Problem Set 2 Flashcards
(23 cards)
Four auxotrophic mutant strains were isolated from wild type yeast. These strains responded to the addition of certain nutritional supplements to minimal culture media either by growth (+) or no growth (-). The table below gives the growth patterns for single-gene mutant strains. Diagram a biochemical pathway that is consistent with the data, indicating where in the pathway each mutant strain is blocked.
Supplements Added B A R T S 1 + - + - - 2 + + + + - 3 + - + + - 4 - - + - -
S—>A—>T—>B—>R would be the chemical pathway.
Mutations create roadblocks at specific steps in the pathway. Adding supplements downstream from the roadblock allow growth to occur. If supplements are added upstream, the roadblock is still in place and growth will not occur. Mutant one would interrupts the enzyme between T and B, while mutant two will affect the enzyme between S and A. Mutant three would create a block between A and T while mutant four
affects the enzyme between B and R.
In your own words, describe the central dogma of genetics.
Varies.
At the molecular level, a gene is defined as a segment of DNA that is used to make a functional product, either an RNA molecule or a polypeptide. The first step in this process is called transcription, which refers to the process of synthesizing RNA from a DNA template. When a protein-encoding gene is transcribed, the first product is an RNA molecule known as messenger RNA (mRNA). During polypeptide synthesis—a process called translation—the sequence of codons within the mRNA determines the sequence of amino acids in a polypeptide.
One or more polypeptides then assemble into a functional protein. DNA replication provides a
mechanism for copying that information so it can be transmitted to new daughter cells and from
parent to offspring.
Compare DNA and RNA with regard to structure, function, location, and stability
These include the difference in the
sugar, the presence/absence of 2’ OH which distinguishes ribose from deoxyribose.
RNA has uracil while DNA carries thymine.
DNA is double stranded and helical. RNA is usually single stranded but internal base pairing allows the formation of secondary structural elements.
In eukaryotes, the DNA is found in the nucleus or mitochondria or chloroplast. RNA is largely found in the cytoplasm of eukaryotes but is synthesized inside the nucleus or other organelles, allowing it to be found there as well.
Double stranded DNA is much more stable than single stranded RNA.
The different types of RNA have different levels of stability. The most stable RNA is rRNA…. whose function is to provide the machinery for translation. RNA are a bit less stable, and serve the purpose of carrying amino acids to the ribosome and decoding the information in mRNA to determine the order of the amino acids in the synthesized protein. The least stable are the mRNA…. Which carry the coded instructions for the assembly of amino acids to
What is the substrate of transcription?
The substrate would be NTP.
What parts of DNA make up a transcription unit? Draw a typical prokaryotic transcriptional unit and identify its parts.
The prokaryotic gene is made of two strands of DNA, template and nontemplate. Transcription begins at +1 (the transcriptional start site, TSS), copying base information from the template into the RNA being made. The TSS is identified because of its proximity to the promoter. The promoter includes two consensus sequences, the -10 and -35 boxes that are recognized by the sigma factor, which binds with great specificity and enhanced affinity. Transcription continues until the terminator, which always includes an inverted repeat in the DNA which results in the formation of a stem loop when copied into the RNA
What is the role of the following in prokaryotic transcription: promoter, RNA polymerase, rho factor, messenger RNA, sigma factor, holoenzyme, terminator loop?
Promoter is the DNA sequence upstream of the TSS that is recognized as the beginning point for the RNA polymerase holoenzyme to bind during initiation. The core RNA polymerase is the enzyme responsible for pulling apart the two strands of DNA and using one strand as a template. The polymerase makes phosphodiester bonds linking NTP together in a chain to form the new RNA. Rho factor is the protein that aids in termination during rho dependent termination. Some genes require rho to bind to the rut site on the RNA and slide down until transcription pauses (due to terminator). Rho has helicase activity and is able to pull the RNA transcript away from the DNA template, ending transcription. mRNA is one flavor of RNA
produced from the transcription of structural genes. mRNA carries the codons that will ultimately get translated. Sigma factor is a protein that binds together with the core polymerase to form the holoenzyme. Because of the sigma factor, the holoenzyme has greater affinity for DNA than just the core and it has greater specificity, finding to the promoter exclusively. The terminator loop forms in the RNA after the inverted repeat at the end of the gene is transcribed. This secondary structure prevents the RNA from being able to exit through the exit channel of the polymerase, tying a knot that results in pausing of transcription. This begins the termination process regardless of whether rho is needed or not to finish termination.
Why does sigma factor protein have to fit into the major groove in order to carry out its function?
Parts of sigma factor must fit into the major groove so it can recognize a base sequence of a promoter, especially the -35 box sequence.
Researchers have identified a mutant form of sigma factor that binds more tightly to the core polymerase enzyme compared to a normal sigma factor. Would the rate of transcription be faster or slower in cells that carry this mutant sigma factor? Explain why
If the sigma factor is more likely to be bound with the core polymerase, there should be more copies of the holoenzyme ready to bind to the promoter and initiate transcription. Therefore, the rate of initiation should be greater. Alternatively, this mutation may inhibit
transcription because it would inhibit the transition to the elongation phase, preventing the removal of sigma.
Look at the examples of promoter sequences shown below and compare them to the consensus sequence, which is the most efficient promoter sequence. Based on this comparison, which positions of the –35 sequence (i.e., first, second, third, fourth, fifth, and/or sixth) are least tolerant of changes? Do you think that those positions play a more or less important role in the binding of pi (sigma factor)? Explain why.
Consensus Sequence: TTAGCA Gene 1 TTGGCA Gene 2 TTAAGA Gene 3 TCGGCA Gene 4 TTAAGA Gene 5 CTAGCA Gene 6 TTGAGA Gene 7 TTAGCG Gene 8 TTGGCA Gene 9 TTTGGA Gene 10 TTAGCA
The most highly conserved positions (i.e., those that are least tolerant of substitutions) are the first, second, and sixth. In general, when particular sites within promoter sequences are conserved, they are more likely to be important for binding. That explains why changes are not found at these positions; if a mutation altered a conserved position, the promoter would probably not work very well. By comparison, changes are tolerated occasionally at the fourth position and frequently at the third and fifth positions. The positions that tolerate changes are less important for binding by sigma factor.
What would be the most likely effect of a mutation at the following locations in an E. coli gene? A. -8 B. -35 C. -20 D. +1
Exchanging a T for A at -8 is not likely to have much of an effect but substituting a G for a T may make it more difficult to pull the two strands apart. At -35, base substitutions will make it more difficult for sigma factor to recognize the promoter. Base substitutions at -20 will have no effect as this falls in between the two consensus sequences. However, an indel at this position may disrupt the spatial distribution of the consensus sequences, inhibiting transcription. Mutations at +1 will not have an effect on the initiation of transcription but will alter the first nucleotide encoded into the RNA.
Changes in the sequences of bacterial promoters may increase or decrease the level of gene transcription. Promoter mutations that increase transcription are called up promoter mutations, whereas those that decrease transcription are down promoter mutations. Figure 12.5 shows the sequence of the –10 region of the promoter for the lac operon, which is TATGTT. The consensus sequence at -10 for all genes is TATAAT, which is the most efficient promoter sequence. Would you expect the following mutations to be up promoter or down promoter mutations?
A. TATGTT to TTTGTT
B. TATGTT to TATATT
C. TATGTT to TATGAT
Mutations that make a sequence more like the consensus sequence are likely to be up promoter mutations, while mutations that cause the promoter to deviate from the consensus sequence are likely to be down promoter mutations. Also, in the –10 region, AT pairs are favored over GC pairs, because the role of this region is to form the open complex. AT pairs are
more easily separated, because they form only two hydrogen bonds compared to GC pairs, which form three hydrogen bonds.
A. Down promoter
B. Up promoter
C. Up promoter
For the following DNA sequences, what is the consensus sequence? If two or more bases are equally frequent at a given position, put those bases in parentheses.
TGACTGCCAT TGTCAGCCAT AGTCACGCAT TGGGAGCAAA TGTCAGCCTA TGTCAGGCAA
TGTCAGCCA(T or A)
For the following DNA sequences, what is the consensus sequence? If two or more bases are equally frequent at a given position, put those bases in parentheses. TGCCATTGAC AGCCATTGTC ACGCATAGTC AGCAAATGGG AGCCTATGTC AGGCAATGTC
AGCCA(T or A)TGTC
A strain of bacteria possess a temperature sensitive mutation in the gene that encodes the rho subunit of RNA polymerase. At high temepratures rho is not functional. When these bacteria are raised at elevated temperatures, which of the following effects would you expect to see? For each, explain why or why not.
A. Transcription does not take place
B. All RNA molecules are longer than normal
C. Some RNA molecules are longer than normal
D. Some RNA molecules are longer than normal
E. RNA is copied from both strands
transcription will still take place, because a mutation in rho will not influence initiation or elongation stages. Not all RNA will be longer than normal since some genes use rho independent termination and are not influenced by mutant rho at all. Those genes that are rho dependent for termination will produce RNA that are longer than normal. No RNA will be
shorter than normal because the mutant rho is not causing early termination. RNA is copied from only one strand at a time. A mutant rho would not.
The following DNA nucleotides are found near the end of a bacterial transcription unit. Find the terminator in this sequence.
3’ – AGCATACAGCAGACCGTTGGTCTGAAAAAAGCATACA – 5’
Where will transcription terminate. Is this a rho independent or dependent terminator?
When transcribed, the sequence highlighted in blue will internal pair with the sequence highlighted in green. The GUU in-between will form the loop of the stem loop structure from this inverted repeat. The string of A’s in the DNA will be paired with U’s in the RNA, making this a rho independent terminator. Transcription will end at the end of the string of A/U base pairs.
Compare eukaryotic and prokaryotic initiation of transcription. How are they similar and how are they different?
They are similar in that both require the binding of RNA polymerase to the promoter sequence in order for initiation to occur. Both use other proteins to guide the polymerase into place either the sigma factor or general transcription factors.
They are different in that eukaryotes use a
different polymerase for each type of RNA (mRNA, tRNA, rRNA) and the initiation process is a bit
different for each, with their own unique promoter sequences. Eukaryotic promoters are larger and
more complicated than prokaryotic promoters are. Prok promoters are limited to the -35 and -10
consensus sequences while euk promoters have many different consensus sequences that make up the core promoter and many others that make up the regulatory promoter plus enhancer sequences.
Draw a typical eukaryotic gene, pre-mRNA, and mature mRNA. Assume the gene contains three exons.
Identify the following items. A. AAUAAA consensus sequence B. Promoter C. Transcription start site D. Introns E. Exons F. Poly A tail G. 5’ cap
A. AAUAAA consensus sequence This sequence is found towards the end of the primary RNA transcript and serves as the cleavage signal. Once it emerges from the polymerase a series of proteins bind to this sequence and a sequence a little downstream. A cut is made (cleavage) and one of the proteins then begins to synthesize the poly A tail on the end of the
mRNA
B. Promoter Sequences towards the 5’ end of the gene that serve as the assembly point of the
RNA polymerase. Composed to two main parts in euks, the core promoter and the regulatory promoter. General transcription factors bind to consensus sequences in the core promoter to direct the polymerase to bind at the promoter site. However this basal transcriptional apparatus rarely initiates on its own. Other proteins (activators) must bind
to the regulatory promoter or enhancer sequences (even further upstream) and make contact with the mediator complex to initiate transcription.
C. Transcription start site The first nucleotide copied into the RNA from the DNA template
D. Introns Intervening sequences that are found within the primary transcript but do not carry
relevant information to determine the order of amino acids. During splicing the snurps will recognize the 5’ and 3’ splice sites and cut the introns out of the transcript and discard them.
E. Exons Expressed sequences. These sequences are retained and glued together by the snurps at the end of splicing. They will make up the mature mRNA at the end of splicing and will contain the ORF needed for translation.
F. Poly A tail Found at the 3’ end of the mature mRNA. Added after cleavage of the transcript away from the gene. Used for ribosomal recycling.
G. 5’ cap Also added to the mature mRNA after transcription. A guanosine residue is added to
the 5’ end of the mRNA using an odd covalent bond. Methyl groups are typically added to nucleotides at the end of the RNA as well. This cap will serve as an important binding point during the initiation of translation.
Your goal is to achieve a high level of transcription—a level similar to its in vivo levels—of a eukaryotic gene in an in vitro transcription system. You add the DNA template for this gene, which contains the following, to the in vitro transcription system: a TATA box sequence located –25 to –30 bp upstream of transcription start site, the sequence for the gene, and terminator sequences. How would you interpret the result if you do not get high levels of RNA transcription, and what corrective measures would you take to achieve your goal?
You may not have enough promoter sequence to get high levels of transcription.
Some general transcription factors bind further upstream than the TATA box and help to guide the
polymerase onto the promoter. The lack of a regulatory promoter and enhancer may also inhibit
transcription as there is no place for activators to bind and stimulate transcription. The solution would
be to include more promoter elements to increase the rate of transcription.
Gene X encodes a pre-mRNA that has 6 exons and 5 introns. A mutation in gene X removes the 3’ splice site from intron 3. After splicing is completed, what exons would be found in the mature mRNA?
This mutation would cause exon skipping. In this case, exon 4, which is downstream from the 3’ splice site of intron 3, would be spliced out. So the mRNA would have exons 1-2-3-5-6.
A geneticist discovers that two different proteins are encoded by the same gene. One protein has 56 amino acids and the other has 82 amino acids. Provide a two possible explanations for how the same gene can encode both of these proteins.
Either alternative splicing (putting exons together in different combinations) or alternative cleavage (recognizing two different cleavage signals resulting in mRNA of different lengths and different sequences could also produce mRNA of different lengths and different content.
Use the mRNA sequence shown below to answer the following questions.
mRNA: 5’-ACUGGACAGGUAAGAAUACAACACAGUCGGCACCACG 3’
a. Underline the 5 and 3 splice sites, then write the sequence of the spliced mRNA in the space provided. b. Predict what would happen if the G in the 5 splice site were mutated to a C.
a. ACUGGACAGUCGGCACCACG
b. Longer RNA since splicing is ignored
Besides initiation of transcription, compare and contrast the process of transcription in prokaryotes and eukaryotes.
Differences include that eukaryotes go through many more types of processing, including the addition of the MG cap, the cleavage towards the 3’ end of mRNA, and the addition of the poly A tail to eukaryotic mRNA. Eukaryotes use a variety of mechanisms to bring about termination of transcription, depending upon what type of RNA is being synthesized. Euk tRNA terminates after a string of U’s are transcribed, sort of like rho independent termination in proks…. But without the need of the stem loop created by the inverted repeat. Euk rRNA termination depends upon a termination protein that binds to a sequence in the RNA transcript… sort of like rho but without the inverted repeat. Euk mRNA terminates after the transcript has been cleaved away. The remaining RNA is degraded by an enzyme that attacks RNA that lack 3 phosphates on their 5’ end. You could also mention that eukaryotic transcription occurs inside the nucleus or the mito/chloro. Prokaryotic transcription occurs in the cytoplasm of the cell. This allows transcription and translation to be coupled in prokaryotes (occur simultaneously)
