chapter 7 questions bio 214
which of the following equations are true for the percentage of nucleotides in double-stranded DNA? 1) (A + G)/(C + T)= 1.0 2) (A + T)/(G + C)= 1.0 3) (A)/(T) = (G)/(C) 4) (A)/ (C) = (G) / (T) 5) (A) / (G) = (T) (C)
1 TRUE 2 FALSE 3 TRUE 4 FASLE 5 TRUE
A sample of double-stranded DNA is found to contain 27% cytosine. determine the percentage of adenine in the sample
23%
what is the complementary DNA strand shown? 3'-ATTGCAGTACC-5'
5'- TAACGTCATGG -3
How does RNA participate in DNA replication? A) RNA serves as a primer for elongation by DNA polymerase. B) RNA prevents reannealling of separated DNA strands. C) RNA relaxes supercoilling of DNA strands. D) RNA transports free nucleotide triphosphates to the phosphodiester bond formation site of DNA polymerase III
A
If a strain of E. coli acquired a mutation that inactivated DNA polymerase III function, would the cell be able to replicate its DNA? Why or why not? A) The cell would not be able to replicate its DNA because it lacks the enzyme responsible for the bulk of DNA synthesis. B)The cell would not be able to replicate its DNA because it lacks the enzyme responsible for removing the RNA primer. C) The cell would be able to replicate its DNA, though at a much lower rate. D) The cell would be able to replicate its DNA because there are analogs of DNA polymerase III in the cell.
A
What were the results of their experiment for two DNA replication cycles? A) After one round of replication all of the DNA was 14N/15N; after two rounds half the DNA was 14N/15N and half was 14N/14N. B) After one round of replication all of the DNA was 14N/15N; after two rounds half the DNA was 14N/15N and half was 15N/15N. C) After one round of replication all of the DNA was 14N/14N; after two rounds half the DNA was 14N/15N and half was 15N/15N. D) After one round of replication all of the DNA was 14N/14N; after two rounds all of the DNA was 15N/15N. E) After one round of replication all of the DNA was 14N/15N; after two rounds all of the DNA was 14N/14N. F) After one round of replication all of the DNA was 14N/15N; after two rounds one third of the DNA was 14N/15N, one third was 14N/14N and one third was 15N/15N
A
What is the functional role of telomeres? Select the two correct statements. A) Telomeres provide a mechanism for replication of the ends of linear chromosomes. B) Telomeres provide a protective "cap" on the ends of linear chromosomes, that distinguishes normal chromosome ends from ends generated by double-stranded chromosome breaks. C) Telomeres attract replication enzymes and thereby serve as the replication origins. D) Telomeres allow chromosome end-to-end fusion which is essential during metaphase of mitosis. E) Telomeres prevent topoisomerase enzymes from sliding out of the DNA double-helix.
A,B
Which of the following equations are true for the percentages of nucleotides in double-stranded DNADNA? Select all that apply. A). (A)/(G) = (T)/(C) B). (A)*(C)=(G)*(T) C). (A-G)/(T-C)=1 D). (A)*(T)=(C)*(G) E) (2A-T)/(2G-C)=1
A,B,C
two viral genomes are sequences, and the following percentages of nucleotides are identified genome 1= A=28% C=22% G=28% T=22% genome 2= A=22% C =28% G =28% T=22% Are the DNA molecules in each genome single stranded or double stranded?
Answer: Part A: Double-stranded DNA Explanation: Thymine is present, so it is DNA. If the genome follows Chargaff's rule, it is double- stranded. The rule says that no. purines is equal to the no. of pyrimidines. Otherwise, the genome would be single-stranded. A+G = T+C or A (28%)= T(28%) G (22%) = C (22%), follows rule so double stranded
If there are five molecules of DNA containing the target region at the beginning of a PCR reaction, how many copies of the target will be present after three rounds of amplification? A) 15 B) 40 C) 500 D) 125
B (Correct. The number of target sequences is doubled with each replication cycle.)
One strand of a fragment of duplex DNA has the sequence 5'-ATCGACCTGATC-3'. b) What is the name of the bond that joins one nucleotide to another in the DNA strand? c) is the bond in part (b) a covalent or non covalent bond? d) which chemical groups of nucleotides reaction to form the bond in part (b) e) what enzymes catalyze the reaction in part (d)
B) bond that joins one nucleotide to another in DNA strand is called a 3-5 phosphodiester linkage c)yes the bond is covalent (phosphodiester linkage are strong colvanet linkage) d) hydroxyl group and phosphate group e) DNA polymerases (catalyzed for the formation of phosphodiest bond in DNA)
Why is telomerase usually active in germ-line cells but not in somatic cells? A) In somatic cells there is an alternative mechanism for replication of ends of linear chromosomes. B) During replication, the 3' end of each DNA strand in somatic cell is occupied by SSBs and therefore cannot bind telomerase RNA. C) Germ-line cells divide many times, whereas many somatic cells are capable of a limited number of cell divisions. D) Telomerase action is required in rolling circle replication, which only takes place in germ-line cells
C
which polymerase removes the RNA primer and fills in gaps with DNA
DNA polymerase
which polymerase is responsible for the bulk of synthesis of DNA on the leading strand and is responsible for the bulk synthesis of DNA for the lagging strand
DNA polymerase III
explain how RNA participates in DNA replication
RNA is synthesized and serves as a primer for elongation by DNA polymerase (can only extend an existing strand)
consider the sequence 3'-ACGCTACGTC-5' a) what is the double-stranded sequence?
a. double stranded sequence : 3'-ACGCTACGTC-5' 5'-TGCGATGCAG-3' A complements T , C complements G and vice versa.
One strand of a fragment of duplex DNA has the sequence 5'-ATCGACCTGATC-3'. I) what term is used to describe the polarity of two DNA strand in a duplex?
antiparallel (the term used to describe polarity of two DNA strands in a duplex)
Mathew meselson and Franklin Stahl demonstrated that DNA replication is semiconservative in
bacteria
telomeres are found at the ends of
eukaryotic chromosomes
Why do the genomes of eukaryotes, such as Drosophila, need to have multiple origins of replication, whereas bacterial genomes, such as that of E. coli, have only a single origin?
eukaryotic chromosomes are presumed to replicated different genomic regions at different times during S phase of the cell cycle, Eukaryotic DNA polymerase functions about 10 mixtures slower than bacterial DNA polymerase and eukaryotic genomes are much larger than bacterial genomes
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 SSB, DNA polymerase III, ligase, helicase, primase, DNA polymerase I,
helicase, SSB, primase, DNA polymerase III, DNA polymerase I, ligase
The following diagram shows the parental strands of a DNA molecule undergoing replication. Draw the daughter strands present in the replication bubble, indicating The polarity of daughter strands The leading and lagging strands Okazaki fragments The locations of RNA primers
insert image
which of the following equalities is not true for double-stranded DNA? G+T = A + C G+C=A+T G+A=C+T
second option
Which of the following statements best represents the central conclusion of the Hershey-Chase experiments? A) DNA is the identity of the hereditary material in phage T2. B) Phage T2 is capable of replicating within a bacterial host. C) Some viruses can infect bacteria. D) When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates radioactive label.
A (Because phage DNA and not proteinwas associated with bacteria at the end of the experiment, it could be concluded that DNA - not protein - must be the genetic material.)
which of the following would result from a third round of replication using the methods of meselson and stahl A) One light band and one intermediate band B) One light band C) One heavy band, one light band, and one intermediate band D) One heavy band
A (Correct. Of the molecules generated in the third round, 75% are completely light, 25% are intermediate.)
Immediately after the primers have annealed to the target sequence, _______. A) the temperature is raised so that taq polymerase can extend the primers B) the temperature is raised to cause denaturation C) the annealing temperature is maintained until polymerase has finished extension of the new strands D) the temperature is lowered so that taq polymerase can extend the primers
A (Correct. The temperature is raised to 70-75∘∘C, the temperature over which taq polymerase is optimally active.)
Which of the following statements about manual Sanger sequencing is true? A) One sequencing reaction is performed. B) The DNA sequence obtained is complementary to the template strand. C) The DNA sequence is read from the top of the gel to the bottom. D) Each of the four terminating ddNTPs is labeled with a different fluorescent dye.
B (The DNA fragments produced in sequencing reactions are synthesized by DNA polymerase to be complementary to the template strand. Return to Assignment)
What is the complementary DNA sequence to 5′ ATGCTTGACTG 3′? View Available Hint(s)for Part C A) 5′ ATGCTTGACTG 3′ B) 5′ CAGTCAAGCAT 3′ C) 5′ TACGAACTGAC 3′ D) 5′ ACTCTACGTAG 3′
B (this sequence is complementary and in the correct orientation)
What is the sequence composition of telomeres? A) the repeated sequence 5'-TAAAGG-3' or a variant thereof B) 3'-TTATCCACA-5' C) the repeated sequence 5'-TTAGGG-3' or a variant thereof D) 5'-TTATCCACA-3' E) 13-18 of consecutive adenosine nucleotides F) 5'-GATCTATTTATTT-3'
C
In 1928, Frederick Griffith established that _______. A)mice could be infected with bacteria B) proteases have no effect on DNA C) heat-killed bacteria harbor the constituent(s) necessary to convey genetic properties to living bacteria D)mouse DNA could be transferred into bacterial cells
C (Because some of the nonvirulent bacteria acquired properties of the virulent bacteria, instructions for this transformation must be carried by the virulent bacteria.)
which statement about the polarity of DNA strands is true? A) the 3' end has a free phosphate group B) the 5' end has a free OH group C) the 3' end has a free OH group
C (Nucleotides are joined together by a phosphodiester linkage, which requires a free 3' OH group on the sugar molecule, and an alpha phosphate of an entering dNTP.)
bacterial DNA polymerase I and DNA polymerase III perform different functions during DNA replication. a) identify the principal function of each molecule
DNA pol I is required to remove the RNA prier and fill the gap with DNA. DNA pol III is responsible for the bulk synthesis of DNA on the leading and lagging strands
telomeres are found at the ends of eukaryotic chromosomes d) why is telomerase usually active in germ-line cells but not in somatic cells?
Telomerase is required to ensure complete the replication of chromosome ends (telomeres), ensuring that every cell division produces two daughter cells with complete chromosomes. In the absence of telomerase, chromosomes shorten with every cell division, eventually resulting in loss of telomeres and nearby genes. Germ-line cells divide many times; therefore, they require telomerase. Somatic cells are capable of a limited number of cell divisions (some are unable to divide at all); therefore, they do not require telomerase. It is also thought that the lack of telomerase in somatic cells helps prevent indefinite cell division because loss of telomeres activates DNA damage responses that stop cell division and lead to cell death. This response would help protect the organism from the spread of cancerous cells.
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.
consider the sequence 3'-ACGCTACGTC-5' b) what is the total number of covalent bonds joining the nucleotides in each strand?
b. In each strand there are 10 phosphodiester bonds. phosphodiester bonds are formed between adjacent nucleotides forming a linear strand.
One strand of a fragment of duplex DNA has the sequence 5'-ATCGACCTGATC-3'. a) What is the sequence of the other strand in the duplex?
3'-TAGCTGGACTAGS-5'
Which of the following outcomes would be most likely if the Hershey-Chase experiments were repeated without the step involving the blender? for Part C A) Neither preparation of infected bacteria would exhibit radioactivity. B) The phage would fail to infect bacteria. C) Both preparations of infected bacteria would exhibit radioactivity. D) Both preparations of infected bacteria would contain both P32 and S35.
C (Instead of being removed from the preparation, the "ghosts" would be retained. Because both bacterial preparations would include ghosts as well as viral DNA, both would be radioactive, one with P32, one with S35.)
the following dideoxy DNA sequencing gel is produced in a laboratory. what is the double-stranded DNA sequence of this molecule? label the polarity of each strand
add image question 29
what would be the effects on DNA replication if mutation of DNA pol III caused it to lose each of the following activities 5'- to 3' polymerase activity
loss of the 5-3' polymerase activity leads to no replication. the 5' to 3' polymerase activity is critical for this enzyme
telomeres are found at the ends of eukaryotic chromosomes a) what is the sequence composition of telomeres?
telomeric DNA is composed of repetitive, short DNA sequence. in many organisms the repeated sequence is 5'-TTAGGG-3'
consider the sequence 3'-ACGCTACGTC-5' c) what is the total number of covalent bonds joining the nucleotides of complementary strands
there are 4 A-T and 6 G-C pairs . A-T pair contains 2 H bonds while G-C pair contains 3 H bonds. Hence, there are 4(2) + 6(3) = 8 + 18 = 26 H bonds are present in total.
A sample of double-stranded DNA is found to contain 27% cytosine. determine the percentage of thymine in the sample
23%
A sample of double-stranded DNA is found to contain 27% cytosine. determine the percentage of guanine in the sample
27%
What is the sequence of DNADNA template strand? Express your answer as a sequence of nucleotides separated by dashes from 3′3′ to 5′5′ (example 3′-AGTC-5′3′-AGTC-5′).
3'-ATGACTACGCTACGATTCG-5'
for the following fragment of DNA, determine the number of hydrogen bonds and the number of phosphodiester bonds present? 5'-ACGTAGAGTGCTC-3' 3'-TGCATCTCACGAG-5'
33 hydrogen bonds, 24 phosphodiester bonds
What is the sequence of DNADNA strand being synthesized during sequencing? Express your answer as a sequence of nucleotides from 5′5′ to 3′3′ (example 5′-TCAG-3′5′-TCAG-3′).
5'-TACTGATGCGATGCTAAGC-3'
The principles of complementary base pairing and antiparallel polarity of nucleic acid strands in a duplex are universal for the formation of nucleic acid duplexes. What is the chemical basis for this universality? A) It is the alignment of bases in the two complementary antiparallel DNA strands that allows the formation of a stable number of hydrogen bonds. B) It is the repulsion of negative charges between the bases of the two complementary antiparallel DNA strands that favors the formation of nucleic acid duplexes. C) It is the alignment of deoxyribose rings in the two complementary antiparallel DNA strands that favors the formation of phosphodiester bonds. D) It is the alignment of deoxyribose rings in the two complementary antiparallel DNA strands that allows all possible hydrogen bonds to form.
A
If you were to try and pair a thymine with a cytosine (a non Watson-Crick base pairing), then would you expect to see any stability with respect to the hydrogen bonding (assuming the geometrical configurations of both bases were favorable to each other)? If yes, then how many hydrogen bonds could form between these two bases? A) Yes, two hydrogen bonds could form between thymine and cytosine. B) No, hydrogen bonds cannot form between thymine and cytosine. C) Yes, one hydrogen bond could form between thymine and cytosine. D) Yes, three hydrogen bonds could form between thymine and cytosine.
A (One hydrogen bond could form between the C4 carbonyl group on thymine (a hydrogen bond acceptor) and the C4 amino group on cytosine (a hydrogen bond donor). Another hydrogen bond could form between N3 of thymine (a hydrogen bond donor) and the N3 of cytosine (a hydrogen bond acceptor). Note that the C2 carbonyl groups found on both bases are both hydrogen bond acceptors and therefore a hydrogen bond cannot be formed between them.)
Which of the following statements about ddNTPs is true? A) They have a hydrogen at the 3′ carbon of the sugar. B) DNA polymerase can add a new dNTP to a 3′ ddNTP. C) They have a free 3′‑hydroxyl group on the sugar. D) They have an oxygen at the 2′ carbon of the sugar.
A (ddNTPs terminate synthesis because there is no 3′‑hydroxyl group onto which DNA polymerase can add nucleotides.)
Hershey and Chase selected the bacteriophage T2 for their experiment assessing the role of DNADNA in heredity. Why was T2 a good choice for this experiment? Select the two correct statements. A)T2 contains DNDN and protein but no RNARNA. B) The entire T2 virus particle does not enter the bacterial cell. C) The protein shell of the T2 virus contains large amounts of sulphur and phosphorus. D) The entire T2 virus particle does not contain protein molecules. E) T2 bacteriophage does not have to infect bacterial cells to reproduce.
A, B
DNA replication in early Drosophila embryos occurs about every 5 minutes. The Drosophila genome contains approximately 1.8×108 1.8×10^8 base pairs. Eukaryotic DNA polymerases synthesize DNA at a rate of approximately 40 nucleotides per second. Approximately how many origins of replication are required for this rate of replication?
Approximately 7500 origins of replication. Five minutes=300seconds. seconds.Working bidirectionally, each origin generates (300sec.)(40nt.)(2)=24,000 nucleotides per second, requiring 1.8×10^8/2.4×10^4=0.75×10^4 origins.
Why do the genomes of eukaryotes, such as Drosophila, need to have multiple origins of replication, whereas bacterial genomes, such as that of E. coli, have only a single origin? Select the three correct statements. A) Eukaryotic genomes are much larger than bacterial genomes. B) Eukaryotic DNA polymerase functions about 10 times slower than bacterial DNA polymerase. C) Replication in eukaryotic cells is unidirectional, whereas replication in bacterial cells is bidirectional. D) Bacterial chromosomes are circular, whereas eukaryotic chromosomes are linear. E) Eukaryotic chromosomes are presumed to replicate different genomic regions at different times during S phase of the cell cycle. F) Replication in eukaryotic cells is dispersive, whereas replication in bacterial cells is conservative.
A,B,E
How does telomerase assemble telomeres? Drag the appropriate labels to their respective targets. some may be used more than once telomerase uses a segment of its _A_ as the template to add multiple copies of a simple sequence to the _B_ of each strand of _C_ on a linear chromosome. This strand, which corresponds to the _D_ for _E_ synthesis, is copied by the normal mechanism of _F_ synthesis after it is extended by telomerase. WORD BANK RNA 3' end template strand lagging strand DNA 5' end
A= RNA B= 3' end C=DNA D= template strand E= lagging strand F= lagging strand
DNA polymerase III is the main DNA-synthesizing enzyme in bacteria. Describe how it carries out its role of elongating a strand of DNA. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all. DNA polymerase III determines which free _A_ is _B_ to the base being copied. DNA polymerase III catalyzes _C_ bond formation between the _D_ of the incoming _E_ and the _F_ group of the last nucleotide added to the strand. WORD BANK nucleotide triphosphate nucleotide diphosphate nucleotide monophosphate parallel antiparallel complementary identical hydrogen phosphodiester 5' phosphate 3' phostphte 3' hydroxyl 5' hydroxyl
A=nucleotide triphosphate B=complementary C=phosphodiester D=5' phosphate E=nucleotide triphosphate F=3' hydroxyl
Which enzyme catalyzes the addition of nucleotides to a growing DNA chain? A) Helicase B) Telomerase C) Primase D) DNA polymerase
D (DNA polymerase catalyzes the addition of nucleotides to a growing DNA chain.)
Choose the correct description of their experiment. A) Meselson and Stahl initially cultured E. coli in medium containing only 15N until all cells contained only 15N/15N DNA. They then cultured the 15N/15N E. coli in medium containing both 14N and 15N and collected samples after one, two, and three rounds of DNA replication. B) Meselson and Stahl initially cultured E. coli in medium containing only 15N until all cells contained only 15N/15N DNA. They then cultured the 15N/15N E. coli in normal medium (14N) and collected samples after one, two, and three rounds of DNA replication. C) Meselson and Stahl initially cultured E. coli in medium containing both 14N and 15N until all cells contained only 14N/15N DNA. They then cultured the 14N/15N E. coli in normal medium (14N) and collected samples after one, two, and three rounds of DNA replication. D) Meselson and Stahl initially cultured E. coli in medium containing both 14N and 15Nuntil all cells contained only 14N/15N DNA. They then cultured the 14N/15N E. coli in medium containing only heavy nitrogen (15N) and collected samples after one, two, and three rounds of DNA replication.
B
If a mutation inactivated DNA polymerase I in a strain of E. coli, would the cell be able to replicate its DNA? If so, what kind of abnormalities would you expect to find in the cell? A) The cell would be able to replicate its DNA. The mutation will result in a newly replicated RNA chain instead of a DNA chain. B) The cell would be able to replicate its DNA. The mutation will result in newly replicated DNA containing small segments of RNA. C) The cell would be able to replicate its DNA. The mutation would not cause any abnormalities. D) The cell would not be able to replicate its DNA.
B
What is the explanation of why Avery, MacLeod, and McCarty's in vitro transformation experiment showed that DNA, but not RNA or protein, is the hereditary molecule? A) The experiment showed that the transforming principle can destroy RNA and protein but not DNA. B) The experiment showed that enzymes that destroyed RNA and protein did not destroy the transforming principle, whereas enzymes that destroyed DNA did. C) The experiment showed that DNA is much more stable and prone to transformation than RNA and protein. D)The experiment showed that heat did not destroy the transforming principle, whereas enzymes that destroyed DNA did.
B
after observing the results of one round of replication, the scientist obtained results from a second round. the purpose of one addition round of replication was to A) distinguish between conservative and dispersive replication B) distinguish between semi-conservative and dispersive replication C) confirm that replication is conservative D) distinguish between conservative and semi-conservative replication
B (After one round of replication, the results of these two possibilities are indistinguishable. A second round was required to distinguish between these two possibilities.)
The Hershey and Chase experiments involved the preparation of two different types of radioactively labeled phage. Which of the following best explains why two preparations were required? for Part A A) Establishing the identity of the genetic material required observation of two phage generations. B) It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment. C) Each scientist had his own method for labeling phage, so each conducted the same experiment using a different isotope. D) The bacteriophage used in the experiments was a T2 phage.
B (Because it was concluded that the component associated with bacteria at the end of the experiment must be the genetic material, it was critical that the component be identifiable as either DNA or protein.)
the role of the primers in PCR (polymerase chain reaction) is A) solely to define the target region B) to define the target region and provide a 3' end that can be extended by taq polymerase C) to denature the template DNA and define the target region D) the annealing temperature is maintained until polymerase has finished extension of the new strands
B (Correct. Primers bind to end of the target DNA strands, thentaq polymerase synthesizes a new strand using the target DNA as a template.)
The minor groove contains less information about the identity of base pairs than the major groove because of the A) glycosidic bond angles and structure of purine bases. B) glycosidic bond angles and structure of pyrimidine bases. C) glycosidic bond angles and geometry of base pairings. D) geometry of base pairing and structure of ribose.
B (The geometry of a base pair is such that the two glycosidic bonds each point outward from their respective bases, thereby creating a narrow (120°) angle on one side of the base pair and a broad (240°) angle on the other side. Moreover, because the glycosidic bonds are found on the N1 of each pyrimidine, only three functional groups can be found in each minor (narrow angle) groove, whereas four functional groups are found in the corresponding major (broad angle) groove.)
how does rolling circle replication differ from bidirectional replication?
Bidirectional replication is a process of replication of genome involves replicating DNA in two direction at the same resulting in formation of one new strand as leading strand and the other as the lagging strand. and Rolling circle replication is a process to replicate the genome (nucleic acid) and synthesize multiple copies of circular molecules of DNA or RNA . It is a unidirectional process of replication in which 3' end of nicked DNA act as a primer and the new strand starts replicate from this end using unnicked DNA as a leading strand template . Actually at the starting of this process of replication circular dsDNA will be nicked (means cleavage of one Phosphidiester bond) of one DNA strand and this DNA strand is known as Nicked DNA and the remaining other Strand is known as Unnicked DNA. The nicked DNA seperate as circular ssDNA after the replication of leading strand completed on the unnicked strand template and now through okazaki Frangments the replication of another strand takes place in this seperate circular ssDNA .
Identify how the alternative models of DNA replication were excluded by the data. A) The conservative model predicted that after the first round of replication there would be three types of DNA, which remain throughout. The dispersive model predicted that after each round of replication, there would be only one form of DNA. B) The conservative model predicted that after each round of replication, there would be only one form of DNA. The dispersive model predicted that after the first round of replication there would be three types of DNA, which remain throughout. C) The conservative model predicted that the original 15N/15N DNA would remain throughout along with the new 14N/14N DNA. The dispersive model predicted that after each round of replication, there would be only one form of DNA, which would become less and less dense. D) The conservative model predicted that after the first round of replication there would be two types of DNA. The dispersive model predicted that the original 15N/15N DNA would remain throughout.
C
to help achieve proper base pairing and hence form a double helix, which condition must be met? A) a purine base must pair with a purine base B) a pyrimidine base must pair with pyrimidine base C) a purine base must pair with a pyrimidine
C (Chargaff's rules state that there is a 1:1 ratio of purines to pyrimidines. More specifically the amount of adenine is equal to the amount of thymine, and the amount of cytosine is equal to the amount of guanine. Two purines would place the bases too far apart for hydrogen bonds to form. Two pyrimidines would place the bases too close together, making the bond unstable due to steric repulsion.)
What were the results of the meselson-stahl experiments relied on all of the following except A) a means of distinguishing among the distribution patterns of newly synthesized and parent molecule DNA possible B) that a heavy isotope of nitrogen could be incorporated into replicating DNA molecules C) the fact that DNA is the genetic material D)a cesium chloride gradient
C (Correct. This fact had already been established and was not of any consequence in these experiments.)
To be certain that the extract prepared from virulent cells still contained the transforming principle that was present prior to lysis, Avery _______. A) incubated virulent cells with the complete extract B)destroyed proteins, polysaccharides, DNA, and RNA contained in the extract C) incubated nonvirulent cells with the complete extract D) injected mice with the extract
C (The complete extract possessed the same ability to induce transformation in IIR bacteria as whole heat-killed IISbacteria.)
If a DNA-binding protein "reads" a short stretch of DNA and detects the following "second" genetic code provided by the functional groups located on each base as H-HD-CH3-HA-HA-HA-HA-HD, then what is the corresponding sequence of bases? A) C-T-A-G B) C-G-G-A C) C-T-G-A D) C-A-G-A
C (The first two functional groups (nonpolar hydrogen and exocyclic amino group) indicate cytosine, the second two (methyl and carbonyl groups) indicate thymine, the next two (N7 and carbonyl group) indicate guanine, and the last two functional groups (N7 and exocyclic amino group) indicate the presence of an adenine)
which of the following statements about DNA structure is true A) nucleic acids are formed through phosphodiesteer bonds that link nucleotides together B) hydrogen bonds formed between the sugar-phosphate backbones of the two DNA chains help to stabilize DNA structure C) the nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions D) the pentose sugar in DNA is ribose
C (This statement is true; the 5′-3′ orientation of each chain runs in opposite directions.)
DNA replication in early Drosophila embryos occurs about every 5 minutes. The Drosophila genome contains approximately 1.8×1081.8×108 base pairs. Eukaryotic DNA polymerases synthesize DNA at a rate of approximately 40 nucleotides per second. Approximately how many origins of replication are required for this rate of replication? A) 45000 B) 15000 C) 1.2 X 10^6 D) 7500 E) 1500
D
What results from the experiments of Frederick Griffith provided the strongest support for his conclusion that a transformation factor is responsible for heredity? A) The fact that injection of living SIII bacteria into mice produced illness and death. B) The fact that injection of a mixture of heat-killed SIII and living RII strains of Pneumococcus into mice did not induce illness. C) The fact that injection of "heat-killed" SIII bacteria into mice did not induce illness. D) 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.
D
If Avery had observed transformation using only the extracts containing degraded DNA, degraded RNA, and degraded protein, but NOT the extract containing degraded polysaccharides, he would have concluded that _______. A) mice with diets rich in polysaccharides are resistant to bacterial infection B) RNA is the genetic material C) the preparations were contaminated D) polysaccharides are the genetic material
D (Failure to transform suggests that the chemical degraded in that preparation is the one responsible for transformation, in this case polysaccharides.)
Which of the following statements about DNA replication is true? A) Okazaki fragments are DNA fragments synthesized on the leading strand. B) DNA gyrase unwinds the DNA double helix. C) DNA polymerase adds dNTP monomers in the 3′-5′ direction. D) Single‑strand binding proteins stabilize the open conformation of the unwound DNA.
D (Once helicase unwinds the double helix, single‑strand binding proteins bind to the open DNA and prevent it from winding together again.)
Bloom syndrome (OMIM 210900) is an autosomal recessive disorder caused by mutation of a DNA helicase. Among the principal symptoms of the disease are chromosome instability and a propensity to develop cancer. Explain these symptoms on the basis of the helicase mutation.
DNA helicases unwind dsDNA during DNA replication and repair. Bloom syndrome is characterized by chromosome instability and an increased rate of cancer. Chromosome instability is evident when chromosomes are lost from cells, typically because of a failure during mitosis. Mitosis fails to occur properly if chromosomes are not completely replicated. Cancer is a disease caused by accumulation of somatic mutations, which will accumulate at an elevated rate if DNA repair by DNA replication is defective. Based on the information provided, it is reasonable to speculate that lack of the DNA helicase encoded by the Bloom syndrome gene results in incomplete replication during S phase and during repair of DNA damage. Failure to completely replicate chromosomes could result in a failure to pass chromosomes on to progeny cells during mitosis, which would result in chromosome instability. Failure to repair DNA damage would also lead to an increased rate of somatic mutation, which would lead to cancer.
DNA polymerase III is the main DNA-synthesizing enzyme in bacteria. describe how it carries out its role of elongating a strand of DNA.
DNA pol III determine which free nucleotide triphosphate is complementary to the base being copied. DNa pol III catalyzes phosphodiester bond formation between the alpha phosphate of the incoming nucleotide triphosphate and the 3' hydroxyl group of the last nucleotide added to the strand
In a dideoxy DNA sequencing experiment, four separate reactions are carried out to provide the replicated material for DNA sequencing gels. Reaction products are usually run in gel lanes labeled A, T, C, and G. c) Why is incorporation of a dideoxynucleotide during DNA sequencing identified as a "replication-terminating" event?
Dideoxynucleotides contain a hydrogen group instead of a hydroxyl group on their 3′ carbon. When a dideoxynucleotide is incorporated into a growing DNA strand, there is no 3′ hydroxyl group present to allow phosphodiester bond formation with the next nucleotide to be added; therefore, no additional nucleotides are added to this DNA strand, and thus synthesis of this strand is terminated.
In a dideoxy DNA sequencing experiment, four separate reactions are carried out to provide the replicated material for DNA sequencing gels. Reaction products are usually run in gel lanes labeled A, T, C, and G. b) how does the PCR play a role in dideoxy DNA sequencing?
Dideoxysequencing uses DNA synthesis to generate labeled DNA fragments of different lengths, which are then resolved by gel electrophoresis or column chromatography. To visualize the products of DNA synthesis in traditional dideoxysequencing, relatively high levels of template were necessary. The use of PCR allows detectable levels of DNA synthesis from much lower levels of template DNA.
bacterial DNA polymerase I and DNA polymerase III perform different functions during DNA replication. c) if a strain of E.coli acquired a mutation that inactivated DNA polymerase III function, would the cell be able replicate its DNA? why or why not?
E.coli mutant without a functional DNA pol III will be unable to replicate its DNA because It lacks the enzyme responsible for the bulk of DNA synthesis during replication
there is a problem completing the replication of linear chromosomes at their ends b) what is the function of telomerase, and how does it operate to synthesize telomeres?
End-lengthening by telomerase counteracts end-shortening due to lagging strand. DNA synthesis, resulting in the maintenance of chromosomes length in the cells expressing telomerase. Telomerase is a reverse transcriptase that synthesize the DNA using an endogenous RNA as a template. Telomerase synthesize DNA on to the 3' end of both DNA strand of a chromosome, creating a long single stranded DNA that is then replicated by lagging strand synthesis.
DNA fragments that are 600 bp long will migrate more quickly through a sequencing gel than fragments that are 150 bp long. TRUE OR FALSE
FALSE (Small DNA fragments have less hindrance in moving through the gel, so they migrate more quickly than larger fragments)
a sample of double-stranded DNA is found to contain 20% cytosine. determine the percentage of the three other DNA nucleotides in the sample ?
In any DNA the amount of cytosine is always equal to amount of guanine and amount of adenine is equal to the amount of thymine.We know that C= 20% , C = G and A=T ...............equation 1from this we know that amount of guanine is 20%A+T+G+C = 100%A+A+20+20 =100 ...............from equation 12A+ 40 =100A = 100-40/2A=30therefore the amount of thymine is 30%
there is a problem completing the replication of linear chromosomes at their ends a) describe the problem and identify why telomeres shorten in each replication cycle
Linear chromosomal DNA is replicated by bidirectional replication, which involve leading and lagging strand synthesis. The problem replicating the ends of linear chromosome is due to lagging strand synthesis, which is responsible for replicating one DNA strand at each chromosomal end. A lagging strand synthesis initiates at RNA primer and move inwards, away from the chromosomal end. Because of this, one DNA strand at each end of chromosome is not replicated all the way to the end. This happens every time a linear chromosome is replicated, which cause ends of the chromosome shorten with each round of replication.
what would be the effects on DNA replication if mutation of DNA pol III caused it to lose each of the following activities 3' to 5' exonuclease activity
Loss of 3' to 5' exonuclease activity leads to issues with the fidelity of replication. In other words, the polymerase would not correct its mistakes but it would still perform the replication with errors
telomeres are found at the ends of eukaryotic chromosomes b) how does telomerase assemble in telomeres
Telomerase uses a segment of its RNA as the template to add multiple copies of a simple sequence to the 3' end of each strand of DNA on a linear chromosome. This strand, which corresponds to the template for lagging strand synthesis, is copied by the normal mechanism of lagging strand synthesis after it is extended by telomerase
telomeres are found at the ends of eukaryotic chromosomes c) what is the functional role of telomeres?
Telomeres are thought to provide two functions, one in chromosome replication and the other in chromosome protection. Telomeres provide a mechanism for replication of the ends of linear chromosomes. Without telomeres, lagging strand synthesis would fail to extend to the chromosome ends, leaving a gap at each end after each round of replication. This would shorten the chromosome and, after many rounds of replication, would result in loss of important DNA sequences (genes). Telomeres are repetitive DNA, which prevents loss of important DNA sequences if shortening occurs. Telomeres are also the binding site for telomerase, which extends the lagging strand template to compensate for sequences lost during incomplete lagging strand synthesis. Telomeres also provide a protective "cap" on the ends of linear chromosomes; this cap distinguishes normal chromosome ends from ends generated by double-stranded chromosome breaks (DNA damage). Without telomeric DNA and the proteins that bind telomeric DNA, the ends of chromosomes are recognized as broken chromosomes and are fused together by DNA repair enzymes. Such breakage can create chromosome end-to-end fusions, which then create dicentric chromosomes that can be broken during the next cell division, creating new breaks and new fusions in an endless cycle known as the bridge-break-fusion cycle.
Hershey and Chase selected the bacteriophage T2 for their experiment assessing the role of DNA in heredity because T2 contains protein and DNA, but not RNA. Explain why T2 was a good choice for this experiment.
The entire T2 virus particle does not enter the bacterial cell.T2 contains DNA and protein but no RNA.
A PCR reaction begins with one double-stranded segment of DNA. How many double-stranded copies of DNA are present after the completion of 10 amplification cycles? After 20 cycles? After 30 cycles?
The number of double stranded copies of DNA after a given number of cycles (say n) = 2n So, after 10 cycles , it will be 210 After 20 cycles, it will be 220 After 30 cycles,it will be 230 So, after 10 cycles , it will be 210 = 1024
In a dideoxy DNA sequencing experiment, four separate reactions are carried out to provide the replicated material for DNA sequencing gels. Reaction products are usually run in gel lanes labeled A, T, C, and G. a)Identify the nucleotides used in the dideoxy DNA sequencing reaction that produces molecules for the A lane of the sequencing gel.
The reaction would have equal concentrations of deoxycytidine triphosphate, deoxythymidine triphosphate, and deoxyguanidine triphosphate. It would also have a mixture of deoxyadenosine triphosphate and dideoxyadenosine triphosphate.
in which strand or both strand is synthesized 5' to 3'
both strands (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.)
Joel Huberman and Arthur Riggs used pulse-chase labeling to examine the replication of DNA in mammalian cells. Briefly describe the Huberman-Riggs experiment, and identify how the results exclude a unidirectional model of DNA replication.
cells were incubated in medium 3H-thymine for a short period of time (pulse) and then transferred to medium containing an excessive of unlabeled thymine (the chase). the cells were then collected and their DNA was perpetrated for electron microscopy which can detect replication structures of DNA and for autoradiography which reveals the location of 3H thymine incorporation into DNA. the results showed DNA replication bubbles that contained regions of label on both ends of the bubble. since bidirectional replications produced a replication bubble with DNA synthesis occurring at both ends, whereas unidirectional replication results in a replication bubble with DNA synthesis occurring at one end, these results excluded unidirectional DNA replication and supported bidirectional replication
One strand of a fragment of duplex DNA has the sequence 5'-ATCGACCTGATC-3'. f) identify the bond that joins one strand of a DNA duplex to the other strand g) is the bond in part (f) a covalent or a non covalent bond?
f) hydrogen bonds (joins one strand of DNA duplex to the other strand in which two hydrogen bond between adenine nitrogen base and thymine nitrogen base, three hydrogen bond between guanine and nitrogen base to cytosine nitrogen base) g) hydrogen bond is not a covalent bond
The data obtained from the Meselson-Stahl experiment after one generation of replication eliminated the dispersive model of DNA replication. true or false
false (The data obtained from the Meselson-Stahl experiment after one generation was consistent with both the semiconservative and the dispersive model of DNA replication. The conservative model of DNA replication was eliminated because it predicted that there would be two bands representing the original DNA at one density and the newly replicated DNA at a different density.)
One strand of a fragment of duplex DNA has the sequence 5'-ATCGACCTGATC-3'. h) what term is used to describe the pattern of base pairing between one DNA strand and its partner in a duplex
h) complementary base pairing (between on DNA strand and its partner in a duplex occur between adenine-thymine and guanine-cytosine)
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. what protein binds at the replication fork and breaks H-bonds between bases
helicase (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.)
list the order in which the following proteins and enzymes are active in E.coli DNA replication: DNA pol I, SSB, helixes, DNA pol III and primase
helicase, SSB, primase, DNA pol III, DNA pol I, ligase
figure 1.6 presents simplified depictions of nucleotides containing deoxyribose, a nucleotide base, and a phosphate group. use this simplified method of representation to illustrate the sequence 3'-AGTCGAT-5' and its complementary partner in a DNA duplex b) what kind of bonds join the C in one strand to the G in the complementary strand?
hydrogen bonds
in which strand or both are multiple primers needed, the daughter strand elongates away from replication from and is made in segments
lagging strand (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.)
in which strand or both is only one primer needed, daughter strand elongates toward the replication fork and is made continuously
leading strand (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.)
figure 1.6 presents simplified depictions of nucleotides containing deoxyribose, a nucleotide base, and a phosphate group. use this simplified method of representation to illustrate the sequence 3'-AGTCGAT-5' and its complementary partner in a DNA duplex a) what kind of bond joins the C to the G within the single strand?
phosphodiester bonds
bacterial DNA polymerase I and DNA polymerase III perform different functions during DNA replication. b) if mutation inactivate DNA polymerase I in a strain of E.coli, would the cell be able to replicate its DNA? if so, what kind of abnormalities would you expect to find in the cell?
the absence of DNA pol I will not prevent the bulk of DNA replication but will result in newly replicated DNA containing small segments of RNA and nicks at the junctions of polymerase III synthesized DNa and the 5' end of the RNA primers
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.) pol I binds to 5' end of primer A, pol I replaces primer A with DNA, pol III moves 5' to 3' adding DNA nucleotides to primer B pol II binds to 3' end of primer B->
pol II binds to 3' end of primer B->pol III moves 5' to 3' adding DNA nucleotides to primer B->pol I binds to 5' end of primer A->pol I replaces primer A with DNA (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.)
Matthew Meselson and Franklin Stahl demonstrated that DNA replication is semiconservative in bacteria. Briefly outline their experiment and its results for two DNA replication cycles, and identify how the alternative models of DNA replication were excluded by the data.
recall call that meselson stahl experiment and consider how the results exclude the alternatives to the semiconservative model for DNA replication. they initially culture e.coli in medium contain N15 (heavy nitrogen) unitil all cells contained only N15/N15. they then cultured the N15/N14 E.coli in normal medium (N14) and collected samples after, one, two, and three rounds of DNA replication. these results showed that before transfer to N14 medium, only N15/N15 DNA was present. after one round of replication in N14 medium all of the DNA was N15/N14 after two rands of replications, half the DNA was N15/N14 and half was N14/N14 and after the third round of replication 1/4 of the DNA was N15/N14 and 3/4 was N/14/N14. The conservative model predicted that the original N15/N15 DNA would remain throughout and therefore the results rule out the conservative model after one round of replication. the dispersive model predicted that each round of replication there would be only one form of DNA, which would become less and less dense. although this model was not rule out after one round of replication, there persistence of the N15/N14 DNA and the presence of two classes of DNA (N15/N14 and N14/N14) after rounds two and three ruled out the dispersive model
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. what protein prevents H bonds between bases and binds after the replication fork
single strand binding protein (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.)
the principles of complementary base paring and antiparallel polarity of nucleic acid strands in a duplex are universal for the formation of nucleic acid duplexes. what is the chemical basis for this universality?
the chemical bonds that form base pairs in double stranded dna are weak noncovlaent bonds called hydrogen bonds. hydrogen bonds involve two atoms sharing a hydrogen nucleus, and the distance between the atoms sharing the hydrogen nucleus is critical for hydrogen bonds to form. the bases in the two complementary antiparallel DNA strands are aligned such that each of the atoms that share a hydrogen nuclei (N and O or N and N) in each base are positioned next to each other at a distance that allows all possible hydrogen bonds to form. the bases in the complementary but parallel strands are not aligned in this manner therefore the atoms that could form hydrogen bonds do no align and are not close enough together to allow hydrogen bonding between all possible and necessary chemical groups
Raymond Rodriguez and colleagues demonstrated conclusively that DNA replication in E. coli is bidirectional. Explain why locating the origin of replication on one side of the circular chromosomes and the terminus of replication on the opposite side of the chromosome supported this conclusion.
the copy strands we're produced by formation of two replication forks that proceeded along the circular chromosomes in opposite directions away from the origin. this resulted in the replication of each strand of the chromosome by each replication fork. these forks final merged at a specific point on the chromosome exactly opposite to the organ of terminus orf replication.
A family consisting1 of a mother (I-1), a father (I-2), and three children (II-1, II-2, and II-3) are genotyped by PCR for a region of an autosome containing repeats of a 10-bp sequence. The mother carries 16 repeats2 on one chromosome and 21 on the homologous chromosome. The father carries repeat numbers of 18 and 26. c)What genetic term best describes the pattern of inheritance of this DNA marker? Explain your choice.
the genetic temp that best describes the pattern of DNA inheritance is codominance. co dominance is a relationship between two versions of a gene in which individual received a single allele from each parent if the received alleles are dominant a hybrid character is expressed
Explain why Avery, MacLeod, and McCarty's in vitro transformation experiment showed that DNA, but not RNA or protein, is the hereditary molecule.
the key results were those showing that enzymes that destroyed RNA and protein did not destroy the transforming principle, whereas enzymes that destroyed DNA did. the transforming principal was considered to be genetic material. the most reasonable interpretation of those results that DNA was the only essential component of the transforming principle and therefore the genetic material
A family consisting1 of a mother (I-1), a father (I-2), and three children (II-1, II-2, and II-3) are genotyped by PCR for a region of an autosome containing repeats of a 10-bp sequence. The mother carries 16 repeats2 on one chromosome and 21 on the homologous chromosome. The father carries repeat numbers of 18 and 26. b)Identify all the possible genotypes of children of this couple by specifying PCR fragment lengths in each genotype.
the possible genotypes in children are (16,18) (16,26) (21, 18) (21 26) the possible length of DNA fragments are 34,42,39,47
figure 1.6 presents simplified depictions of nucleotides containing deoxyribose, a nucleotide base, and a phosphate group. use this simplified method of representation to illustrate the sequence 3'-AGTCGAT-5' and its complementary partner in a DNA duplex c) how many phosphodiester bonds are present in this DNA duplex ?
there are 12 phosphodiester bonds in the molecule
figure 1.6 presents simplified depictions of nucleotides containing deoxyribose, a nucleotide base, and a phosphate group. use this simplified method of representation to illustrate the sequence 3'-AGTCGAT-5' and its complementary partner in a DNA duplex d) how many hydrogen bonds are present in this DNA duplex?
there are 17 hydrogen bonds in the DNA molecule
Explain how the Hershey and Chase experiment identified DNA as the hereditary molecule.
they prepared T2 particles whose protein was labeled with the radioactive sulfur S35 and whose DNA was labeled with radioactive phosphorus P32. the used the labeled T2 to infect bacteria and then separated the infected bacteria from the employ phage shells using a blender. they found that essential al the P32 labeled T2 DNA but little to none of the S35 labeled T2 protein was in the infected bacterial cells. sine T2 genetic material must be inside the infected cells to direct new virus particle synthesis. these results pointed to DNA as the genetic material of phage T2
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. what protein binds ahead of the replication fork and breaks covalent bonds in DNA backbone
topoisomerase (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.)
guanine and adenine are purines found in DNA true or false
true (Guanine and adenine are indeed purines found in DNA; thymine and cytosine are the pyrimidines found in DNA.)
