Gen Gen Test Uno
In a DNA nucleotide, to which carbon on deoxyribose is the nitrogenous base attached?
1'
The letters on the gel are indicating bands (the letter is marking the band BELOW the letter). What is the size of the band indicated by the letter (B)?
8,000 base pairs
Examine the diagrams of cells from an organism with diploid number 2n=62n=6, and identify what stage of MM phase is represented.
A) Anaphase I of meiosis. B) Metaphase of mitosis. C) Telophase of mitosis. D) Anaphase of mitosis.
Based on the results of the copolymer experiment, what triplet code can definitely be assigned to valine?
Both copolymers have the GUG codon and only have Valine in common.
A dihybrid has the genotype AaBb. The diagram depicts the chromosome composition of different cells from the animal. Which of the cells could represent a mature gamete?
Cell G
What consensus sequences are detected in the mammalian ββ -globin gene promoter
GC-rich region TATA box CAAT box
Which amino acids are encoded by the UG copolymer?
The two codons present in the UG copolymer are UGU which codes for cysteine and GUG which codes for valine.
Dark G bands that appear along chromosomes contain ...
heterochromatic regions.
Once elongation is underway, tRNAs involved in the process occupy a series of sites on the complexed ribosome. The occupation of sites occurs in the following order.
A Site, P Site, E Site tRNAs associate with sites on the ribosome in the order listed.
What conclusion(s) could Griffith draw from his experiment? Select all that apply.
A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent. From this experiment, Griffith concluded that a transforming factor from the heat-killed S strain transformed the R strain, making the R strain virulent. The experiment showed that there was a transforming factor, but it didn't demonstrate the nature of the transforming factor.
Which of the following statements is INCORRECT regarding how agarose gel electrophoresis works? See Section 25.1 (Page 781) .
Agarose gel electrophoresis can separate fragments that differ in length by as few as 10 bp. Correct. Agarose gel electrophoresis is a laboratory method for separating proteins or nucleic acid molecules or fragments using electrical current in a gel matrix. For a standard agarose gel electrophoresis, a 0.8% agarose gel gives good separation or resolution of large 5-10kb DNA fragments, while a 2% agarose gel gives good resolution for small 0.2-1kb fragments.
The letters on the gel are indicating bands (the letter is marking the band BELOW the letter). Which DNA band has the lowest molecular weight?
C
Which of the following cells could be created from a mitotic division but would not be created during a normal meiosis?
Cell K
What is the first order of chromatin packing?
Coiling around nucleosomes
Identify the features that bacterial ribosomes and eukaryotic ribosomes have in common.
Common features of bacterial chromosomes and eukaryotic ribosome Consist of RNA and protein Consist of two subunits Have three tRNA binding sites Both have two subunit. Each of the subunits is made up of protein and RNA components. The small ribosomal subunit is made up of one ribosomal RNA and approximately twenty-one proteins in bacteria and approximately thirty-three proteins in eukaryotes. They have three tRNA binding sites.
Which of the following molecules is not required for a PCR reaction?
D. Ligase
What is the purpose of raising the temperature to 90-95°C at the beginning of each cycle of PCR?
D. To separate the double‑stranded DNA
This diagram shows an early stage of DNA replication. Can you name the protein represented by each icon in the diagram? Then, for each protein, can you identify how DNA replication would be affected if that protein were nonfunctional? First drag the name of the correct protein into each bin. Then drag the correct "If nonfunctional" description into each bin.
DNA helicase breaks the hydrogen bonds between DNA strands, separating them. Single-stranded binding proteins bind to the single-stranded DNA and prevent it from reannealing. Primase synthesizes a short RNA primer that is necessary for DNA polymerase III to begin synthesizing DNA. Topoisomerase relieves the supercoiling caused by unwinding the DNA.
DNA fragments that are 600 bp long will migrate more quickly through a sequencing gel than fragments that are 150 bp long.
FALSE Small DNA fragments have less hindrance in moving through the gel, so they migrate more quickly than larger fragments.
Identify the locations of the following structures on DNA and polypeptide. Drag the appropriate labels to their respective targets.
IN pic
Below is a diagram of a replication bubble. Label the three types of DNA fragments; then label the 5' and 3' ends of the indicated pieces of DNA. Drag the blue labels onto the diagram to identify the types of DNA fragments. Then use the pink labels to identify the 5' and 3' ends; pink labels can be used more than once.
New strands of DNA are synthesized from the 5' end to the 3' end. The leading strand synthesizes DNA towards the replication fork and the replication is continuous. The lagging strand is synthesized away from the replication fork and replication occurs in fragments called Okazaki fragments. An RNA primer is found at the 5' end of all newly synthesized DNA; this primer is required for DNA synthesis.
Which of the following would result from a third round of replication using the methods of Meselson and Stahl in semi-conservative replication?
One light band and one intermediate band: Of the molecules generated in the third round, 75% are completely light, 25% are intermediate.
CODIS analysis using STR markers can be used to exclude potential suspects from a crime. How many of the 20 CODIS STRs failing to match between a suspect and the crime scene are needed to EXCLUDE that suspect as a potential suspect? See Section 25.1 (Page 781) .
Only one mismatch is needed to exclude the suspect from the crime scene
A sample of double-stranded DNA is found to contain 19% cytosine. Determine the percentage of guanine in the sample Determine the percentage of adenine in the sample. Determine the percentage of thymine in the sample.
Out of 50% SO G=19 A and T=31
How does RNA participate in DNA replication?
RNA serves as a primer for elongation by DNA polymerase
Order the following events from first to last in initiation of translation in eukaryotes.
Recall that in the initiation of translation in eukaryotes, the 80S ribosome does not bind the mRNA as an entire complex, but rather it is assembled on the mRNA. The small 40S subunit with its corresponding rRNA are critical for the complex to recognize the the proper AUG start codon before the 80S ribosome is fully assembled.
What does the term "STR" stand for? See Section 25.1 (Page 781) .
The term "STR" stands for short tandem repeat. These are clusters of variable numbers of repetitive DNA sequences of less than 10 base pairs each that are variable among individuals and can be used as genetic markers
Which codons are present in the UGG copolymer?
UGGUGGUGGUGGUGG The possible codons are determined by repeating the copolymer. Within a UGGUGGUGG sequence of RNA, it is possible to have a UGG codon, a GGU codon and a GUG codon. It is not possible to have a UGU codon, because the copolymer is repeated UGG.
Which codons are present in the UG copolymer?
UGU GUG The possible codons are determined by repeating the copolymer. Within a UGUGUGUGUG sequence of RNA, it is possible to have a UGU codon and a GUG codon. It is not possible to have a UGG codon, because the copolymer is repeated UG, and there are never two G's together.
Which one of the following amino acids is encoded by both copolymers?
VALINE The UG copolymer has UGU and GUG codons. The UGG copolymer has UGG, GGU and GUG codons. Both copolymers have the GUG codon in them, which encodes valine.
Within each replication bubble, replication is ...
bidirectional and semi-conservative
How is the cap attached to the 5′ end of eukaryotic mRNAs
by an unusual 5′-5′ triphosphate linkage
In the process of eukaryotic pre-mRNA splicing, how is the lariat intermediate formed?
by joining the 5′ end of the intron to the 2′ carbon of the branch site Submit
After one round of replication, a single DNA band was observed in the cesium chloride gradient. Which replication model could be ruled out based in this data?
conservative: Correct. A single band after one round of replication SUPPORTED BOTH the semi-conservative and dispersive models, but DISPROVED the conservative model. Conservative replication would produce two different daughter DNAs - one made entirely of the original, parental DNA (both strands contain heavy 15N) and one made entirely of newly made, daughter DNA (both strands lighter 14N). Each of these DNA molecules would migrate at a different position in the cesium chloride gradient producing TWO bands after the first round.
During meiosis I, when does homologous chromosome pairing and recombination occur?
prophase I
Look at the sequence of the nucleic acids in the video. How are the sequences of the mRNA and the coding (non-template) DNA related?
same sequence with U instead of T same polarity
What is the name of the structure formed when the 10-nm fibers of chromatin form a cylindrical filament of coiled nucleosomes
solenoid structure
Determine which of the following sequences and structures represent part of mature eukaryotic mRNA
stop codon poly-A tail 3'-UTR start codon 5'-UTR 5'-cap AAUAAA
In the process of translational initiation in bacteria, where does the initiator tRNA first bind to the ribosome?
to the partial P site on the small ribosomal subunit
Mitosis is a process of cell division that results in
two identical daughter cells
3'-AUCGUCAUGCAGA-5' Where is the promoter region for this gene located?
upstream of the 5' end of the coding strand
If a male with Klinefelter syndrome undergoes spermatogenesis (without nondisjunction), how many chromosomes would be present in the sperm?
23 and 24 During meiosis I in an individual with Klinefelter syndrome, two sex chromosomes will migrate to one daughter cell and one sex chromosome will migrate to the other daughter cell. As a result, following meiosis II, two gametes will contain 2n+1 chromosomes, and the other two gametes will contain n chromosomes.
Complete the mRNA codons.
5'-A-A-C-A-U-A-U-G-U-G-A-A-G-G-C-G-A-G-A-A-U-G-A-A-C-G-A-3'
When do homologous chromosomes separate during meiosis
Anaphase I
Complete the polypeptide sequence using three letter abbreviations.
Asn-Ile-Cys-Glu-Gly-Glu-Asn-Glu-Arg Submit
Which of the following represents the contents of a cell in G2 phase (before cell division begins)?
Cell M
In the Hershey-Chase experiment, what was labeled by growing bacteriophage in 32P-containing medium?
DNA
What role do promoters play in transcription?
They determine the site and frequency of transcription initiation.
The second order of chromatin packing occurs when nucleosomes coil together to form a fiber that is 300 nm in diameter
false: The second order of chromatin packing occurs when nucleosomes coil together to form a solenoid fiber that is 30 nm in diameter.
What are the three parts of a DNA nucleotide?
phosphate group, deoxyribose, base
What type of bond is formed between the hydroxyl group of one nucleotide and the phosphate group of an adjacent nucleotide, forming the sugar-phosphate backbone of DNA?
phosphodiester bond
A sufficient amount of a small DNA fragment is available for dideoxy sequencing. The fragment to be sequenced contains 20 nucleotides following the site of primer binding: 5' - ATCGCTCGACAGTGACTAGC- [primer site]-3'Dideoxy sequencing is carried out, and the products of the four sequencing reactions are separated by gel electrophoresis. Locate the bands you expect will appear on the gel from each of the sequencing reactions. Drag the label "BAND" to its appropriate targets on the gel. Leave the other targets empty. Do not use the label "X"
stupid
Which type of chromosome has no p arms?
telocentric
In the Meselson-Stahl a heavy isotope of nitrogen (15N) was incorporated into the DNA of E. coli. Which part of the DNA incorporated the 15N?
the base
Which of the following proteins unwinds the DNA at the replication fork?
the enzyme helicase
The results of the Meselson-Stahl experiments relied on all of the following except
the fact that DNA is the genetic material
A tRNA anticodon is 5' GAA 3'. Answer the following questions. Which one of the following codons is recognized by this tRNA?
5' UUC 3' Anticodons and codons are present in RNA molecules that base-pair in an anti-parallel orientation just like DNA.
What is the anticodon triplet sequence of the next tRNA to interact with mRNA?
5'-GGC-3'
3'-AUCGUCAUGCAGA-5' What is the sequence of the template DNA strand for this mRNA sequence?
5'-TAGCAGTACGTCT-3' The template and coding strands are complementary to each other. The template is read in a 3' to 5' direction by the polymerase to make the mRNA.
In the electrophoresis gel below, what letter indicates the correct direction of DNA migration?
A
Which histone helps stabilize the solenoid structure?
H1: Note: 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.
Complete the polypeptide sequence using one-letter abbreviations.
N-I-C-E-G-E-N-E-R
Is coupling of transcription and translation possible in single-celled eukaryotes such as yeast?
NO
How many different RNA polymerase enzymes are required to transcribe the various classes of RNA in eukaryotic cells?
THREE
What results from the experiments of Frederick Griffith provided the strongest support for his conclusion that a transformation factor is responsible for heredity?
The fact that injection of a mixture of heat-killed SIII and living RII strains of Pneumococcus into mice caused the mice to die from pneumonia.
Which amino acids are encoded by the UGG copolymer?
The three codons present in the UGG copolymer are UGG which codes for tryptophan, GGU which codes for glycine and GUG which codes for valine.
Based on the following replication bubble, which of these statements is true?
W and Y are leading strands, X and Z are lagging strands
After observing the results of one round of replication, the scientists obtained results from a second round. The purpose of one additional round of replication was to _______.
distinguish between semi-conservative and dispersive replication: Correct. After one round of replication, the results of these two possibilities are indistinguishable. A second round was required to distinguish between these two possibilities.
Complete the tRNA anticodons.
3'-U-U-G-U-A-U-A-C-A-C-U-U-C-C-G-C-U-C-U-U-A-C-U-U-G-C-U-5'
The following portion of the 16S rRNA base pairs with the Shine-Dalgarno sequence found in mRNA.5'...CCUCCU...3'Which mRNA sequence would have perfect complementarity with this rRNA?
5'...AGGAGG...3' The 16S rRNA base pairs with the Shine-Dalgarno sequence 5'...AGGAGG...3' (in this example) to position the 70S ribosome at the start codon.
What is the common structure of a bacterial promoter with respect to consensus sequences
-35 sequence TTGACA -10 sequence TATAAT Promoters the sequence on bacterial cell which is recognised by RNA polymerase enzyme. RNA polymerase then comes and binds to the promoter to initiate transcription of the DNA. In bacteria the two most common promoters are TATA box which is also called as - 10 element or pribnow box and the other promoter is - 35 element.
In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B. Drag the labels to their appropriate locations in the flowchart below, indicating the sequence of events in the production of fragment B. (Note that pol I stands for DNA polymerase I, and pol III stands for DNA polymerase III.)
1-pol III binds to 3' end of primer B 2-pol III moves 5' to 3', adding DNA nucleotide to primer B 3-pol I binds 5' end of primer A 4-pol I replaces primer A with DNA 5-DNA ligase links fragments A and B
Which of the following statements about ddNTPs is true?
They have a hydrogen at the 3′ carbon of the sugar. ddNTPs terminate synthesis because there is no 3′‑hydroxyl group onto which DNA polymerase can add nucleotides.
Locate the following sequences and structures on the mRNA
in pic
As in most areas of biology, the study of mitosis and the cell cycle involves a lot of new terminology. Knowing what the different terms mean is essential to understanding and describing the processes occurring in the cell.
1. DNA replication produces two identical DNA molecules, called sister chromatid(s), which separate during mitosis. 2. After chromosomes condense, the centromere(s) is the region where the identical DNA molecules are most tightly attached to each other. 3. During mitosis, microtubules attach to chromosomes at the kinetochore(s). 4. In dividing cells, most of the cell's growth occurs during interphase. 5. The mitotic spindle(s) is a cell structure consisting of microtubules, which forms during early mitosis and plays a role in cell division. 6. During interphase, most of the nucleus is filled with a complex of DNA and protein in a dispersed form called chromatin. 7. In most eukaryotes, division of the nucleus is followed by cytokinesis, when the rest of the cell divides. 8. The centrosome(s) are the organizing centers for microtubules involved in separating chromosomes during mitosis.
Now can you predict nucleotide quantities in a hypothetical scenario with more than four different types of nucleotides? An alien species was discovered that has DNA comprised of 6 different bases: G binds to C; A binds to T: and X binds to Y. Approximately 16% of this alien genome is comprised of nucleotides containing C, and 22% is comprised of nucleotides containing Y. Given this information, calculate the percentage of the alien genome that is comprised of nucleotides containing G, T, A, and X. (a) _____% of the alien genome is comprised of G.(b) _____% of the alien genome is comprised of T. (c) _____% of the alien genome is comprised of A. (d) _____% of the alien genome is comprised of X. Enter your answer as four numbers, separated by commas (example: 12, 26, 32, 10).
16, 12, 12, 22 Nucleotide bases on one strand of DNA are complementary to the corresponding nucleotide bases on the other strand of DNA. For this alien organism, C binds with G, A binds with T, and X binds with Y. Because of this complementarity, you can calculate the amount of any nucleotide in a genome when the amounts of only two nucleotides are known. 16+12+22=50
Now that you have come up with an equation that describes the relationship between amounts of different nucleotide bases in DNA, can you use it to predict the amounts of all four nucleotide bases when you only know the amount of one type of base? Approximately 21% of the human genome is comprised of nucleotides containing C. Given this information, calculate the percentage of the human genome that is comprised of nucleotides containing G, T, and A. (a) _____% of the human genome is comprised of G. (b) _____% of the human genome is comprised of T. (c) _____% of the human genome is comprised of A. Enter your answer as three numbers separated by commas (example: 12, 26, 32).
21, 29, 29 Nucleotide bases on one strand of DNA are complementary to the corresponding nucleotide bases on the other strand of DNA. Cytosine (C) binds with guanine (G), and adenine (A) binds with thymine (T). Because of this complementarity, you can calculate the amount of any nucleotide in a genome when the amount of only a single nucleotide is known.
A tRNA anticodon is 5' GAA 3' The first base at the 5' end of the anticodon has relaxed base-pairing rules allowing some tRNAs to bind multiple codon sequences. This is called the third base "wobble" hypothesis. How many codons could a tRNA with the anticodon 5'-GAA-3' recognize?
2: The first base at the 5' end of the anticodon has relaxed base-pairing rules allowing some tRNAs to bind multiple codon sequences. This is called the third base "wobble" hypothesis. The 5' nucleotide of this tRNA anticodon is G, a purine (double-ring structure). Base-pairing rules requires a purine-pyrimidine or pyrimidine-purine match. This means that G could pair with either C or U, which are both pyrimidines. Therefore, there are 2 codona that this tRNA could recognize.
During transcription, RNA polymerase reads the template DNA in the ________ direction and makes the new RNA strand in the _____________ direction.
3' to 5'; 5' to 3'
Complete the template strand of DNA.
3'-T-T-G-T-A-T-A-C-A-C-T-T-C-C-G-C-T-C-T-T-A-C-T-T-G-C-T-5'
The extraordinary accuracy of the DNA polymerase III enzyme lies in its ability to "proofread" newly synthesized DNA, a function of the enzyme's
3-to-5 exonuclease activity.
Complete the coding strand of DNA.
5'-A-A-C-A-T-A-T-G-T-G-A-A-G-G-C-G-A-G-A-A-T-G-A-A-C-G-A-3'
Name the three processes that commonly modify eukaryotic pre-mRNA.
5′ capping, 3′ splicing and polyadenylation, and RNA splicing to remove introns before being transported to the cytoplasm where they are translated by ribosomes.
A gene is depicted in the image below. The polarity of DNA strands, transcription start site and direction of transcription are indicated on the figure. Which letter in the image is indicating the promoter region? Notice how the promoter you chose relates to the start site and direction of transcription.
A
What is the Shine-Dalgarno sequence?
A short conserved nucleotide sequence upstream of the AUG start codon that serves to align the mRNA on the bacterial small ribosomal subunit The Shine-Dalgarno sequence is a purine-rich sequence of six nucleotides localized three to nine nucleotides upstream of the AUG start codon.
Genes A and B are on different chromosomes. A dihybrid with the genotype AaBb undergoes meiosis. Which of the following depicts a possible genotype in a gamete?
AB
Which of the following forms the backbone of a polynucleotide chain?
Alternating sugar and phosphate groups joined by phosphodiester bonds form the sugar-phosphate backbone of a polynucleotide chain.
alternative promoters
Alternative promoter is that which can be important for the location of the promotor either upstream or to the downstream. So the meaning of the term alternative promoter also include a sequence that is used to direct transcription initiation at a different site in a gene.
The diagram below shows a replication bubble with synthesis of the leading and lagging strands on both sides of the bubble. The parental DNA is shown in dark blue, the newly synthesized DNA is light blue, and the RNA primers associated with each strand are red. The origin of replication is indicated by the black dots on the parental strands.
As soon as the replication bubble opens and the replication machinery is assembled at the two replication forks, the two primers for the leading strands (primers a and h) are produced. The production of the first primers on the lagging strands (those closest to the origin of replication, b and g) is delayed slightly because the replication forks must open up further to expose the template DNA for the lagging strands. After completion of the first segments of the lagging strands, additional template DNA must be exposed before the second primers (c and f) can be produced. And after completion of the second segments, additional template DNA must be exposed before the third primers (d and e) can be produced. In summary, because of the way the replication bubble expands, the lagging strand primers near the origin of replication were produced before the primers near the replication forks.
As DNA replication continues and the replication bubble expands, the parental double helix is unwound and separated into its two component strands. This unwinding and separating of the DNA requires three different types of proteins: helicase, topoisomerase, and single-strand binding proteins.
At each replication fork, helicase moves along the parental DNA, separating the two strands by breaking the hydrogen bonds between the base pairs. (This makes the two parental DNA strands available to the DNA polymerases for replication.) As soon as the base pairs separate at the replication fork, single-strand binding proteins attach to the separated strands and prevent the parental strands from rejoining. As helicase separates the two parental strands, the parental DNA ahead of the replication fork becomes more tightly coiled. To relieve strain ahead of the replication fork, topoisomerase breaks a covalent bond in the sugar-phosphate backbone of one of the two parental strands. Breaking this bond allows the DNA to swivel around the corresponding bond in the other strand and relieves the strain caused by the unwinding of the DNA at the helicase.
In the electrophoresis gel below, what letter indicates the location of the anode (positive electrode)?
B
The thermostability of Taq polymerase is required during the annealing phase of PCR.
B. False The annealing phase takes place at the lowest temperature of PCR. Taq polymerase is derived from bacteria that live in hot springs, so the enzyme is thermostable, meaning that its enzymatic properties can withstand the high temperatures needed for denaturation.
A gene is depicted in the image below. The polarity of DNA strands, transcription start site and direction of transcription are indicated on the figure. Which letter in the image is indicating the coding strand (non-template)?
B: Notice the polarity of the coding strand in relation to the direction of transcription. Remember that the top strand will not always be the coding strand.
The diagram below shows a length of DNA containing a bacterial gene. Drag the labels to their appropriate locations in the diagram to describe the function or characteristics of each part of the gene. Not all labels will be used.
Bacterial transcription is a four-stage process.1. Promoter recognition: RNA polymerase is a holoenzyme composed of a five-subunit core enzyme and a sigma (σσ) subunit. Different types of σσ subunits aid in the recognition of different forms of bacterial promoters. The bacterial promoter is located immediately upstream of the starting point of transcription (identified as the +1 nucleotide of the gene). The promoter includes two short sequences, the -10 and -35 consensus sequences, which are recognized by the σσ subunit. 2. Chain initiation: The RNA polymerase holoenzyme first binds loosely to the promoter sequence and then binds tightly to it to form the closed promoter complex. An open promoter complex is formed once approximately 18 bp of DNA around the -10 consensus sequence are unwound. The holoenzyme then initiates RNA synthesis at the +1 nucleotide of the template strand.3. Chain elongation: The RNA-coding region is the portion of the gene that is transcribed into RNA. RNA polymerase synthesizes RNA in the 5′ → 3′ direction as it moves along the template strand of DNA. The nucleotide sequence of the RNA transcript is complementary to that of the template strand and the same as that of the coding (nontemplate) strand, except that the transcript contains U instead of T. 4. Chain termination: Most bacterial genes have a pair of inverted repeats and a polyadenine sequence located downstream of the RNA-coding region. Transcription of the inverted repeats produces an RNA transcript that folds into a stem-loop structure. Transcription of the polyadenine sequence produces a poly-U sequence in the RNA transcript, which facilitates release of the transcript from the DNA.
Which of the following statements is NOT consistent with the DNA fragments shown in the gel?
Band 1 has a lower molecular mass than Band 2
As the two parental (template) DNA strands separate at a replication fork, each of the strands is separately copied by a DNA polymerase III (orange), producing two new daughter strands (light blue), each complementary to its respective parental strand. Because the two parental strands are antiparallel, the two new strands (the leading and lagging strands) cannot be synthesized in the same way.
Because DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand and because the two parental DNA strands are antiparallel, synthesis of the leading strand differs from synthesis of the lagging strand. The leading strand is made continuously from a single RNA primer located at the origin of replication. DNA pol III adds nucleotides to the 3' end of the leading strand so that it elongates toward the replication fork. In contrast, the lagging strand is made in segments, each with its own RNA primer. DNA pol III adds nucleotides to the 3' end of the lagging strand so that it elongates away from the replication fork. In the image below, you can see that on one side of the origin of replication, a new strand is synthesized as the leading strand, and on the other side of the origin of replication, that same new strand is synthesized as the lagging strand. The leading and lagging strands built on the same template strand will eventually be joined, forming a continuous daughter strand.
A tetrad is composed of one pair of homologous chromosomes at synapsis of prophase I.
Chromosomes are duplicated during interphase; at synapsis of prophase I, one chromosome (with two chromatids) in a tetrad is paternally inherited while the other is maternally inherited.
During which stage of prophase I does crossing over take place?
Crossing over occurs during pachynema when bivalents are closely paired.
What enzyme(s) is/are responsible for removal of RNA primers and joining of Okazaki fragments?
DNA polymerase I and DNA ligase
DNA replication always begins at an origin of replication. In bacteria, there is a single origin of replication on the circular chromosome, as shown in the image here. Beginning at the origin of replication, the two parental strands (dark blue) separate, forming a replication bubble. At each end of the replication bubble is a replication fork where the parental strands are unwound and new daughter strands (light blue) are synthesized. Movement of the replication forks away from the origin expands the replication bubble until two identical chromosomes are ultimately produced. In this activity, you will demonstrate your understanding of antiparallel elongation at the replication forks. Keep in mind that the two strands in a double helix are oriented in opposite directions, that is, they are antiparallel.
DNA polymerase III can only add nucleotides to the 3' end of a new DNA strand. Because the two parental DNA strands of a double helix are antiparallel (go from 3' to 5' in opposite directions), the direction that DNA pol III moves on each strand emerging from a single replication fork must also be opposite. For example, in the replication fork on the left, the new strand on top is being synthesized from 5' to 3', and therefore DNA pol III moves away from the replication fork. Similarly, the new strand on the bottom of that same replication fork is being synthesized from 5' to 3'. But because the bottom parental strand is running in the opposite direction of the top parental strand, DNA pol III moves toward the replication fork. In summary, at a single replication fork, one strand is synthesized away from the replication fork, and one strand is synthesized toward the replication fork. When you look at both replication forks, note that a single new strand is built in the same direction on both sides of the replication bubble.
This diagram shows a late stage of DNA replication. Can you name the protein represented by each icon in the diagram? Then, for each protein, can you identify how DNA replication would be affected if that protein were nonfunctional? First drag the name of the correct protein into each bin. (The first one is done for you.) Then drag the correct "If nonfunctional" description into each bin.
DNA polymerase III synthesizes most of the DNA during DNA replication. DNA polymerase I removes the RNA primers and replaces them with DNA. DNA ligase forms a phosphodiester bond in the DNA backbone to fill in the nicks left there from removing the RNA primers
The diagram below shows a double-stranded DNA molecule (parental duplex). Semi-conservative DNA replication
During DNA replication, each strand in the parental duplex serves as the template for the production of a daughter strand by complementary base pairing. Therefore, one cycle of replication will produce two daughter duplexes, each with one parental strand and one newly synthesized strand. During a second cycle of replication, all four strands in the two duplexes will serve as templates, resulting in four duplexes (eight strands of DNA).
Maternal nondisjunction, the failure of homologous chromosomes or sister chromatids to separate properly, is associated Down syndrome and other types of aneuploidy in humans. Maternal age is associated with nondisjunction, although no age effect is seen in males. How might these findings be explained with respect to gametogenesis?
During ovum production, primary oocytes are arrested for years in meiosis I, increasing the likelihood of components involved in chromosome segregation to break down.During spermatogenesis, germ cells are produced daily, while during oogenesis, germ cells are produced prenatally. The difference in the "age" of the gametes at the time of fertilization may contribute to the observed effect. Although scientists do not know with certainty why nondisjunction increases with advanced maternal age, both the age of the ovum at the time of fertilization and the fact that oocytes are arrested in meiosis I for years are thought to contribute to this effect. Oogenesis begins during fetal development and oocytes are arrested in prophase I by birth. During puberty, ovulation begins and meiosis is reinitiated in one egg during each ovulatory cycle. As a result, each ovum that is released has been arrested in meiosis I for one month longer than the one released during the preceding cycle. Therefore, women in their 30's and 40's are producing ova that are much older than those which were ovulated when puberty began, which may contribute to nondisjunction.
Females with only one X chromosome do not develop; this condition is lethal.
False Females with only one X chromosome are viable but have Turner syndrome, which is characterized by underdeveloped ovaries. Males that lack an X chromosome do not develop; this condition is lethal.
The cells that form between Meiosis I and Meiosis II are diploid, because they contain 2 copies of each piece of DNA in the form of sister chromatids.
False: In meiosis I, homologous chromosomes separate, while in meiosis II, sister chromatids separate. Meiosis II produces 4 haploid daughter cells, whereas meiosis I produces 2 diploid daughter cells. Genetic recombination (crossing over) only occurs in meiosis I
Can you identify the bases that will be added to this parent strand during DNA replication? Drag the labels to the appropriate targets to identify the sequence and orientation of the daughter strand. Blue labels can be used once, more than once, or not at all.
For complementary base-pairing in DNA, purines bind with pyrimidines. A (a purine) binds with T (a pyrimidine) through the formation of two hydrogen bonds. G (a purine) binds with C (a pyrimidine) through the formation of three hydrogen bonds. The two strands are considered to be antiparallel because they are orientated in different directions, from 5' to 3'. Thus, the 5' end of one strand aligns with the 3' end of the opposite strand.
Intrinsic termination of transcription in bacteria requires which of the following?
Formation of a hairpin in the nascent RNA followed by a string of uracil nucleotides
Now that you have identified the sequence of the daughter strand based on the sequence on the parent strand, can you come up with an equation that explains the relative proportions of each base in a DNA macromolecule? (Note that in the equation below, "G" represents the percentage of bases that are guanine, "T" represents the percentage of bases that are thymine, "A" represents the percentage of bases that are adenine, and "C" represents the percentage of bases that are cytosine.) Drag the labels to the appropriate targets to complete the equation. Labels can be used once, more than once, or not at all.
G+A=C+T The key to solving this problem is to recall that there will be the same number of A's as T's, and there will be the same number of G's as C's. This is because of complementary base pairing: G binds to C, and T binds to A.
Which enzyme is required to initiate 5 capping of eukaryotic mRNA transcripts by removing the terminal phosphate group?
Guanyl transferase initiates 5'
A cell can form 10 nm chromatin fibers, but not 30 nm fibers. Which molecule has likely been removed or mutated in this cell?
H1
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 positive charge that allow them to bind to DNA. Acetylation of histones, decreases their positive charge and weakens the histone-DNA interaction. Note: 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.
Compare and contrast properties of sister chromatids and homologous chromosomes.
Homologous chromosomes are a pair of chromosomes, one maternal and one paternal, that come together during fertilization. They have the same centromere position, and the same genetic loci (genes); however the DNA sequence is not identical. Crossing over between homologous chromosomes during prophase I of meiosis generates genetic diversity in the offspring. Homologous chromosomes pair up along the midline during metaphase I of meiosis, and move apart during anaphase I. Sister chromatids are the result of the replication of a single chromosome. They are identical in DNA sequence (apart from mutation or crossing over with a chromatid from a homologous chromosome). During metaphase of mitosis, chromosomes line up down the midline and sister chromatids separate during anaphase. In meiosis, sister chromatids separate during meiosis II.
Three independently assorting STR markers (A, B, and C) are used to assess the paternity of a colt recently born to a quarter horse mare. Blood samples are drawn from the mare, her colt, and three possible male sires (S1, S2, and S3). DNADNA at each marker locus is amplified by PCR, and a DNADNAelectrophoresis gel is run for each marker. Amplified DNADNAbands are visualized in each gel by ethidium bromide staining. Gel results are shown below for each marker. Evaluate the data and determine which of the potential sires can be excluded.
Horses S1 and S3 both lack the nonmaternal allele for Marker C that is carried by the colt, therefore, S1 and S3 can be excluded from consideration.
A family consisting of a mother (I-1), a father (I-2), and three children (II-1, II-2, II-3) are genotyped by PCR for a region of an autosome containing repeats of a 10-bp sequence. The mother carries 16 repeats on one chromosome, and 21 on the homologous chromosome. The father carries repeat numbers of 18 and 26.
Identify all the possible genotypes of children of this couple by specifying PCR fragement lengths in each genotype. Select the four correct genotypes. 16,18-----16,26----21,18-----21,26
Three genes identified in the diagram as AA, BB, and CCare transcribed from a region of DNA. The 5'-to-3' transcription of genes AA and CC elongates mRNA in the right-to-left direction, and transcription of gene BBelongates mRNA in the left-to-right direction.
Identify the coding strand for gene A,B,C A: LOWER B: UPPER C: LOWER
The following eukaryotic structural gene contains two introns and three exons. The table below shows four possible mRNA products of this gene. Use the labels to explain what may have caused each mRNA. Drag the correct label to each location in the table. Labels may be used once, more than once, or not at all.
If a mutation alters a splicing signal sequence of an intron, that intron will not be removed accurately during the splicing reaction. This will result in the production of an abnormally spliced mature mRNA.Mutations in promoter sequences will affect transcription initiation and are likely to result in no mRNA being produced.
A diploid cell with 4 chromosomes (one pair metacentric, the other telocentric) begins meiosis, and nondisjunction occurs during the first meiotic division.
If nondisjunction occurs during meiosis I, homologous chromosomes fail to separate. This produces abnormal gametes that contain two members of the affected chromosome or none at all.
A diploid cell with 4 chromosomes (one pair metacentric, the other telocentric) begins meiosis, and nondisjunction occurs during the second meiotic division.
If nondisjunction occurs during meiosis II, sister chromatids fail to separate. In this case, 50% of the gametes that are produced are normal haploid gametes, 25% of gametes have an extra chromosome, and 25% are missing a chromosome.
In DNA replication in bacteria, the enzyme DNA polymerase III (abbreviated DNA pol III) adds nucleotides to a template strand of DNA. But DNA pol III cannot start a new strand from scratch. Instead, a primer must pair with the template strand, and DNA pol III then adds nucleotides to the primer, complementary to the template strand. Each of the four images below shows a strand of template DNA (dark blue) with an RNA primer (red) to which DNA pol III will add nucleotides.
In the example above, DNA pol III would add an adenine nucleotide to the 3' end of the primer, where the template strand has thymine as the next available base. You can tell which end is the 3' end by the presence of a hydroxyl (-OH) group. The structure of DNA polymerase III is such that it can only add new nucleotides to the 3' end of a primer or growing DNA strand (as shown here). This is because the phosphate group at the 5' end of the new strand and the 3' -OH group on the nucleoside triphosphate will not both fit in the active site of the polymerase.
How does the use of alternative promoters affect transcription?
It results in a different 5' end than that seen when the known promoter is used. The consequence of this is synthesis of an mRNA that has a different 5' end than that seen when the known promoter is used. The alternative promoters may be used under different conditions in a given cell type or may be used in different cell types.
Describe the three processes that commonly modify eukaryotic pre-mRNA.
Its in the picture
Determine the order in which the following proteins and enzymes are active in E. coli DNA replication: Rank the proteins and enzymes from the first one to the last one.
LOOK AT THE PICTURE
After which stage or phase of the cell cycle does cytokinesis occur?
M phase
The cell cycle represents the coordinated sequence of events in the life of a cell from its formation to its division into two daughter cells. Most of the key events of the cell cycle are restricted to a specific time within the cycle. In this exercise, you will identify when various events occur during the cell cycle. Recall that interphase consists of the G1, S, and G2 subphases, and that the M phase consists of mitosis and cytokinesis.
Many organisms contain cells that do not normally divide. These cells exit the cell cycle before the G1 checkpoint. Once a cell passes the G1 checkpoint, it usually completes the cell cycle--that is, it divides. The first step in preparing for division is to replicate the cell's DNA in the S phase. In the G2 phase, the centrosome replicates. In early M phase, the centrosomes move away from each other toward the poles of the cell, in the process organizing the formation of the mitotic spindle. At the end of the M phase when mitosis is complete, the cell divides (cytokinesis), forming two genetically identical daughter cells.
A couple has a daughter with Turner syndrome, a condition in which only a single copy of the X chromosome is present. This results from nondisjunction, the failure of the X chromosome to segregate properly during meiosis. During which meiotic division, and in which parent, could nondisjunction have occurred to produce a child with this condition? Hint: It may be helpful to sketch a diagram depicting the outcome of nondisjunction in meiosis I and II of both parents to solve this problem.
Meiosis I in the mother Meiosis II in the mother Meiosis I in the father Meiosis II in the father During meiosis I, homologous chromosomes separate. If nondisjunction occurs, both homologous chromosomes migrate into the same daughter cell instead of different daughter cells. If this happens in the mother, both X chromosomes end up in one cell, with no X chromosome in the other cell. If this happens in the father, both the X and Y chromosome end up in one cell, with no sex chromosome in the other cell. In both cases it is possible for one parent to not give a sex chromosome to the child. If the other parent gives an X chromosome, the child will be XO and have Turner syndrome. During meiosis II, sister chromatids separate. If nondisjunction occurs, both sister chromatids migrate into same daughter cell, instead of different daughter cells. If this happens in the mother, two identical copies of an X chromosome end up in one cell, with no X chromosome in the other cell. If this happens in the father, either two identical copies of the X chromosome end up in one cell, with no sex chromosome in the other cell - or - two identical copies of the Y chromosome end up in one cell, with no sex chromosome in the other cell. In all cases, it is possible for one parent to not give a sex chromosome to the child. If the other parent gives an X chromosome, the child will be XO and have Turner syndrome.
A couple has a son with Klinefelter syndrome, a condition in which an extra copy of the X chromosome is present. This condition also results from nondisjunction. During which meiotic division, and in which parent, could nondisjunction have occurred to produce a child with this condition?
Meiosis I in the mother Meiosis II in the mother Meiosis I in the father To produce a child with XXY, either mom gave two X chromosomes and dad gave Y OR mom gave one X chromosome and dad gave both X and Y. There are two ways mom could give two X chromosomes - either she gave both of her X's (meiosis I nondisjunction), or she gave two copies of one of her X's (meiosis II nondisjunction). In both these cases if dad gives a Y chromosome, the child is XXY and has Klinefelter's. There is only one way dad could give X and Y, and that's if X and Y did not separate in meiosis I. Then if mom gives an X chromosome, the child is XXY and has Klinefelter's. Nondisjunction in meiosis II in dad would give both sister chromatids - either XX or YY, which would produce either a XXX girl or a XYY boy... not a XXY boy.
A couple has a son with XYY syndrome, a condition in which an extra copy of the Y chromosome is present. This condition also results from nondisjunction. During which meiotic division, and in which parent, could nondisjunction have occurred to produce a child with this condition?
Meiosis II in the father The key to solving this problem is to realize that the child got two Y chromosomes. Both of those Y chromosomes had to have come from the father, as the mother has no Y chromosomes. In order to get two Y chromosomes, nondisjunction would have to have occurred in meiosis II - identical Y sister chromatids migrated together into one cell. During meiosis II, sister chromatids should separate. If nondisjunction occurs, both sister chromatids migrate into same daughter cell, instead of different daughter cells. If this happens in the father, either two identical copies of the X chromosome end up in one cell, with no sex chromosome in the other cell - or as in this case two identical copies of the Y chromosome end up in one cell, with no sex chromosome in the other cell. If the sperm containing two copies of the Y chromosome fertilizes an egg containing an X chromosome, the child will be XYY.
Rank the following levels of chromatin compaction in eukaryotes from the least compact to the most compact.
Naked DNA, nucleosome, solenoid, loop domains, chromatid, metaphase chromosome Note: 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).
The incidence of Down syndrome, also known as trisomy 21, increases with increasing maternal age. Which of the following errors most likely produces this condition?
Nondisjunction during either meiosis I or II in the female gamete Nondisjunction during either meiosis I or II can produce a gamete that will result in a trisomic zygote. Since the incidence of Down syndrome increases with maternal age, it is likely that this error occurs in the female gamete.
Select the properties shared by RNA polymerase and DNA polymerase.
Polymerizes nucleotides in a 5'-to-3' direction. Dependent on a DNA sequence template. Catalyzes phosphodiester bond formation.
Which of the following statements best describes the structure of RNA molecules in general?
RNA molecules are single-stranded, but they commonly form localized secondary structures by base pairing between regions of the molecule.
Order the following events from first to last in initiation of translation in prokaryotes.
Recall that in the initiation of translation in prokaryotes, the 70S ribosome does not bind the mRNA as an entire complex, but rather it is assembled on the mRNA. The small 30S subunit and its 16S rRNA are critical for the complex to recognize the the proper AUG start codon before the 70S ribosome is fully assembled.
In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B.
Synthesis of the lagging strand is accomplished through the repetition of the following steps. Step 1: A new fragment begins with DNA polymerase III binding to the 3' end of the most recently produced RNA primer, primer B in this case, which is closest to the replication fork. DNA pol III then adds DNA nucleotides in the 5' to 3' direction until it encounters the previous RNA primer, primer A. Step 2: DNA pol III falls off and is replaced by DNA pol I. Starting at the 5' end of primer A, DNA pol I removes each RNA nucleotide and replaces it with the corresponding DNA nucleotide. (DNA pol I adds the nucleotides to the 3' end of fragment B.) When it encounters the 5' end of fragment A, DNA pol I falls off, leaving a gap in the sugar-phosphate backbone between fragments A and B. Step 3: DNA ligase closes the gap between fragments A and B. These steps will be repeated as the replication fork opens up. Try to visualize primer C being produced to the right (closest to the replication fork). Fragment C would be synthesized and joined to fragment B following the steps described here.
DNA is composed of two strands that are bound together, resembling a rope ladder with rigid rungs. This DNA ladder is twisted, forming what is called the double helix. The structure of the DNA double helix depends on the complementary pairing of bases between the two strands. Replication of DNA requires that the two strands of the helix separate, as shown in the image below. New daughter molecules are constructed by the sequential addition of nucleotides and the formation of base pairs between the new strand and the parent (template) strand. The replication of the double helix results in two daughter molecules, each composed of one parent strand and one new strand. The enzymes that accomplish the replication of DNA are called DNA polymerases.
The DNA double helix is constructed from two strands of DNA, each with a sugar-phosphate backbone and nitrogenous bases that form hydrogen bonds, holding the two strands together. Each DNA strand has two unique ends. The 3' end has a hydroxyl (-OH) group on the deoxyribose sugar, whereas the 5' end has a phosphate group. In the double helix, the two strands are antiparallel, that is, they run in opposite directions such that the 3' end of one strand is adjacent to the 5' end of the other strand.
Which of the following statements about manual Sanger sequencing is true?
The DNA sequence obtained is complementary to the template strand. The DNA fragments produced in sequencing reactions are synthesized by DNA polymerase to be complementary to the template strand.
Which of the following steps are similar in the semi-conservative and conservative models of DNA replication?
The DNA strands of the parental duplex separate Each parental strand acts as a template for replication In the second round of replication the duplex strands produced in the first round separate and each strand acts as a template for replication.
The virulent form of the bacteria S. pneumoniae is called the S strain because it is surrounded by a polysaccharide coat that makes it appear smooth under a microscope. Sometimes the S strain mutates into a non-virulent form (called the R strain), which lacks the polysaccharide coat and appears rough. Frederick Griffith performed an experiment in which he injected mice with different combinations of these bacterial strains. Drag the labels to indicate whether the mice live or die after each injection and the explanation for each test result.
The S strain of the pneumococcal bacteria is virulent and kills mice. When the S strain is heat-killed, it is no longer virulent and doesn't kill mice. The R strain is non-virulent. However, when the heat-killed S strain is mixed with the live R strain, a transformed virulent R strain is created.
This nucleotide would be found in RNA. How do you know?
The base is a uracil, which is found only in RNA. There is an -OH group on the 2'-carbon of the ribose.
What is meant by the "beads on a string" model of chromatin
The beads are the nucleosomes, and the string is the linker DNA.
Hershey and Chase selected the bacteriophage T2 for their experiment assessing the role of DNA in heredity. Why was T2 a good choice for this experiment? Select the two correct statements.
The entire T2 virus particle does not enter the bacterial cell. T2 contains DNA and protein but no RNA.
Mitosis unfolds through a sequence of stages marked by specific events in the cell. The structural changes in the cell are brought about by a series of tightly coordinated underlying mechanisms. Sort each process into the appropriate bin to indicate the stage of mitosis in which it occurs. If a process occurs in more than one stage, sort it to the stage when it first occurs.
The micrographs in Part A show some of the cellular processes that occur during the stages of mitosis. In prophase, the microtubules of the spindle apparatus begin to assemble from individual tubulin subunits. As the identical chromatids of each pair of sister chromatids condense during this stage, they are held together by cohesin proteins. Prometaphase is marked by fragmentation of the nuclear envelope, expansion of the spindle into the nuclear region, and attachment of some spindle fibers to the chromosomes via the kinetochores. Metaphase, marked by the alignment of chromsomes along the metaphase plate, is brought about by kinetochores aligning and then remaining motionless relative to the poles of the cell. In anaphase, the cohesin proteins are cleaved, and the kinetochores move toward the poles of the cell, separating the sister chromatids. As telophase proceeds, the kinetochore microtubules of the spindle disassemble. As the chromosomes reach the poles of the cell, the nuclear envelopes of the two new daughter nuclei form.
What makes up the protein component of a nucleosome core
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. Two tetramers of histone proteins
How do STRs differ from VNTRs? See Section 25.1 (Page 781) .
The repeat sequences of STRs are very similar to those of variable number tandem repeats (VNTRs). They differ in that STR repeats are less than 10 base pairs (bp) long, whereas VNTRs contain repeats that are longer than 10 bp each. Like VNTRs, the inheritance of STRs follows a codominant pattern. STR repeats are less than 10 base pairs (bp) long whereas VNTRs contain repeats that are longer than 10 bp each.
A tRNA anticodon is 5' GAA 3' Which amino acid does this tRNA carry?
The tRNA anticodon is 5'-GAA-3'. Considering the wobble hypothesis, that means the anticodon could bind to 3'-CUU-5' or 3'-UUU-5'. If you write these codons in the standard convention from 5' to 3', you have UUC or UUU. Both of these codons encode phenylalanine.
Compare and contrast spermatogenesis and oogenesis in human cells.
There are a number of important differences between the production of male and female gametes, even though the events that occur during meiosis are similar in all cells. Spermatogenesis, which takes place in the testes, produces four sperm cells with the haploid number of chromosomes and equal amounts of cytoplasm from one undifferentiated diploid germ cell called a spermatogonium. Because humans can reproduce year-round, spermatogenesis occurs throughout the life of the mature male. Oogenesis, which takes place in the ovaries, differs in several important ways from spermatogenesis. Oogenesis results in the production of one egg cell and other haploid products called polar bodies from a diploid germ cell called an oogonium. The egg cell contains the majority of cytoplasm which contributes to zygote development during fertilization, while the polar bodies contain less cytoplasm. Unlike spermatogenesis, much of oogenesis is completed before birth, with oocytes remaining arrested in prophase I until many years later, when meiosis is continued just prior to ovulation. Additionally, while females produce one gamete per month during their mature reproductive years, males produce millions of sperm every day.
Why are eukaryotic promoters more variable than bacterial promoters?
There are multiple RNA polymerases in eukaryotes, each of which binds to a different consensus sequence. There are multiple transcription factors in eukaryotes, which are specific to each type of polymerase, and these bind to different promoter sequences. Nearby sequences in addition to the promoter consensus sequences in eukaryotes are thought to be important in modulating the level (rate) of transcription initiation
The diagram below depicts translation at the moment the fourth amino acid is added to the polypeptide chain. Label the following structures, amino acids and anticodons and determine the orientation of mRNA being translated. Drag the appropriate labels to their respective targets.
This one was a doozy
Many antibacterial agents and chemotherapy agents act as topoisomerase inhibitors. Why would inhibition of topoisomerase cause cell death in bacterial and eukaryotic cells?
Topoisomerases minimize torsional stress catalyzing a controlled cleavage and rejoining of DNA, thus enabling over-twisted strands to unwind. The accumulating stress could break the DNA template at random locations, potentially leading to a breakdown of DNA replication and, ultimately, to cell death.
Drag the labels into the flowchart to show the order of events as they are thought to occur during eukaryotic transcription involving RNA polymerase II (RNA pol II).
Transcription by RNA pol II in eukaryotes begins when TFIID recognizes and binds to the TATA box. The bound TFIID helps recruit TFIIA, TFIIB, TFIIF, and RNA pol II. Once those subunits of the minimal initiation complex are bound, TFIIE and TFIIH bind to form the complete initiation complex. Assembly of the complete initiation complex releases RNA pol II, which begins synthesizing the RNA transcript in the 5′ → 3′ direction. After the first 20-30 nucleotides have been synthesized, a cap consisting of a methylated guanine is added to the 5′ end of the pre-mRNA. Intron removal occurs as RNA pol II continues to elongate the pre-mRNA. When the polyadenylation signal has been transcribed, a poly-A tail is added to the 3′ end of the pre-mRNA. Polyadenylation is usually coupled with the termination of transcription.
Which of the following statements about the products produced when nondisjunction occurs during meiosis I is true?
Two products of the second meiotic division have both the maternal and paternal chromosomes of a set and the other two products have none for that set. Nondisjunction during meiosis I yields two gametes that are missing a particular chromosome after meiosis II.
Gene A in humans is associated with a disease in humans. You have identified a partial genomic sequence for Gene A in mice; the sequence of ONE strand of mouse Gene A is shown below. You plan to use fluourescence in situ hybridization (FISH) to study Gene A. 5'...AGCTTGGGAGCGGCGGTGAGGCGGGAGGCGTCCTCTCTCCCCCAGGGCCCACCAGCTCTG...3' Use this sequence to answer the following questions.
What information can you obtain by using FISH to study Gene A? (The chromosomal location of Gene A in mice) How does FISH identify the chromosomal location of a gene? (A short DNA or RNA probe is designed that is complementary to the gene in question. The probe contains a fluorescently labeled tag that helps identify only the genomic location that hybridizes to the probe.) Which of the following DNA probes could you use to detect the genomic location of Gene A in mice? The partial sequence of Gene A is shown below. Select all answers that apply. 5'...AGCTTGGGAGCGGCGGTGAGGCGGGAGGCGTCCTCTCTCCCCCAGGGCCCACCAGCTCTG...3' (5' TTGGGAGCGG 3' 3' AACCCTCGCC 5) Although the sequence of only one strand is given, the gene is present in the genome as double-stranded DNA. Therefore a probe can be made to hybridize to either strand. The probe must be complementary and antiparallel. Which of the following DNA probes could you use to detect the genomic location of Gene A in mice? The partial sequence of Gene A is shown below. Select all answers that apply. 5'...AGCTTGGGAGCGGCGGTGAGGCGGGAGGCGTCCTCTCTCCCCCAGGGCCCACCAGCTCTG...3' (5' CCAGGGCCCA 3' 3' GGTCCCGGGT 5') Although the sequence of only one strand is given, the gene is present in the genome as double-stranded DNA. Therefore a probe can be made to hybridize to either strand. The probe must be complementary and antiparallel
This dideoxy DNA-sequencing gel is produced in a laboratory.
What is the sequence of DNA strand being synthesized during sequencing? (5' TAGTCTCGATATCGTCACG3') What is the sequence of DNA template strand? (3' ATCAGAGCTATAGCAGTGC 5')
Avery, MacLeod, and McCarty performed an experiment to narrow down what type of molecule the transforming factor is. They added enzymes to the heat-killed S strain to target different types of molecules in each test. If they had then injected the different mixtures into mice, what would the results have been? Drag the labels to indicate the appropriate explanation and conclusion from each test result.
When DNA is destroyed with DNase, the heat-killed S strain of bacteria is no longer able to transform the R strain into a virulent strain. Destroying RNA (with RNase) or protein (with protease) does not prevent the S strain from transforming the R strain into a virulent strain. This experiment showed that DNA--not RNA or protein--must be the transforming factor.
Consider the distribution of sex chromosomes that would occur when a male with Klinefelter syndrome (X1X2Y) undergoes spermatogenesis. Which chromosome combinations are possible?
X1Y and X2 X2Y and X1 X1X2 and Y During meiosis I in an individual with Klinefelter syndrome, two sex chromosomes will migrate to one daughter cell and one sex chromosome will migrate to the other daughter cell. At meiosis II, sister chromatids separate. There are three possible ways in which the sex chromosomes could separate in an XXY individual. Both X chromosomes migrate away from the Y chromosome:
What is required for DNA polymerase to initiate DNA strand synthesis?
a short RNA primer synthesized by the enzyme primase
When do sister chromatids separate during meiosis
anaphase II
Look at the sequence of the nucleic acids in the video. How are the sequences of the mRNA and the template DNA related?
antiparallel polarity complementary sequence
In Part C you identified the portion of the mRNA sequence (5'...AGGAGG...3') that base pairs with the 16S rRNA. Where do you expect to find the start codon in relation to this sequence?
downstream The 16S rRNA makes contact with the Shine-Dalgarno mRNA sequence near the exit site (E Site) of the 30S subunit of the ribosome. This positions the downstream start codon into the peptidyl site (P Site) in order for the initiator tRNA to base pair with the start codon. Once the initiator tRNA is bound, the 50S subunit binds the 30S subunit.
Light G bands that appear along chromosomes contain ...
euchromatic regions.
Give descriptions for the following terms. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. is a small, basic polypeptide that is positively charged and bind tightly to negatively charged DNA.
histone protein
What types of bonds are formed between complementary DNA bases?
hydrogen bonds
For the following fragment of DNA, determine: 5'-TAGAGTGCTC-3' 3'-ATCTCACGAG-5' the number of hydrogen bonds present. the number of phosphodiester bonds present.
hydrogen bonds: G&C=3 for each pair so 5 x 3 = 15 T&A=2 for each pair so 5 x 2 = 10 10 + 15 = 25 H-Bonds phosphodiester bonds: Total number of base pairs: 10 2 sequences? so 20 total. (n-2) = (20-2) = 18 P Bonds
When a peptide bond is formed between two amino acids, one is attached to the tRNA occupying the P site and the other _______.
is attached to the tRNA occupying the A site
3'-AUCGUCAUGCAGA-5' During transcription, was the adenine at the left-hand side of the sequence the first or the last nucleotide added to the portion of mRNA shown?
last
Which type of RNA is translated within a cell?
mRNA
Novel combinations of genes can arise from _______. Select more than one answer.
reciprocal exchange of DNA between homologs during prophase I the alignment of chromosomes at Metaphase I
Which of the following are classified as pyrimidines?
thymine and cytosine
