Genetics problem set 5
3) Consider the following segment of DNA, which is part of a much longer molecule constituting a chromosome: 5' - ...ATTCGTACGATCGACTGACTGACAGTC... - 3' 3' - ...TAAGCATGCTAGCTGACTGACTGTCAG... - 5' If the DNA polymerase starts replicating this segment from the right: a) Which will be the template for the leading strand? b) Draw the molecules when the DNA polymerase is halfway along this segment. c) Draw the two complete daughter molecules.
A- if replication is proceeding such that the DNA on the right is replicated first, then top strand is template for leading strand. B- draw C-Draw
2) A molecule is replicated in a solution of nucleotides. Specifically, the adenosine nucleotide has been labeled with 32P (no other nucleotides are radioactive). For the two molecules given below, answer for EACH: Will both daughter molecules be radioactive? A) 5' - AAAAAAAAAA - 3' 3' - TTTTTTTTTTT - 5' B) 5' - ATATATATAT - 3' 3' - TATATATATA - 5'
A- only one daughter would be radioactive; the molecule using the top strand as a template will only incorporate T, which is not radioactive. B- Both daughters would be radioactive, as both daughter strands will incorporate both A and T.
A cloned fragment of DNA was sequenced by using the Sanger Sequencing method. A part of the autoradiogram of the sequencing gel is represented here. a. Deduce the nucleotide sequence of the DNA nucleotide chain synthesized from the primer. Label the 5′ and 3′ ends. b. Deduce the nucleotide sequence of the DNA nucleotide chain used as the template strand. Label the 5′ and 3′ ends. c. Draw a the DNA molecule corresponding to band X, indicate which molecules are radioactively labeled with an astrisk (*). Label 5' and 3' ends. d. List the components of the reaction mixture that produced band Y. e. How long is the molecule corresponding to band Y? X Y
A- the gel can be read from bottom to top in 5' to 3' direction 5' TTCGAAAGGTGACCCCTGGACCTTTAGA 3' B-By complementarity, template was 3'AAGCTTTCCACTGGGGACCTGGAAATCT 5' C- 5' primer- TTCGAAAGGT*-3' D- DNA polymerase, buffer, template DNA, primer, dATP, dCTP, d GTP, radioactive ddCTP E- 22 nt plus primer. since we don't know the length of primer we cannot answer accurately.
In the following drawing, the top strand is the template DNA, and the bottom strand shows the lagging strand prior to the action of DNA polymerase I. The lagging strand contains three Okazaki fragments. The RNA primers have not yet been removed. (DNA nucleotides are indicated "-", RNA nucleotides are indicated by "**"). The top strand is the template DNA. A) Which Okazaki fragment was made first, the one on the left or the one on the right? B) Which RNA primer would be the first one to be removed by DNA Polymerase I, the primer on the left or the primer on the right? For this primer to be removed by DNA Polymerase I and for the gap to be filled in, is it necessary for the Okazaki fragment in the middle to have already been synthesized? Explain. C) Let's consider how DNA ligase connects the left Okazaki fragment with the middle Okazaki fragment. After DNA Polymerase I removes the middle RNA primer and fills in the gap with DNA, where does DNA ligase function? See the arrows on either side of the middle RNA primer. Is ligase needed at the left arrow, at the right arrow, or both?
A-The right Okazaki fragment was made first. it is the farthest away from the replication fork. the fork would be to the left of the three Okazaki fragments, and moving from right to left. B-The RNA primer in the right Okazaki fragment would be removed first. DNA polymerase would begin by elongating thte DNA strand of the middle Okazaki fragment and remove the right RNA primer with 5' to 3' exonuclease activity. DNA polymerase I would use 3' end of DNA of the middle Okazaki fragment as a primer to synthesize DNA in region where the right RNA primer is removed. If the middle fragment was not present, DNA polymerase could not fill in this DNA (bc it needs a primer). C- you need DNA ligase only at the right arrow. DNA polymerase I begins at the end of the left okazaki fragment and synthesizes DNA to fill in the region as it removess the middle RNA primer. at the left arrow, DNA pol I is simply extending the length of the left okazaki fragment. No ligase is needed here. When DNA poly I has extended the left Okazaki fragment through the entire region where the RNA primer has been removed, it hits the DNA of middle Okazaki fragment. This occurs at the right arrow. At this point, DNA of the middle Okazaki fragment has a 5' end that is a monophosphate. DNA ligase is needed to connect this monophosphate with the 3' end of the region where middle RNA primer has been removed.
Mutations that occur at the end of a gene may alter the sequence of the gene and prevent transcriptional termination. A. What types of mutations would prevent ρ-independent termination? B. What types of mutations would prevent ρ-dependent termination? C. If a mutation prevented transcriptional termination at the end of a gene, where would gene transcription end? Or would it end?
A. Mutations that alter the uracil-rich region by introducing Gs and Cs, and the mutations that prevent the formation of the stem-loop structure. B- mutations that alter the termination sequence and mutations that alter the p recognition site C- eventually, somewhere downstream from the gene, another transcriptional termination sequence would be found, and transcription would terminate there. this second termination sequence might be found randomly or at the end of an adjacent gene.
5) The compound known as nitrous acid is a reactive chemical that replaces amino groups (- NH2) with keto groups (=O). When nitrous acid reacts with the bases in DNA, it can change cytosine to uracil and change adenine to hypoxanthine. A DNA double helix has the following sequence: 5' - TTGGATGCTGG - 3' 3' - AACCTACGACC - 5' A) What would be the sequence of this double helix immediately after reaction with nitrous acid? Let the letter H represent hypoxanthine and U represent uracil. B) Let's suppose this DNA was reacted with nitrous acid. The acid was then removed, and the DNA was replicated for two generations. What would be the sequences of the DNA products after the DNA had replicated twice? Your answer should contain the sequences of four double helices. Note: During DNA replication, hypoxanthine hydrogen bonds with cytosine.
A. TTGGHTGUTGG HHUUTHUGHUU B. TTGGHTGUTGG HHUUTHUGHUU ↓ TTGGHTGUTGG CCAAACACCAA AACCCACAACC HHUUTHUGHUU ↓ TTGGHTGUTGG TTGGGTGTTGG CCAAACACCAA CCAAACACCAA AACCCACAACC AACCCACAACC GGTTTGTGGTT HHUUTHUGHUU
Although the capture and trading of great apes has been banned in 112 countries since 1973, it is estimated that about 1000 chimpanzees are removed annually from Africa and smuggled into Europe, the US and Japan. this illegal trade is often disguised by simulating births in captivity. Until recently, genetic identity tests to uncover these illegal activities were not used because of the lack of highly polymorphic markers (marks that vary from one individual to the next) and the difficulties of obtaining chimpanzee blood samples. A study was reported in which DNA samples were extracted from freshly plucked chimpanzee hair roots and used as templates for PCR. the primers used in these studies flank highly polymorphic sites in human DNA that result from variable numbers of tandem nucleotide repeats. Several offspring and their putative parents were tested to determine whether the offspring were "legitimate" or the product of illegal trading. The data are shown in the following Southern Blot. Examine the data carefully and explain which, if any, of the offspring are legitimate.
As diploid eukaryotes, each chimp contains two copies of this polymorphic locus. However, each of those loci may contain more or fewer copies of the repeat. The father chimp has one copy of the locus that is 148 nt long and another copy that is 136 nt long. The mother chimp also has has one copy that is 136 nt long and another copy that is 120 nt long. Chimp A (lane 3) appears to have two copies that are 136 nt long. This is possible from the parents (one 136 nt copy each). Chimp B (lane 4) appears to have one copy at 136 nt and another copy at 140 nt. The smaller copy could have come from either parent, but the longer copy could not have come from either of these parents. Therefore Chimp B is illegitimate. Chimp C (lane 5) appears to have two copies that are each 140 nt long. Neither of these could have come from the putative parents and therefore Chimp C is illegitimate.
A eukaryotic protein-encoding gene contains two introns and three exons: exon 1-intron 1- exon 2-intron 2-exon 3. The 5ʹ splice site at the boundary between exon 2 and intron 2 has been eliminated by a small deletion in the gene. Describe how the pre-mRNA encoded by this mutant gene would be spliced. Indicate which introns and exons would be found in the mRNA after splicing occurs. What are the ramifications to the translated peptide?
Only the first intron would be spliced out. The mature RNA would be: exon 1-exon 2- intron 2-exon 3. Intron 2 will be translated. This could include a stop codon (or not); it will likely throw off the frame of the rest of the protein.
What is the primary function of the sigma factor? Is there a protein in eukaryotes analogous to the sigma factor?
Sigma factor, as part of the RNA polymerase holoenzyme, recognizes and binds to the -35 and -10 regions of bacterial promoters. It positions the holoenzyme to correctly initiate transcription at the start site. In eukaryotes, TBP (TATA binding protein) and other GTFs (general transcription factors) have an analogous function.
10) Northern blotting can be used to detect RNA transcribed from a particular gene. In this method, a specific RNA is detected using a short segment of cloned DNA as a probe. The DNA probe, which is labeled, is complementary to the RNA that the researcher wishes to detect. After the probe DNA binds to the RNA, the RNA is visualized as a labeled (dark) band after the RNA is run on a gel. As shown here, the method of Northern blotting can be used to determine the amount of a particular RNA transcribed in a given cell type. If one type of cell produces twice as much of a particular mRNA as occurs in another cell, the band will appear twice as intense. Also, the method can distinguish if alternative RNA splicing has occurred to produce an RNA that has a different molecular mass. Lane 1 is a sample of RNA isolated from nerve cells. Lane 2 is a sample of RNA isolated from kidney cells. Nerve cells produce twice as much of this RNA as do kidney cells. Lane 3 is a sample of RNA isolated from spleen cells. Spleen cells produce an alternatively spliced version of this RNA that is about 200 nucleotides longer than the RNA produced in nerve and kidney cells. Let's suppose a researcher was interested in the effects of mutations on the expression of a particular protein-encoding gene in eukaryotes. The gene has one intron that is 450 nucleotides long. After this intron is removed from the pre-mRNA, the mRNA transcript is 1100 nucleotides in length. Diploid somatic cells have two copies of this gene. Make a drawing that shows the expected results of a Northern blot using mRNA from the cytosol of somatic cells, which were obtained from the following individuals: Lane 1: A normal individual Lane 2: A homozygote for a deletion that removes the -50 to -100 region of the gene that encodes this mRNA Lane 3: A heterozygote in which one gene is normal and the other gene has a deletion that removes the -50 to -100 region Lane 4: A homozygote for a mutation that introduces an early stop codon into the middle of the coding sequence of the gene Lane 5: A homozygote for a three-nucleotide deletion that removes the AG sequence at the 3ʹ splice site
The 1100-nucleotide band would be observed from a normal individual (lane 1). A deletion that removed the -50 to -100 region would greatly diminish transcription, so the homozygote would produce hardly any of the transcript (just a faint amount, as shown in lane 2), and the heterozygote would produce roughly half as much of the 1100-nucleotide transcript (lane 3) compared to a normal individual. A nonsense codon would not have an effect on transcription; it affects only translation. So the individual with this mutation would produce a normal amount of the 1,100-nucleotide transcript (lane 4). A mutation that removed the splice acceptor site would prevent splicing. Therefore, this individual would produce a 1,550-nucleotide transcript (actually, 1,547 to be precise, 1,550 minus 3). The Northern blot is shown here:
4) The DNA polymerases are positioned over the following DNA segment (which is part of a much larger molecule) and moving from right to left. If we assume that an Okazaki fragment is made from this segment, what will be the fragment's sequence? Label it's 5' and 3' ends. 5' - ...CCTTAAGACTAACTACTTACTGGGATC... - 3' 3' - ...GGAATTCTGATTGATGAATGACCCTAG... - 5'
The bottom strand will serve as a template for the okazaki fragment, so the fragment's sequence will be: 5' - ...CCTTAAGACTAACTACTTACTGGGATC... - 3'
6) A diagram of a linear chromosome is shown here. The end of each strand is labeled with an A, B, C, or D. Which ends could not be replicated by DNA polymerase? Why not? How are the replicated instead?
The ends labeled B and C could not be replicated by DNA polymerase. DNA polymerase makes a strand in the 5' to 3' direction using a remplate strand that is running 3' to 5' direction. also, DNA polymerase requires a primer. at the ends labeled B and C, there is no place for a primer to be made. telomerase will replicate these ends.
) In DNA footprinting, when a protein binds a region of DNA, it will protect the DNA in that region from digestion by DNase I. To carry out a DNA footprinting experiment, a researcher has a sample of a cloned DNA fragment. The fragments are exposed to DNase I in the presence and absence of a DNA-binding protein. Regions of the DNA fragment not covered by the DNA-binding protein will be digested by DNase I, producing a series of bands on a gel. Regions of the DNA fragment not digested by DNase I (because a DNA-binding protein is preventing DNase I from gaining access to the DNA) will be revealed, because a region of the gel will not contain any bands. In the DNA footprinting experiment shown here, a researcher began with a sample of cloned DNA 300 bp in length. This DNA contained a eukaryotic promoter for RNA polymerase II. For the sample loaded in lane 1, no proteins were added. For the sample loaded in lane 2, the 300-bp fragment was mixed with RNA polymerase II plus TFIID and TFIIB. A. How long of a region of DNA is "covered up" by the binding of RNA polymerase II and the transcription factors? B. B. Describe how this binding would occur if the DNA was within a nucleosome structure. (Note: The structure of nucleosomes is described in Chapter 10.) Do you think that the DNA is in a nucleosome structure when RNA polymerase and transcription factors are bound to the promoter? Explain why or why not.
The region of the gel from about 250 bp to 75 bp does not contain any bands. This is the region being covered up; it is about 175 base pairs long. B. In a nucleosome, the DNA is wrapped twice around the core histones; a nucleosome contains 146 bp of DNA. The region bound by RNA polymerase II plus TFIID and TFIIB would be slightly greater than this length. Therefore, if the DNA was in a nucleosome structure, these proteins would have to be surrounding a nucleosome. It is a little hard to imagine how large proteins such as TFIID, TFIIB, and RNA polymerase II could all be wrapped around a single nucleosome (although it is possible). Therefore, the type of results shown here makes it more likely that the DNA is released from the core histones during the binding of transcription factors and RNA polymerase II.
A gel retardation assay can be used to study the binding of proteins to a segment of DNA. When a protein binds to a segment of DNA, it retards the movement of the DNA through a gel, so the DNA appears at a higher point in the gel (see the following). Lane 1: 900-bp fragment alone Lane 2: 900-bp fragment plus a protein that binds to the 900-bp fragment In this example, the segment of DNA is 900 bp in length, and the binding of a protein causes the DNA to appear at a higher point in the gel. If this 900-bp fragment of DNA contains a eukaryotic promoter for a protein-encoding gene, draw a gel that shows the relative locations of the 900-bp fragment under the following conditions: Lane 1: 900 bp plus TFIID Lane 2: 900 bp plus TFIIB Lane 3: 900 bp plus TFIID and TFIIB Lane 4: 900 bp plus TFIIB and RNA polymerase II Lane 5: 900 bp plus TFIID, TFIIB, and RNA polymerase II/ TFIIF
When the 900 bp fragment is mixed with TFIID (lane 1), it would be retarded because TFIID would bind. When mixed with TFIIB (lane 2), it would not be retarded because TFIIB cannot bind without TFIID. Compared to lane 1, the 900 bp fragment would be retarded even more when mixed with TFIID and TFIIB (lane 3), because both transcription factors could bind. It would not be retarded when mixed with TFIIB and RNA polymerase (lane 4) because you do not have TFIID, which is needed for the binding of TFIIB and RNA polymerase. Finally, when mixed with TFIID, TFIIB, and RNA polymerase/TFIIF, the 900 bp fragment would be retarded a great deal, because all four factors could bind (lane 5).
1) Assume that a certain bacterial chromosome has one origin of replication. Under some conditions of rapid cell division, replication could start from the origin before the preceding replication cycle is complete. How many replication forks would be present under these conditions?
six: the first replication start would have two forks, and now the replicated origins (there are two) would each have 2 forks. total of six.