Genetics test 3 HW

Computer programmers, working with molecular geneticists, have developed computer programs that can identify genes within long stretches of DNA sequences. Imagine that you are working with a computer programmer on such a project. On the basis of what you know about the process of transcription, what sequences should be used to identify the beginning and end of a prokaryotic gene with the use of this computer program? - The start codon (AUG) indicates the first transcribed base. - Sequences with close similarity to the −25 TATA box are found near the beginning of a gene. - Sequences with close similarity to the −35 and −10 sequences upstream of start site. - Sequences with 10 to 20 repeats of A and T nucleotides mark the beginning of most genes. - Short, inverted repeats followed by at least seven T nucleotides indicating potential intrinsic termination sites. - At least five inverted repeats indicate the last transcribed nucleotide that indicates intrinsic termination site. - A stop codon indicates the last transcribed nucleotide.

- Sequences with close similarity to the −35 and −10 sequences upstream of start site. - Short, inverted repeats followed by at least seven T nucleotides mark the beginning of most genes

The enzyme telomerase is part protein and part RNA. What would be the most likely effect of a large deletion in the gene that encodes the RNA part of telomerase, and how would the function of telomerase be affected? - The catalytic center of telomerase, the polymerase, would be altered. - Telomerase would be unable to proofread its DNA product but would still synthesize telomeres. - The telomerase enzyme would be able to make only a few telomeric repeats before falling off. - Telomerase would be unable to correctly associate with telomeres. - Telomerase would lose the ability to synthesize new telomeric sequences to extend the telomere.

- Telomerase would be unable to correctly associate with telomeres. - Telomerase would lose the ability to synthesize new telomeric sequences to extend the telomere.

An analysis is performed to determine the proportions of each of the four nucleotide bases in the DNA of several tissue samples from various species. The results appear in the table. Which conclusions can be drawn from this data?

- The proportion of C is roughly equal to the proportion of G. - The proportion of T is roughly equal to the proportion of A. - The proportion of purines is roughly equal to the proportion of pyrimidines.

Identify the functions of an enhancer in transcription. - a cis‑regulatory element that increases gene transcription in specific tissues or cells - begins gene expression by binding to and disabling repressors - inhibits transcription by binding to the operator - a cis‑acting regulatory sequence of DNA that reduces levels of transcription - regulates transcription by providing a cluster of binding sites where transcription factors can bind

- a cis‑regulatory element that increases gene transcription in specific tissues or cells - regulates transcription by providing a cluster of binding sites where transcription factors can bind

Why can alternative splicing of messenger RNAs (mRNAs) be advantageous for eukaryotic organisms? - decreases the average gene length in eukaryotes - increases the variety of proteins that can be produced - splices together introns instead of exons from newly transcribed mRNA - allows for smaller genomes than bacteria - allows exons from two different genes to be spliced into a new mRNA

- increases the variety of proteins that can be produced

In eukaryotic gene regulation, RNA interference occurs through - the accumulation of mRNA that blocks transcription of the target gene by feedback inhibition. - the accumulation of mRNA that blocks transcription of nearby genes. - the action of microRNAs that block translation of specific mRNA molecules. - the action of RNA-protein complexes that inhibit translation by altering the three dimensional configuration of rRNA molecules.

- the action of microRNAs that block translation of specific mRNA molecules.

The bacteria Streptococcus pneumoniae has a virulent S strain and a nonvirulent R strain. The S strain is lethal to mice. The S strain contains a chemical factor that can transform the R strain to be virulent. The diagram shows a series of experiments conducted by injecting combinations of these strains into mice to identify the transforming factor. If DNA is the transforming factor, match the expected results of each experiment by placing the appropriate mouse image.

A: R, nonvirulent B: S, virulent C: Heat-treated S D: R strain + heat-treated S E: R + ht S with polysaccharides, lipids, RNA, and proteins destroyed F: R + ht S with polysaccharides, lipids, RNA, proteins, and DNA destroyed

What is the key feature of DNA that allows it to be copied? - the arrangement of chromosomes - the sugar-phosphate backbone - Okazaki fragments - complementary base pairing

Complementary base pairing

Consider the two molecules of DNA. Which two molecules of DNA has the lower melting temperature? Why?

DNA 1, because it has a lower percentage of G-C base pairs that stabilize DNA duplexes.

Identify the key structural features of a DNA molecule. Strong ionic bonds and hydrophobic interaction hold DNA together. DNA contains the nucleotide bases adenine, thymine, guanine, cytosine, and uracil. DNA bases are always paired purine with pyrimidine. DNA strands are antiparallel and include a 5′ end and a 3′ end. DNA is most often found as a left-handed helix, commonly referred to as A‑DNA. The backbone of DNA is made of a sugar and a phosphate molecule.

DNA bases are always paired purine with pyrimidine. DNA strands are antiparallel and include a 5′ end and a 3′ end. The backbone of DNA is made of a sugar and a phosphate molecule.

There are two types of nucleic acids, DNA and RNA. Nearly all organisms use DNA, not RNA, as the central repository for genetic information. Choose the statements that explain this phenomenon. DNA contains a hydroxyl group at the 2′ carbon. DNA contains guanine as one of its nitrogenous bases. DNA is more resistant against enzymes that break down nucleic acids. DNA has a double‑stranded structure that ensures an accurate mechanism of duplication. DNA is flexible and forms complex catalytic structures.

DNA is more resistant against enzymes that break down nucleic acids. DNA has a double‑stranded structure that ensures an accurate mechanism of duplication.

The end‑replication problem (telomere problem) exists in eukaryotic chromosomes and is characterized by the chromosomes shortening with each round of DNA replication. Select the statements that best explain why the end-replication problem exists in eukaryotic chromosomes. DNA polymerase synthesizes DNA from the 5′ end to the 3′ end. DNA ligase links the 5′ OH group of one fragment to the 3′ phosphate group of an adjacent fragment. DNA polymerase requires a primer for DNA synthesis. The lagging strand is synthesized from the hydroxyl end to the phosphate end. The RNA primer is removed in a 3′ to 5′ direction.

DNA polymerase synthesizes DNA from the 5′ end to the 3′ end. DNA polymerase requires a primer for DNA synthesis.

Label the diagram with the names of the three components of a nucleotide

Deoxyribose and phosphate group backbone with nitrogenous base

For the full‑length ovalbumin gene shown, where is the most likely location of the 5′ untranslated region on the DNA and RNA molecule? Where would the 3′ untranslated region be located on the DNA and RNA molecule?

Exon 1, Exon 8

Which of the following statements describes the function of the sigma factor in prokaryotic transcription? It catalyzes the removal of the introns to produce mature mRNA to terminate transcription. It facilitates the binding of RNA polymerase to the promoter to initiate transcription. It causes the RNA polymerase to detach from the mRNA to terminate transcription. It catalyzes the synthesis of mRNA in a 5' to 3' direction to implement transcription elongation. It attaches the amino acid to its cognate tRNA to initiate translation.

It facilitates the binding of RNA polymerase to the promoter to initiate transcription

Fredrick Wilson and his colleagues studied members of a large family who had low levels of magnesium in their blood (see the pedigree). They argued that this disorder of magnesium (and associated high blood pressure and high cholesterol) is caused by a mutation in mtDNA (F. H. Wilson et al., 2004, Science 306:1190−1194). What evidence suggests that a gene in the mtDNA is causing this disorder? Could this disorder be caused by an autosomal dominant allele? Select the answer or answers that correctly explain why or why not.

Males do not pass on the trait, but females pass it to offspring at a high frequency. - No, the pattern of females with the disorder passing the trait to their offspring at a relatively high frequency does not support autosomal inheritance. - No, males with the disorder do not pass the condition to their offspring, and therefore, the pattern is not consistent with autosomal inheritance.

Tom Vulliamy and his colleagues examined 15 families with autosomal dominant DKC (T. Vulliamy et al., 2004, Nature Genetics 36:447-449). They observed that the median age of onset of DKC in parents was 37 years, whereas the median age of onset in the children of affected parents was 14.5 years. Thus, DKC in these families arose at progressively younger ages in successive generations, a phenomenon known as anticipation. The researchers measured telomere length of members of these families; the measurements are given in the adjoining table. Telomere length normally shortens with age, and so telomere length was adjusted for age. Note that the age‑adjusted telomere length of all members of these families is negative, indicating that their telomeres are shorter than normal. For age‑adjusted telomere length, the more negative the number, the shorter the telomere. How do the telomere lengths of parents with DKC compare with the telomere lengths of their children with DKC?

Parents, on average, have longer telomeres than their children.

Which statement explains why the telomeres of people with DKC are shorter than normal? Which statement logically explains the anticipation shown by DKC, or why DKC arises at an earlier age in subsequent generations?

People with DKC have shortened telomeres due to a defective telomerase that cannot maintain telomeres during their own lifetimes. Children with DKC inherit shorter than average telomeres to begin with and, subsequently, the normal shortening of telomeres during replication would result in even shorter telomeres at an earlier gage.

Suppose the data in the table is collected for an unknown nucleic acid. Identify the unknown nucleic acid. single‑stranded RNA double‑stranded RNA single‑stranded DNA double‑stranded DNA

Single stranded RNA - Presence of uracil; C and G do not match so single strand

The Meselson and Stahl experiment starts with E. coli containing 15N/15N labeled DNA grown in 14N media. Which result did Meselson and Stahl observe by sedimentation equilibrium centrifugation to provide strong evidence for the semiconservative model of DNA replication?

The first generation has hybrid 15N/14N DNA and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed.

Suppose an explorer discovered a unique new species of plant and collected a sample for analysis by a geneticist. The geneticist then isolated chromatin from the plant sample and examined the chromatin using electron microscopy. The chromatin observed was organized in a way that is similar to beads on a string. A small amount of nuclease was then added, which was observed to cleave the string into individual beads that each contained 280 bp of DNA. After thorough digestion with more nuclease, only a 120 bp fragment of DNA remained attached to a core of histone proteins. Analysis of the histone core revealed the data indicated in the table. Using the data provided, determine the conclusions that the geneticist could make about the likely structure of the nucleosome in the chromatin of this new plant species.

The nucleosome core contains two each of histones H2A, H2B, and H4. The nucleosome core contains one each of histones H1 and H7. - Typical nucleosome comprised of 8 total histone cores: 2 of histones H2A, H2B, H3, and H4, and one H1 - This plant lacks H3 and has new H7 - Small percentage consistent with single histone

The example shows the splicing of a hypothetical, premature mRNA (pre‑mRNA) transcribed from the gene for yellow fluorescent protein (YFP). Two different YFP mRNAs, isoform A and isoform B, are produced by alternative splicing. What are the possible outcomes of the alternative splicing shown?

The protein translated from isoform‑A mRNA possesses an additional functional domain. The protein translated from isoform B is stable but lacks a functional domain. Isoform‑A mRNA is degraded faster than isoform‑B mRNA is.

A strain of bacteria possesses a temperature‑sensitive mutation in the gene that encodes the sigma factor. The mutant bacteria produce a sigma factor that is unable to bind to RNA polymerase at elevated temperatures. What effect will this mutation have on the process of transcription when the bacteria are raised at elevated temperatures? - Transcription will initiate but will not be able to transcribe to completion. - Transcription initiation will not occur normally at the elevated temperature. - Transcription will not be affected at the elevated temperature. - All transcription will immediately cease at the elevated temperature. - Transcription that begins prior to the temperature shift will be completed. - The rate of transcription will increase at the elevated temperature.

Transcription initiation will not occur normally at the elevated temperature. Transcription that begins prior to the temperature shift will be completed.

As DNA is replicated, both continuous and discontinuous replication occur. Discontinuous replication is the result of which specific feature of DNA?

antiparallel strands

Histone proteins - release bound DNA to enable nuclear division during mitosis. - enable mitochondrial DNA to replicate with nuclear DNA before mitosis. - attach to DNA and form compacted DNA-protein associations. - separate DNA into chromosomes at the beginning of mitosis.

attach to DNA and form compacted DNA-protein associations

Which description applies to alternative mRNA splicing? - protein modifications such as addition of a functional group or structural changes such as folding - heritable changes in gene expression that occur without altering the DNA sequence - processing of exons in mRNA that results in a single gene coding for multiple proteins - a gene cluster controlled by a single promoter that transcribes to a single mRNA strand - mRNA modifications such as additions of a 5′‑cap and 3′ poly‑A tail and removal of introns

processing of exons in mRNA that results in a single gene coding for multiple proteins

DNA has unique properties that allow it to accurately retain genetic information, even after multiple rounds of replication. One aspect of DNA that allows it to accurately store genetic information is the base pairing from Chargaff's first rule of the four nucleotide bases. If the A content of a DNA molecule is 22%, what are the percentages of the remaining bases?

t = 22 G and C both 28