See image. If αα-spectrin is indeed responsible for the correct localization of NP2, then removing αα-spectrin would cause the mislocalization of NP2. This in turn should cause the mislocalization of NP1, based on your previous findings.

Now let's examine how the results from this experiment would look on a gel.In this experiment, you use labeled DNA fragments with all regions intact (1 2 3), and various fragments with the same point mutations as in Part B. Fragments with a mutated region are indicated with an "X." (For example, a fragment with intact Region 1, mutated Region 2, and intact Region 3 is noted as (1 X 3) in the diagram below.)You incubate each fragment with protein extracted from heart (H) or lung (L) tissue and compare it to DNA alone (None). You observe bands of differing sizes for the wild-type (1 2 3) fragment. Predict the gel shift assay results for each mutated fragment by dragging the labels to the correct locations. Labels may be used once, more than once, or not at all.

See image. In real-life research that uses these methods, it is also possible to identify bound regulatory proteins by using specific antibodies that bind to them. Adding an antibody that binds to a regulatory protein (that is bound to labeled DNA) causes a "supershift" on a gel, since the combined mass of the antibody + protein + target DNA is greater than that of protein + DNA alone. This approach requires a hypothesis about what regulatory proteins might be binding, and antibodies specific to known regulatory proteins. Comparing the DNA sequence of a binding region to other genes with known regulatory proteins can help generate hypotheses about what regulatory proteins might be present.

You are studying a bacterium that utilizes a sugar called athelose. This sugar can be used as an energy source when necessary. Metabolism of athelose is controlled by the ath operon. The genes of the ath operon code for the enzymes necessary to use athelose as an energy source. You have found the following: The genes of the ath operon are expressed only when the concentration of athelose in the bacterium is high. When glucose is absent, the bacterium needs to metabolize athelose as an energy source as much as possible. The same catabolite activator protein (CAP) involved with the lac operon interacts with the ath operon. Based on this information, how is the ath operon most likely controlled? Drag the labels onto the diagram to identify the small molecules and the states of the regulatory proteins. Not all labels will be used.

See image. Metabolism of the sugar athelose in this hypothetical system is controlled by an operon that exhibits both positive control and negative control. Transcription of the ath operon is turned on when athelose is present (negative control), and sped up when the bacterium runs out of glucose and must rely on athelose for energy (positive control).

You decide to test the three hypotheses from Part A by making expression constructs that combine different DNA regions from the affected child and her older brother, who has normal mRNA expression. As controls, you make a full-length constructs from both of these children. The construct made from the older brother's DNA has normal expression in tissue culture cells, while the construct from the affected child has no expression.For each of the three hypotheses, predict if each construct will express mRNA. Complete the table by dragging the "yes" and "no" labels to the correct locations.Use the "Yes" label if you predict a construct will express mRNA.Use the "No" label if you predict no expression.

See image. Testing the expression from this series of constructs would you to distinguish between the three hypotheses, since the various constructs will show different outcomes depending on where the defect in the affected child is located.

The three major mechanisms of evolution differ in how they work, and as a result often have different effects on a population. Review your understanding of natural selection, genetic drift, and gene flow by sorting the statements below into the correct bins. Drag each statement into the appropriate bin depending on whether it applies to natural selection, genetic drift, or gene flow.

See image. Natural selection, genetic drift, and gene flow cause a population to evolve in different ways. Natural selection is the result of differential survival and reproduction. It is the only mechanism of evolution that consistently causes a population to become better adapted to its environment. Genetic drift describes evolution due to chance events and causes unpredictable fluctuations in allele frequencies. Genetic drift can have a particularly significant effect in a small population, such as during a bottleneck or founder event. Gene flow results when alleles are transferred into or out of a population due to the migration of fertile individuals or their gametes. Gene flow can bring new alleles (beneficial, harmful, or neutral) into a population.

Mutations in region B and region D give the same results - when either of these regions is mutated, the operon is expressed under all conditions (called constitutive expression).You hypothesize that region B is the operator region and that Region D is the repressor protein.To test your hypothesis, you create two partial diploid lines by introducing a F' plasmid with a wild-type lac operon:Strain 1: F' A+ B+ C+ D+ E+ / A+ B- C+ D+ E+Strain 2: F' A+ B+ C+ D+ E+ / A+ B+ C+ D- E+ What experimental results would be predicted by your hypothesis? Complete this table with the expected results by dragging the "+" and "-" signs to the appropriate locations.

See image. The goal of this experiment is to determine which region is the gene for the repressor and which region is the operator. Recall that repressor proteins are trans-acting; proteins are able to diffuse around the cell and interact with any wild-type operator that exists. Introduction of a wild-type repressor gene will restore wild-type repressor protein to the mutated strain, restoring normal regulation.Unlike the repressor protein, operator regions only regulate genes on that piece of DNA, and cannot regulate genes on other pieces of DNA. Introducing a wild-type operator region will not restore normal regulation to strain with an operator mutation. The wild-type operator sequence on the F' plasmid cannot regulate the genes on the bacterial chromosome. If there is an operator mutation on the bacterial chromosome, expression will be constitutive even if a normal operon is introduced on F' plasmid. If region B is the operator, then the genotype F' A+B+C+D+E+/A+B-C+D+E+ (Strain 1) should still exhibit constitutive expression of the enzymes.If region D is the repressor gene, then the genotype F' A+B+C+D+E+/A+B+C+D-E+ (Strain 2) will exhibit wild-type regulation/expression --enzyme genes will be transcribed when signal is absent, and will stop being transcribed when signal is present.

Suppose that you study nervous system development in Drosophila embryos. You have identified a set of 12 cells that differentiate into various neurons and glia later in development. As part of studying these cells, you investigate proteins that interact with αα-spectrin, a component of the F-actin cytoskeleton.Through your studies you have identified two loci encoding novel proteins that are localized near αα-spectrin early in development. You decide to study the function of these proteins by eliminating their expression through RNA interference (RNAi).To accomplish this, you create anti-sense RNA constructs for both genes connected to yeast UAS sequence. The UAS sequence can be controlled through the expression of the yeast transcription factor GAL4, and you have a construct that expresses GAL4 in a subset of your cells of interest. Complete this table comparing RNAi to traditional forward genetics approaches. Indicate whether each statement is true or false about forward genetics and RNAi by dragging the labels to the correct locations in the table. Labels can be used once, more than once, or not at all.

See image. Traditional forward genetics and RNAi are two approaches used for investigating gene function. While both techniques are powerful approaches, each has specific strengths and weaknesses.Forward genetics requires no knowledge of gene sequences, whereas RNAi does. Conversely, RNAi does not require screening for rare mutations (with perhaps unknown phenotypes) since a gene can be targeted directly with its anti-sense RNA.Traditional forward genetics produces mutations that are present in every cell type, since the organism is genetically mutant in its germline. An RNAi construct, on the other hand, needs to be expressed in a cell to knock down its target gene. As such, the temporal and spatial expression pattern of the RNAi construct determines which cells will have a mutant condition.

Nucleosome positioning along the DNA can influence where transcriptional regulatory proteins are able to bind DNA. If a nucleosome is bound to an enhancer sequence, it may outcompete a regulatory protein from binding the same sequence. Conversely, if an enhancer sequence is in the linker DNA where the nucleosome is absent, the regulatory protein does not have to compete with the nucleosome. The position of the nucleosome can alter the accessibility of a sequence of DNA to DNA binding proteins. Suppose that this region of DNA contained several enhancer sequences that bind highly conserved regulatory proteins to stimulate transcription of a gene in this region, and that binding is more efficient when the enhancer sequence is not associated with a nucleosome. Use this information to complete these sentences. Fill in the blanks with the appropriate words. Words may be used more than once. -fewer -fall within linker DNA -more -be bound by nucleosomes

Since sea urchins have longer linker DNA segments than Drosophila melanogaster, when this region of DNA is moved to the sea urchin it is likely that these enhancer sequences will fall within linker DNA. With more enhancer sequences available in sea urchins to bind regulatory proteins that stimulate transcription, the same region of DNA will likely generate more RNA transcripts. In sea urchins, fewer nucleosomes occupy the same region of DNA (362 vs. 512 in Drosophila melanogaster); therefore, there is less competition between nucleosomes and regulatory proteins for the same enhancer sequences. So, when the region of DNA is moved to the sea urchin genome, there would be more binding of regulatory proteins that stimulate transcription, and more RNA transcripts would be expected to be produced.

Which of the following occurs as a result of an abundance of tryptophan in E. coli? -The leader sequence is not transcribed. -Stalling of the ribosome at trp codons in the leader sequence -The 5 trp genes (TrpA - TrpE) are not transcribed. -The 5 trp genes (TrpA - TrpE) are transcribed, but not translated.

The 5 trp genes (TrpA - TrpE) are not transcribed. When trp is abundant, the genes involved in tryptophan synthesis are negatively regulated at the level of transcription.

Which of the following statements best describes the bacteriophage repressor protein? -The repressor is a monomer that binds in the minor groove of DNA. -The repressor is a dimer that binds in the minor groove of DNA. -The repressor is a dimer that binds in the major groove of DNA. -The repressor is a monomer that binds in the major groove of DNA.

The repressor is a dimer that binds in the major groove of DNA. Repressor binds to operator DNA as dimers. Each monomer binds to an operator half-site (A/T-C/G-A/T-A/T). The amino-terminal domains of repressor are responsible for DNA binding and the carboxy-terminal domains (not shown) are primarily responsible for dimerization of the repressor monomers. The amino-terminal domain of each monomer is composed of five alpha-helices connected by short loops.

The MutS protein is composed of two monomers: the mismatch recognition monomer and the non-mismatch-binding monomer. Although the two MutS monomers are structurally identical, they differ in their orientations relative to bound DNA. This means that the mismatch domains (shown here in blue) of the two monomers do not contact DNA in the same way. So although the structure is a homodimer, functionally it acts more like a heterodimer. Using this image as a guide, how do the mismatch domains of the two monomers interact with DNA? -The mismatch recognition monomer makes specific contacts to the mismatched bases via the major groove. The non-mismatch-binding monomer only makes nonspecific contacts with the backbone of the major groove. -The mismatch recognition monomer makes specific contacts to the mismatched bases via the minor groove. The non-mismatch-binding monomer only makes nonspecific contacts with the backbone of the major groove. -The mismatch recognition monomer makes specific contacts to the matched bases via the minor groove. The non-mismatch-binding monomer only makes nonspecific contacts with the backbone of the minor groove. -The mismatch recognition monomer only makes nonspecific contacts with the backbone of the major groove. The non-mismatch-binding monomer makes specific contacts to the mismatched bases via the minor groove.

The mismatch recognition monomer makes specific contacts to the mismatched bases via the minor groove. The non-mismatch-binding monomer only makes nonspecific contacts with the backbone of the major groove. The mismatch domain of the mismatch recognition monomer makes extensive contact with the bases in the region of the mismatch. In contrast, the mismatch domain of the non-mismatch-binding monomer only contacts the DNA backbone.

In humans that are XX/XO mosaics, the phenotype is highly variable, ranging from females who have classic Turner syndrome symptoms to females who are essentially normal. Likewise, XY/XO mosaics have phenotypes that range from Turner syndrome females to essentially normal males. How can the wide range of phenotypes be explained for these sex-chromosome mosaics? -The resulting phenotype depends on the individual's genealogical background because the syndrome has a cumulative effect. -The resulting phenotype depends on the relative percentage of cells with each karyotype at different stages of development. -The resulting phenotype depends on the environmental conditions in which the individual develops.

The resulting phenotype depends on the relative percentage of cells with each karyotype at different stages of development.

A survey of organisms living deep in the ocean reveals two new species whose DNA is isolated for analysis. DNA samples from both species are treated to remove nonhistone proteins. Each DNA sample is then treated with DNase I that cuts DNA not protected by histone proteins but is unable to cut DNA bound by histone proteins. Following DNase I treatment, DNA samples are subjected to gel electrophoresis, and the gels are stained to visualize all DNA bands in the gel. The staining patterns of DNA bands from each species are shown in the figure. The number of base pairs (bp) in small DNA fragments is shown at the left of the gel. Choose the correct interpretation of the gel results in terms of the spacing of nucleosomes in the chromatin of each species. -The nucleosome spacing in species A is 600 bp, whereas the spacing in species B is 400 bp. -The nucleosome spacing in species A and B is 800 bp. -The nucleosome spacing in species A is 400 bp, whereas the spacing in species B is 800 bp. -The nucleosome spacing in species A is 200 bp, whereas the spacing in species B is 400 bp.

The nucleosome spacing in species A is 200 bp, whereas the spacing in species B is 400 bp.

Now that you have determined which are the regulatory regions, identify how this operon is regulated. Is this operon inducible or repressible? Is it under positive control or negative control? -The operon is inducible (the signal induces/starts expression of the enzymes).The operon is under negative control (the regulatory protein is a repressor). -The operon is repressible (the signal represses/stops expression of the enzymes).The operon is under negative control (the regulatory protein is a repressor). -The operon is inducible (the signal induces/starts expression of the enzymes).The operon is under positive control (the regulatory protein is an activator). -The operon is repressible (the signal represses/stops expression of the enzymes).The operon is under positive control (the regulatory protein is an activator).

The operon is repressible (the signal represses/stops expression of the enzymes).The operon is under negative control (the regulatory protein is a repressor). You know from Part A that this operon is repressible; the signal molecule stops transcription.To figure out whether the operon is under negative or positive control, you must determine whether the regulatory protein is a repressor (negative control) or an activator (positive control). There are three regulatory regions for this operon: the promoter sequence, the gene that produces the regulatory protein, and the binding site for the regulatory protein.A mutation in region A prevents transcription of both enzyme genes, suggesting this is the promoter region. Both region B and region D mutations give constitutive expression. Eliminating the regulatory protein or its binding site gives expression under all conditions. This suggests the operon is regulated by a repressor protein binding to an operator sequence.

If the UGG codons in Region 1 of trpL were changed to AGG codons, what effect would this have on expression of the trp operon? -The operon would never be expressed. -The operon would be constitutively expressed. -The operon would be regulated by tryptophan, but attentuation would be triggered at higher levels of tryptophan than in wild type. -The operon would be regulated by tryptophan, but attentuation would be triggered at lower levels of tryptophan than in wild type. -The operon would be regulated by arginine levels; high levels of arginine would attenuate expression of trp genes.

The operon would be regulated by arginine levels; high levels of arginine would attenuate expression of trp genes. The two tryptophan codons (UGG) are responsible for sensing the level of tryptophan. If tryptophan is high, transcription/translation continues quickly because tRNAs charged with tryptophan are readily available. Fast transcription/translation pulls region 1 and 2 into the ribosome quickly, leaving only region 3 and region 4 able to complementary base pair and form a stem-loop. A 3-4 stem loop causes termination of transcription (attenuation). If tryptophan is low, transcription/translation stalls because tRNAs charged with tryptophan are scarce. If transcription/translation is slow, region 2 will be able to bind to region 3, forming an anti-termination loop and transcription continues. If these codons were changed to arginine codons (AGG), the operon would no longer be sensitive to tryptophan levels, but would instead be sensitive to arginine levels. Quick transcription/translation would be dependent on readily available tRNAs charged with arginine.

Which of the following features of the trp operon is likely least essential to the process of attenuation? -The ability of sequences within the leader mRNA to pair with one another -Transcription and translation of the leader sequence occur simultaneously. -Trp codons near the beginning of the leader sequence -The order of the structural genes, E, D, C, B, A

The order of the structural genes, E, D, C, B, A Because each gene encodes an enzyme involved in the trp synthetic pathway, the order in which the genes occur is likely not important in terms of the attenuation process.

Helicases are "motorized" proteins that help unwind DNA by separating duplexed DNA into single-stranded form. They are powered by the energy derived from ATP hydrolysis. Helicases have a polarity that is defined as the direction they travel along single-stranded DNA (i.e., either in a 5'→→ 3' or in a 3'→→ 5' direction). What are the polarities of the RecB and RecD subunits? -The polarity of RecB is 3'→→ 5', wheras the polarity of RecD is 5'→→ 3'. -Both RecB and RecD have a polarity of 5'→→ 3'. -Both RecB and RecD have a polarity of 3'→→ 5'. -The polarity of RecB is 5'→→ 3', whereas the polarity of RecD is 3'→→ 5'.

The polarity of RecB is 3'→→ 5', wheras the polarity of RecD is 5'→→ 3'. The RecC pin splits the DNA duplex into 5'- and 3'-tailed strands. The RecB and RecD helicase subunits bind to different single-stranded tails. The 3'-ending strand is fed to the RecB helicase, and the 5'-ending strand is fed to the RecD helicase.

You test the expression of the constructs from Part B and obtain the following results. Which of the three hypotheses do these results support? -The results support the hypothesis that the defect in the affected child is due to a mutation in the transcribed region that causes mRNA instability. -The results support the hypothesis that the defect in the affected child is due to a defective promoter. -The results support the hypothesis that the defect in the affected child is due to regulatory DNA that will not remodel.

The results support the hypothesis that the defect in the affected child is due to regulatory DNA that will not remodel. The results of the expression construct analysis support the hypothesis that the defect in the affected child is due to regulatory DNA that will not remodel. You decide to further test this hypothesis with a DNAse I sensitivity assay.

Researchers interested in studying mutation and mutation repair often induce mutations with various agents. What kinds of gene mutations are induced by chemical mutagens? -transversions -transitions -frameshifts -all of the above

all of the above

You are investigating a new bacterial operon that contains five regions (A, B, C, D and E) involved in coordinated regulation of transcription. One is the gene for a regulatory protein, one is the binding site for the regulatory protein, two produce structural enzymes (enzyme 1 and enzyme 2), and one is the promoter for the two enzyme genes.The table below shows data collected for five different bacterial strains. A "+" indicates that the structural enzyme is produced and a "-" indicates that the enzyme is not produced. The "signal" is a molecule that either represses or induces the operon. Use the information to determine how the operon is regulated, and the identity of each region. In the wild-type operon (A+ B+ C+ D+ E+), how does the presence of the signal affect expression of the structural enzymes ? -The structural enzymes are never produced. The presence or absence of the signal has no effect on the production of the structural enzymes. -The presence of the signal stops production of the structural enzymes. Without the signal, the structural enzymes are produced. -The structural enzymes are produced under all conditions. The presence or absence of the signal has no effect on the production of the structural enzymes. -The presence of the signal causes the structural enzymes to be produced. Without the signal, no enzymes are made.

The presence of the signal stops production of the structural enzymes. Without the signal, the structural enzymes are produced. If you look at only the wild-type genotype (top row), you can see that enzyme 1 and enzyme 2 are only produced when the signal molecule is absent (-signal). When the signal is present (+signal), the enzymes are not produced.In this operon, the signal stops transcription of the operon; this means that the operon is repressible. In repressible operons, transcription occurs until the level of the final product is high. High levels of the final product suppress further transcription of the operon.

Describe the roles of writers, readers, and erasers in eukaryotic gene regulation. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all.

The proteins called writers and erasers refer to the enzymes which modify histones and are recruited to specific chromosome regions by transcription factors. Specifically, writers add chemical groups such as methyl or acetyl groups to histones, whereas erasers remove those chemical groups. Meanwhile, readers refer to proteins that bind to the modified histones and then cause changes in chromatin structure. These structural changes make it either more compact and therefore less likely to be transcribed, or more open and therefore more likely to be transcribed.

If nondisjunction occurs in meiosis II, how many of the four gametes will be aneuploid? -One -Four -Two -Three

Two Nondisjunction in meiosis II produces one nullosomic gamete and one disomic gamete. The two products produced by the other meiosis I daughter cell will produce two normal haploid gametes.

How are his−his− bacteria used in the Ames test? What mutational event is identified using his−his− bacteria? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

The test detects reverse mutations using his − bacteria, which are bacteria that cannot grow on medium that lacks the amino acid histidine. Reversion of his− back to his+ is detected by plating millions of his− bacteria on medium lacking histidine and counting the number of colonies that grow.

What would be the result of a reciprocal balanced translocation event that takes place between the blue chromosome (designated 1-2) and the yellow chromosome (designated 3-4)? -The translocation would generate 2 new chromosomes: 1-4 and 3-2. -The translocation would generate 4 new chromosomes: 1-2, 1-4, 3-2, and 3-4. -There would be no effect. The chromosomes would remain unchanged.

The translocation would generate 4 new chromosomes: 1-2, 1-4, 3-2, and 3-4. Reciprocal translocations involve the exchange of segments between two nonhomologous chromosomes. Translocation heterozygotes have one normal chromosome and one altered chromosome in each homologous pair.

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

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

Which of the following is characteristic of open promoters? See Section 13.2 (Page 485) . -They tend to contain a TATA box. -They are relatively insensitive to DNase I cleavage. -They are constitutively expressed. -Their expression is regulated by transcription factors, and transcription is blocked until nucleosomes overlying the promoter are removed.

They are constitutively expressed. Open promoters are associated with constitutively active genes, such as housekeeping genes encoding proteins vital for basic cellular functions. See Figure 13.15 for more information.

Which of the following characteristics is NOT associated with transposable genetic elements? See Section 11.7 (Page 425) . -They are noncoding sequences of DNA. -They may lead to the inactivation of functional genes. -They have flanking direct repeats. -They have terminal inverted repeats.

They are noncoding sequences of DNA. Transposable elements are protein coding sequences. They code for transposase, which is the enzyme that generates staggered cuts in the target DNA sequence.

What role do insulator sequences play in the regulation of eukaryotic transcription? See Section 13.1 (Page 479) . -They block communication between enhancers and nontargeted promoters. -They block binding of transcription factors to enhancers. -They block binding of transcription factors to cis-acting promoter elements. -They block binding of transcription repressors to cis-acting promoter elements.

They block communication between enhancers and nontargeted promoters. Insulators are protein-binding sequences that direct enhancers to interact with the intended promoter and that block communication between enhancers and other promoters. See Figure 13.12 for more information.

Why are multiple VNTR probes used in DNA fingerprinting? -They increase the probability of producing a DNA fingerprint that is unique to an individual. -They allow for the detection of multiple alleles at the same VNTR locus. -They allow for the detection of loci that are linked. -They increase the probability that a probe will hybridize to a VNTR present in the individual.

They increase the probability of producing a DNA fingerprint that is unique to an individual. The use of a large number of probes increases the chances that the DNA fingerprint produced will be unique.

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

They may be sterile and unable to produce offspring. Polyploid individuals are sterile if they produce genetically unbalanced gametes; both autopolyploid and allopolyploid individuals may be sterile.

Answer the following questions concerning the accuracy of DNA polymerase during replication. What general mechanism do DNA polymerases use to check the accuracy of DNA replication and identify errors during replication? Select the two correct answers. -They bind ends of double stranded breaks. -They remove incorrect nucleotides immediately after they are added. -They recognize and restores bases of tautomeric shifts. -They ensure that the correct nucleotide is added to the growing DNA strand. -They fill gaps that arises after extension of pyrimidine dimers.

They remove incorrect nucleotides immediately after they are added. AND They ensure that the correct nucleotide is added to the growing DNA strand.

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

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

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

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

Describe the difference between DNADNA transposons and retrotransposons. Drag the appropriate items to their respective bins.

Transposons: -Insert into new locations by excising themselves from one location and inserting into a new location. -Insert into new locations by DNA replication and insertion of the replicated copy into a new location. Retrotransposons -Insert into new locations by first being transcribed to create RNA copies, which are then reverse-transcribed into DNA. Neither transposons nor retrotransposons: -Insert into new locations by first being replicated as DNA, then transcribed to RNA, and finally reverse-transcribed into DNA.

What kinds of gene mutations are induced by radiation energy? -transitions -transversions -all of the above

all of the above

What is the primary function of the SWI/SNF complex? See Section 13.2 (Page 485) . -To remodel histones by replacing H2A with H2A.Z -To arrange nucleosomes so that the DNA is transcriptionally silent -To measure linker DNA and evenly distribute nucleosomes -To open chromatin structure by displacing or ejecting nucleosomes

To open chromatin structure by displacing or ejecting nucleosomes The composition of this complex varies somewhat among eukaryotic species, but in each species it functions to open chromatin structure by displacing or ejecting nucleosomes. These actions expose promoter and other regulatory sequences to allow binding of transcription factors or activators that help initiate transcription. See Figure 13.17 for more information.

What is the difference between a transition mutation and a transversion mutation? Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

Transitions are mutations that replace a purine with a purine and a pyrimidine with a pyrimidine, whereas transversions are mutations that replace a purine with a pyrimidine and a pyrimidine with a purine. For example, a mutation of a G:C base pair to a C:G or a T:A base pair is a transversion, whereas a mutation of a G:C base pair to an A:T is a transition.

Duplications and deletions can be detected during meiosis by the presence of extrachromosomal loops that do not pair properly with their homolog. True False

True During meiosis, the presence of a chromosomal loop signals that a duplication or deletion has occurred.

Lactose is an inducer of the lac operon. True False

True Lactose indirectly induces or stimulates the transcription of genes involved in its metabolism.

What are the examples of radiant energy that cause mutations? Check all that apply. -microwave irradiation -UV radiation -IR radiation -γ radiation

UV radiation AND γ radiation

Give an example (real or hypothetical) of each of the following: Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

Upstream activator sequence : Gal4 protein binds these yeast elements to activate transcription of galactose utilization genes. Insulator sequence : the formation of specific DNADNA loops that protect particular genes from enhancers. Silencer sequence : the yeast Mig1 and Tup1 proteins bind this element during glycolysis to prevent transcription of galactose utilization genes. Enhanceosome : an element, known as Mediator, that assembles at yeast enhancers. It contacts promoter-bound proteins to activate transcription. RNARNA interference : prevention of translation of the Dnmt3 DNADNA methyltransferase transcript by this leads to the development of fertile queens in honeybees.

A man who is color blind and has hemophilia and a woman who is wild type have a daughter with triple X syndrome (XXX) who has hemophilia and normal color vision. What are the genotypes of the parents? Select at least two answers (one genotype for the father and one genotype for the mother). If there is more than one possible genotype for a parent, select all that apply. X CH Y X Ch Y X cH Y X ch Y X CH X CH X Ch XCh X cH X cH X Ch X CH X Ch X cH X ch X ch

X ch Y AND X Ch X CH AND X Ch X cH

What is a chromosome territory? -a small, structural region of the chromosome, such as the centromere -a small region of the interphase nucleus that is occupied by a single chromosome or portion thereof -a dark or light band as visualized by Giemsa staining -a large region of a chromosome, such as the large or small arm

a small region of the interphase nucleus that is occupied by a single chromosome or portion thereof

For the following crosses, determine as accurately as possible the genotypes of each parent, the parent in whom nondisjunction occurs, and whether nondisjunction takes place in the first or second meiotic division. Both color blindness and hemophilia, a blood-clotting disorder, are X-linked recessive traits. Capital letters represent wild-type alleles, lowercase letters represent mutant alleles. In each case, assume the parents have normal karyotypes. A man and a woman who each have wild-type phenotypes have a son with Klinefelter syndrome (XXY) who has hemophilia. What are the genotypes of the parents? Select at least two answers (one genotype for the father and one genotype for the mother). If there is more than one possible genotype for a parent, select all that apply. X H Y X h Y X H X H X H X h X h X h

X H Y AND X H X h

A color-blind man and a woman who is wild type have a daughter with Turner syndrome (XO) who has normal color vision and blood clotting. What are the genotypes of the parents? Select at least two answers (one genotype for the father and one genotype for the mother). If there is more than one possible genotype for a parent, select all that apply. X C Y X c Y X C XC X C X c X c X c

X c Y AND X C XC AND X C X c

A man who is color blind and a woman who is wild type have a son with Jacob syndrome (XYY) who has hemophilia. What are the genotypes of the parents? Select at least two answers (one genotype for the father and one genotype for the mother). If there is more than one possible genotype for a parent, select all that apply. X C Y X c Y X H X H X H X h X h X h

X c Y AND X H X h

P-elements are a type of transposon commonly used in Drosophila (fruit fly) research. A common marker for P-elements is the miniwhite + gene, an engineered version of the white locus. Flies mutant for the natural white have white eyes instead of wild-type red eyes. A miniwhite + gene on a P-element in a white mutant will produce eye pigment, and will act as a dominant marker indicating that the P-element is present. P-elements used for research are nonautonomous, meaning that they do not encode their own transposase. The transposase is supplied using a second locus, where the transposase enzyme is no longer within a transposon and thus is immobilized (meaning that it cannot move itself). For flies with both a nonautonomous P-element and immobilized transposase, the P-element can move within the genome. A researcher is interested in screening for P-elements inserted into a region on chromosome 3. To produce new insertions, she crosses females homozygous for a miniwhite + P-element on the X chromosome with males homozygous for immobilized transposase on chromosome 2. What genotype will the F1 male flies have? (Note that a "+" indicates a wild-type chromosome without the transposase locus or the P {miniwhite +} insertion. The X /Y chromosomes are listed first, followed by the second chromosome.) -YP{miniwhite+}/ X ; transposase / + -Xtransposase/ Y; P{miniwhite+} /+ -XP{miniwhite+}/ Y; transposase / + -XP{miniwhite+}/ Y; transposase / transposase

XP{miniwhite+}/ Y; transposase / + In this cross, the F1 males receive the XP{miniwhite+} chromosome from their mothers and the Y chromosome from their fathers, making them XP{miniwhite+}/ Y. They also receive chromosome 2 bearing transposase from their fathers, and a wild-type chromosome 2 from their mothers, making them (transposase / + ) for chromosome 2. The full genotype of the F1 males is thus XP{miniwhite+}/ Y; transposase / + .The sperm produced by the flies will have P-elements that might move to new locations in the genome, since they have the nonautonomous P-element and active transposase in the same cell. In most cases, however, the P-element will remain at its original location.

A muscle enzyme called ME1 is produced by transcription and translation of the ME1 gene in several muscles during mouse development, including heart muscle, in a highly regulated manner. Production of ME1 appears to be turned on and turned off at different times during development.To test the possible role of enhancers and silencers in ME1 transcription, a biologist creates a recombinant genetic system that fuses the ME1 promoter, along with DNA that is upstream of the promoter, to the bacterial lacZlacZ(ββ-galactosidase) gene. The lacZlacZ gene is chosen for the ease and simplicity of assaying production of the encoded enzyme.The diagram shows the structure of the recombinant, as well as bars that indicate the extent of six deletions the biologist makes to the ME1 promoter and upstream sequences. The blue bar is the site of the promoter whereas the gray bars span potential enhancer/silencer modules.The table displays the percentage of ββ-galactosidase activity in each deletion mutant in comparison to the recombinant gene system without any deletions. Does this information indicate the presence of enhancer and/or silencer sequences in the ME1 upstream sequence? no yes

Yes

Generally speaking, which of the following mutations would most severely affect the protein coded for by a gene? -a base substitution at the beginning of the gene -a base substitution at the end of the gene -a frameshift deletion at the beginning of the gene -a frameshift deletion at the end of the gene

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

A small population of deer living on an isolated island are separated for many generations from a mainland deer population. The populations retain the same number of chromosomes and but hybrids are infertile. One chromosome (shown in the figure) has a different banding pattern in the island population than in the mainland population. Describe how the banding pattern of the island population chromosome most likely evolved from the mainland chromosome. What term or terms describe the difference between these chromosomes? -a pericentric inversion -an unbalanced translocation -a paracentric inversion -None of the above.

a pericentric inversion

Helix three of each repressor monomer binds to the recognition sequence in the major groove. Specific glutamines in the repressor protein make hydrogen bonds to specific bases in DNA. Identify the specific bases in this interaction. -adenine 1, guanine 2, thymine 4 -adenine 1, adenine 3, adenine 4 -guanine 1, adenine 2, cytosine 4 -cytosine 2, adenine 3, adenine 4 -adenine 1, cytosine 2, adenine 3

adenine 1, guanine 2, thymine 4 Helix three of each repressor monomer is primarily responsible for binding to the ACAA recognition sequence in the major groove. As shown for 1 monomer, glutamine 28 forms hydrogen bonds with adenine one of this sequence. Hydrogen bonds also link glutamine 29 of helix three with guanine 2, as well as glutamine 33 with thymine 4.

Classify the nature of the mutation in colony 5. -adenine auxotrophic mutation -temperature sensitive mutation -adenine prototrophic mutation

adenine auxotrophic mutation

The trp repressor binds to the operator region of DNA as a dimer (two repressor protein subunits bound together as a unit).You isolate a new mutant E.coli strain that you call Rnovel . In this strain, the repressor protein cannot attach to DNA (it cannot bind trpO+ ), but it can make dimers. In this strain, how is expression of the trp operon regulated? -noninducible -constitutive -inducible -repressible

constitutive The wild-type trp operon is repressible - expression of the trp operon is blocked by high levels of tryptophan. Binding of tryptophan to the repressor activates the repressor and allows it to bind to the operator and prevent transcription.If the repressor cannot bind to the operator sequence, transcription cannot be blocked. In this mutant strain, the trp operon will be expressed constitutively.

The role of transposase activity in IS element transposition includes _______. -regulating expression of the target sequence -filling in single-stranded gaps after IS insertion -cutting DNA at the target sequence -truncating other proteins

cutting DNA at the target sequence Transposase makes staggered cuts at the site targeted for insertion.

Which type of DNA produces a light band when treated with Giemsa stain? -heterochromatin -DNA that is not actively transcribed -centromere DNA -euchromatin

euchromatin Euchromatin contains actively transcribed genes. Since it is not highly condensed, it does not bind significant amounts of Giemsa stain and appears as a light band.

In an isolated population of 1000 individuals, 444 are genotype MN, 478 are MM, and 78 are NN. What are the frequencies of the M and N alleles in this population? See Section 20.1 (Page 727) . f(M) = 0.5, f(N) = 0.5 f(M) = 0.478, f(N) = 0.078 f(M) = 0.4, f(N) = 0.6 f(M) = 0.7, f(N) = 0.3

f(M) = 0.7, f(N) = 0.3 Because f(M) = [478 + (1/2)(444)]/1000 = 0.7 f(N) = [78 + (1/2)(444)]/1000 = 0.3

Two different mutations are identified in a haploid strain of yeast: The first prevents the synthesis of adenine by a nonsense mutation of the ade-1 gene. In this mutation, a base-pair substitution changes a tryptophan codon (UGGUGG) to a stop codon (UGAUGA).The second affects one of several duplicate tRNA genes. This base-pair substitution mutation changes the anticodon sequence of a tRNATrp from 3'-ACCACC - 5' to 3'-ACUACU - 5'. Do you consider the first mutation to be a forward mutation or a reversion? -reversion -forward mutation

forward mutation

It has been observed that diseases caused by repeat expansions are the result of trinucleotide repeats rather than smaller or larger repeat lengths. If non-trinucleotide repeat expansions were present in the coding region of a gene, which type of mutation would you expect to be the most likely? nonsense missense silent frameshift

frameshift Repeat lengths that are not multiples of three in the coding regions of genes would likely cause frameshift mutations, rendering the protein product nonfunctional. Two mutant alleles may not be developmentally viable; thus these genotypes do not survive to pass on these alleles. While nonsense mutations could still exist with non-trinucleotide expansions, they would be less likely as they are dependent on the sequence in the repeat and the length of the repeat. Frameshift mutations are only dependent on the length of the repeat. Repeat expansions that are multiples of three still allow a protein to be produced (with one or two amino acids repeated) and may have enough function for viability even if two mutant alleles are present. This is the case with Friedreich's ataxia. Alternatively, a single mutant allele in Huntington's disease patients produces a mutant protein with toxic functions. Therefore trinucleotide repeat expansion alleles can have either recessive or dominant effects.

The partial amino acid sequence of a wild-type protein is:. . . Arg-Met-Tyr-Thr-Leu-Cys-Ser . . .The same portion of the protein from a mutant has the sequence:. . . Arg-Met-Tyr-Thr-Pro-Leu-Phe . . . Identify the type of mutation. -substitution mutation -frameshift mutation -none of the above

frameshift mutation

You recently discovered four new chemical mutagens (#1 - #4), but you do not know what type(s) of mutations these chemicals induce. You exposed E. coli to each chemical individually and observed mutant phenotypes. You then exposed this bacteria to a second known mutagen to observe if the mutations from chemicals #1 - #4 could be reversed. Your data is presented below. What type(s) of mutations are likely produced by chemical #1? Select all that apply. transitions transversions frameshifts

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

Huntington's disease occurs when a CAG repeat expansion in the coding region of the HTT gene becomes too long. The CAG repeat is also in frame.(... | CAG | CAG | CAG | ...) Which amino acid is coded for in this repeat? glutamic acid asparagine glutamine proline

glutamine CAG codes for glutamine. Normal individuals typically have 10-34 CAG. Individuals affected by Huntington's disease have 40-200 CAG repeats; therefore they have a longer polyglutamine tract than normal individuals.

I+P+OCI+P+OCZ+Y+Z+Y+ -is lac- and the genes are not transcribed. -is lac + and the genes are not transcribed. -is lac- and the genes are inducibly transcribed. -is lac- and the genes are constitutively transcribed. -is lac+ and the genes are constitutively transcribed. -is lac + and the genes are inducibly transcribed.

is lac+ and the genes are constitutively transcribed.

I−P+O+I−P+O+Z+Y+Z+Y+ -is lac- and the genes are not constitutively transcribed. -is lac + and the genes are not transcribed. -is lac + and the genes are inducibly transcribed. -is lac+ and the genes are constitutively transcribed. -is lac- and the genes are inducibly transcribed. -is lac- and the genes are not transcribed.

is lac+ and the genes are constitutively transcribed.

I+P+O+I+P+O+Z−Y+Z−Y+ -is lac + and the genes are inducibly transcribed. -is lac + and the genes are constitutively transcribed. -is lac- and the genes are inducibly transcribed. -is lac- and the genes are constitutively transcribed. -is lac- and the genes are not transcribed. -is lac + and the genes are not transcribed.

is lac- and the genes are not transcribed.

I+P−O+I+P−O+Z+Y+Z+Y+ -is lac + and the genes are not transcribed. -is lac + and the genes are inducibly transcribed. -is lac + and the genes are constitutively transcribed. -is lac- and the genes are inducibly transcribed. -is lac- and the genes are not transcribed. -is lac- and the genes are constitutively transcribed.

is lac- and the genes are not transcribed.

I+P−O+I+P−O+Z+Y+Z+Y+ -is lac- and the genes are constitutively transcribed. -is lac + and the genes are not transcribed. -is lac+ and the genes are constitutively transcribed. -is lac- and the genes are inducibly transcribed. -is lac + and the genes are inducibly transcribed. -is lac- and the genes are not transcribed.

is lac- and the genes are not transcribed.

For the cross in Part G, in which parent and during what meiotic division might nondisjunction have occurred? Select all that apply. -maternal meiosis I -maternal meiosis II -paternal meiosis I -paternal meiosis II

maternal meiosis II AND paternal meiosis II

DNA mismatch repair can accurately distinguish between the template strand and the newly replicated strand of a DNA duplex. What characteristic of DNA strands is used to make this distinction? -deamination -methylation -acetylation -bulky modification

methylation

If a replication error escapes detection and correction, what kind of abnormality is most likely to exist at the site of replication error? -loss of one of the purines -mismatched base pairs -frameshift mutations -trinucleotide repeat disorders

mismatched base pairs

A geneticist searching for mutations uses the restriction endonucleases SmaI and PvuII to search for mutations that eliminate restriction sites. SmaI will not cleave DNADNA with CpGCpG methylation. It cleaves DNADNA at the restriction digestion sequence ↓↓ 5′-CCC5′-CCCGGG-3′GGG-3′5′-GGG5′-GGGCCC-3′CCC-3′ ↑↑ PvuII is not sensitive to CpGCpG methylation. It cleaves DNADNA at the restriction sequence ↓↓ 5′-CAG5′-CAGCTG-3′CTG-3′5′-GTC5′-GTCGAC-3′GAC-3′ ↑↑ What common features do SmaI and PvuII share that would be useful to a researcher searching for mutations that disrupt restriction digestion? Select the two correct answers. -both recognition sequences includes stop codons and therefore could be located in the coding sequence of genes -neither recognition sequence includes a stop codon and therefore they could be located in the coding sequence of genes -the recognition sequences occur rather frequently -the recognition sequences are very stable

neither recognition sequence includes a stop codon and therefore they could be located in the coding sequence of genes AND the recognition sequences occur rather frequently

A true reversion occurs when the wild-type DNA sequence is restored to encode its original message by a second mutation at the same site or within the same codon. Which of the following isoleucine codon(s) could be mutated once to form a methionine codon, and then mutated at a second site to restore an isoleucine codon? Select all that apply. (Note that two different positions in the codon must be mutated.) AUC AUU AUA none of these codons

none of these codons In order to generate a methionine codon from an isoleucine codon, the #3 position must be changed to a G, giving AUG. A subsequent mutation would then be required at a second site in the AUG codon, but since no isoleucine codons (AUA, AUC, AUU) contain a G in the #3 position, a true reversion can't occur without mutating the #3 position again.

For which codon(s) could a single base change account for this amino acid change? Alanine to Phenylalanine Select all that apply. GCA GCU GCC GCG none of these codons

none of these codons There are four codons for alanine (GCA, GCU, GCG, GCC) and two codons for phenylalanine (UUC, UUU). Neither of the two sets have codons that share any two positions, therefore it is not possible for a single base change to result in a change from an alanine codon to a phenylalanine codon.

Which two repair processes are the most error prone? Select two. -homologous recombination (synthesis-dependent strand annealing) -nonhomologous end joining -translesion DNA synthesis -nucleotide excision repair -base excision repair

nonhomologous end joining AND translesion DNA synthesis Both double-strand breaks and DNA lesions can block replicative DNA polymerases from duplicating the genome which can stall the cell cycle. When other repair mechanisms are unable to restore the genome to an intact state, repair mechanisms that allow for DNA sequence changes can at least restore the genome to a state that replicative DNA polymerase can proceed. These processes are said to be error prone. Nonhomologous end joining repairs double-strand breaks, but in doing so nucleases have to trim back the DNA surrounding the break. Thus DNA sequences are lost, but the break is ligated back together. Translesion DNA synthesis is performed by bypass polymerases. Lesions that block replicative DNA polymerases do not block bypass polymerases. In order to synthesize DNA through lesions, these polymerases lack proofreading activity. Thus, DNA sequence errors can be incorporated into the DNA.

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

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

Identify two DNADNA repair mechanisms that remove UV-induced DNADNA lesions. Select all that apply. -nucleotide excision -nonhomologous end joining -photoreactivation -synthesis-dependent strand annealing

nucleotide excision AND photoreactivation

The trp and lac operons are regulated in various ways. How do bacteria regulate transcription of these operons? Sort the statements into the appropriate bins depending on whether or not each operon would be transcribed under the stated conditions.

operon is not transcribed: trp operon: tryptophan present lac operon: lactose absent operon is transcibed, but not sped up through positive control: trp operon: tryptophan absent lac operon: lactose present, glucose present operon is transcribed quickly through positive control: lac operon: lactose present, glucose absent The trp operon is regulated through negative control only. When tryptophan is present, the operon genes are not transcribed. The lac operon is regulated through both negative control and positive control. Negative control:When lactose is absent, the repressor protein is active, and transcription is turned off.When lactose is present, the repressor protein is inactivated, and transcription is turned on. Positive control:When glucose is absent, another regulatory protein (CAP) binds to the promoter of the lac operon, increasing the rate of transcription if lactose is present.

A pair of homologous chromosomes in Drosophila has the following content (single letters represent genes): Chromosome 1 RNMDHBGKWU Chromosome 2 RNMDHBDHBGKWU What term best describes this situation? -partial duplication -partial deletion -unequal crossover -None of the above.

partial duplication

For the cross in Part E, in which parent and during what meiotic division might nondisjunction have occurred? Select all that apply. -maternal meiosis I -maternal meiosis II -paternal meiosis I -paternal meiosis II

paternal meiosis I AND paternal meiosis II

For the cross in Part C, in which parent and during what meiotic division might nondisjunction have occurred? Select all that apply. -maternal meiosis I -maternal meiosis II -paternal meiosis I -paternal meiosis II

paternal meiosis II

Which repair process in E. coli uses visible light to repair thymine dimers? -base excision repair -homologous recombination (synthesis-dependent strand annealing) -nucleotide excision repair -photoreactivation repair -nonhomologous end joining

photoreactivation repair Photoreactivation utilizes the enzyme photolyase to break the bonds formed during pyrimidine dimerization. A thymine dimer produced by UV irradiation is bound by photolyase. Visible light energy is absorbed by photolyase and is redirected to break the bonds forming the dimer. In addition to bacteria, this DNA repair mechanism is also found in single-celled eukaryotes, plants, and some animals such as Drosophila.

Identify two mechanisms that can correct the kind of abnormality resulting from the circumstances identified in part (c). Select the two correct answers. -proofreading by DNA polymerase -nucleotide base excision repair -double-strand break repair -photoreactive repair -mismatch repair -nucleotide excision repair

proofreading by DNA polymerase AND mismatch repair

Ultraviolet (UV) radiation is mutagenic. What kind of DNADNA lesion does UV energy cause? -induction of DNADNA strand breaks -pyrimidine dimerization -deamination -frameshift mutation

pyrimidine dimerization

Which type of chromosome alteration results in the formation of a tetravalent (cross-like) structure at synapsis? -large duplication -reciprocal balanced translocation -paracentric inversion -pericentric inversion

reciprocal balanced translocation

Do you consider the second mutation to be a forward mutation or a reversion? -forward mutation -true reversion -intragenic reversion -second-site reversion (suppressor)

second-site reversion (suppressor)

What can you say about colony 4? Select the two correct answers. -temperature sensitive mutation -temperature resistive mutation -adenine auxotrophic mutation -adenine prototrophic mutation

temperature sensitive mutation AND adenine auxotrophic mutation

The mRNA that is transcribed from the trpL region can alternately fold to form three possible stem-loop structures. Different stem-loop structures have varying effects on transcription. Which stem-loop structure is responsible for terminating transcription? the 1-2 stem loop the 2-3 stem loop the 3-4 stem loop

the 3-4 stem loop The formation of stem loops of the trpL region is directly related to the continuation or termination of transcription. In trpL, region 1 is complementary to region 2, region 2 is complementary to region 3, region 3 is complementary to region 4. The 3-4 stem loop is the termination stem loop, signaling transcription termination. Formation of the 3-4 stem loop halts RNA polymerase in the leader region before it reaches the structural genes. Region 4 is followed by a poly-uracil sequence; this configuration is the same as that in intrinsic termination of transcription in bacteria.

Which of the following stem loop structures of the trpL attenuator region is directly involved in transcriptional termination of the trp operon? -the 1-2 stem loop -formation of- the 1-4 stem loop -formation of- the 2-3 stem loop -the 3-4 stem loop

the 3-4 stem loop This structure is followed by a polyU tail, which promotes separation of the mRNA transcript from the DNA template.

The role that p53 protein plays in suppressing inappropriate progression through the cell cycle depends on all of the following EXCEPT _______. -p53 stability increases in the presence of unrepaired DNA lesion -sp53 activates transcription of WAF1 -p53 indirectly blocks G1 to S transition in the cell cycle -the ability of p53 to bind DNA lesions

the ability of p53 to bind DNA lesions p53 protein does not bind DNA lesions.

Two populations of deer, one of them large and living in a mainland forest and the other small and inhabiting a forest on an island, regularly exchange members who migrate across a land bridge that connects the island to the mainland. If you compared the allele frequencies in the two populations, what would you expect to find? -the allele frequencies of both populations are completely different. -the allele frequencies of both populations are likely to be similar. -the allele frequencies of both populations are exactly the same.

the allele frequencies of both populations are likely to be similar.

Why does deletion D effectively eliminate transcription of lacZ? -the deletion removes the promoter -the deletion removes a silencer -the deletion removes an enhancer element -the deletion removes a part of coding region

the deletion removes the promoter

Two different mutations affect PREPRE. Mutant 1 decreases transcription from the promoter to 10% of normal. Mutant 2 increases transcription from the promoter to tenfold greater than the wild type. How will the mutation affect the determination of the lytic or lysogenic life cycle in mutant 1 phage strains? -the mutation will make it difficult to reverse lysogeny -the mutation will inhibit lysogeny establishment -the mutation will have no effect

the mutation will inhibit lysogeny establishment

The reason some cells respond to the presence of a steroid hormone while others do not is that _______. -only certain cells contain the gene that is targeted by a given steroid hormone -chaperone proteins block the hormone response elements (HREs) in some cells -the specific HRE is present only in certain cells -the receptors necessary for regulation differ among cells of various types

the receptors necessary for regulation differ among cells of various types The specificity of steroid hormone regulation is due to the presence or absence of the receptor in different cell types.

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

the smaller of the two reciprocal products of translocated chromosomes is lost After several cell divisions, only the larger of the two translocated chromosomes remains. Its reciprocal is lost.

How many different secondary structures (hairpins) may be formed by the trp leader mRNA? one two three four

three The trp leader mRNA may form 3-4, 2-3, and 1-2 stem loops. See Figure 14.16.

Describe the purpose of the Ames test. -to determine whether ultraviolet light can act as a mutagen -to determine whether a chemical can act as a mutagen -to determine whether a transposon can act as a mutagen -to determine whether a metal can act as a mutagen

to determine whether a chemical can act as a mutagen

What is the primary function of histone H1? -to stabilize the histone octamer -to stabilize the 30-nm solenoid -to stabilize the 10-nm fiber -to attach chromatin loops to the nuclear scaffold

to stabilize the 30-nm solenoid

If the kind of abnormality identified in part (c) is not corrected before the next DNA replication cycle, what kind of mutation occurs? -transition mutation -deletion mutation -frameshift mutation -insertion mutation

transition mutation

Following the spill of a mixture of chemicals into a small pond, bacteria from the pond are tested and show an unusually high rate of mutation. A number of mutant cultures are grown from mutant colonies and treated with known mutagens to study the rate of reversion. Most of the mutant cultures show a significantly higher reversion rate when exposed to base analogs such as 2-aminopurine. What does this suggest about the nature of the chemicals in the spill? -transversion-induced mutagens -frameshift-induced mutagens -base-pair substitution-inducing mutagens -transition-inducing mutagens

transition-inducing mutagens

What type(s) of mutations are likely produced by chemical #2? Select all that apply. frameshifts transitions transversions

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

The purple spots on colorless background of corn kernels observed by Barbara McClintock were the result of _______. -mutation in the transposase gene -transposition of the Ds elements out of the color gene -environmental factors -transposition of the Ds elements into the color gene

transposition of the Ds elements out of the color gene Upon loss of Ds from the color gene "C" sequence, wild-type (purple) phenotype results in cells in which the loss has occurred.

What type(s) of mutations are likely produced by chemical #4? Select all that apply. transversions frameshifts transitions

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

If you were to make a partial diploid with the genotype trp R+ trpOc / trpRs trpO+ , what would happen? What is the dominance relationship between trpOc and trpRs ? -trp genes will not be expressed; trpO c is dominant to trpR s. -trp genes will not be expressed; trpR s is dominant to trpOc . -trp genes will be expressed constitutively; trpRs is dominant to trpO c. -trp genes will be expressed constitutively; trpO c is dominant to trpRs . -trp genes will be expressed normally (repressible); trpO c and trpRs cancel each other out.

trp genes will be expressed constitutively; trpO c is dominant to trpRs . The trpR s protein is a repressor that binds to the operator even when tryptophan levels are low. The trpOc mutation disrupts the sequence of the operator region so that it cannot be recognized by the repressor. Regardless of the conditions that trigger repressor binding (high typtophan in wild-type strains or any tryptophan levels in an Rs mutant strain), the repressor recognizes a specific sequence of DNA, the operator, to bind to DNA and stop transcription. If the operator sequence is mutated, repressor will be unable to bind regardless of how its regulated by tryptophan. In a trpR s trpOc mutant, expression will be constitutive because the repressor is unable to recognize the operator. Therefore, the trpOc mutation is dominant to the trpR s mutation.

What term best describes the unusual structure that forms during pairing of these chromosomes? -partial duplication -partial deletion -unpaired loop -None of the above.

unpaired loop