Mod 2: DNA & RNA Structure quiz+ Wksht 1 + quiz 1(45min)

QUIZ STARTS HERE

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Within a nucleotide, sugar position where nitrogenous base binds:

1'

R-group

1. An R-group is a molecule's side chain containing either simply a Hydrogen(H), a Carbon or multiple Carbons, a ring structure, or another combination of H, N, O elements. The R group helps with distinguishing the name and type of structure. & importantly: R-group chemical composition influences biochemical behavior of the amino acid, and determines higher order conformation of the polypeptide

Number of hydrogen bonds between complementary A & T nucleotides:

2

Number of hydrogen bonds between complementary G & C nucleotides:

3

Within a nucleotide, sugar position where phosphate group binds:

5'

RNA sequence complementary to DNA sequence: 5' - AATTGGCC - 3'

5' - GGCCAAUU - 3'

Polarity of a polynucleotide refers to:

5' phosphate-3' hydroxyl

Peptide bond

A peptide bond is a special strong covalent bond that forms bond between amino and carboxyl groups of neighboring amino acids. The result is a strong connection between the two molecules that can only be broken by the elimination of water also known as hydrolysis.

Translation

After the RNA strands have been generated, translation occurs where a molecule called ribosome comes by and its duty is to latch onto the free-floating strand of RNA. Once this RNA-ribosome complex forms, another molecule called a tRNA swings by onto the complex where it performs codon (three nucleotides) pattern reading to generate a polypeptide strand, thus forming a protein. A polypeptide chain formed by amino acids, which are codons made up of nucleotides.

Which of the following is not true about Southern blotting:

Autoradiogram results are indicative of gene expression pattern

What is the purpose of including ddNTPs in the Sanger sequencing procedure?

Cause termination of polymerization to generate different sized DNA fragments which can be separated by gel electrophoresis

A 13-nucleotide section of an autoradiogram from a Sanger sequencing experiment is depicted in the image below. Based on the band pattern observed here, write out the sequence of both the complementary strand generated during the experiment, and the template strand that is being analyzed. Be sure to clearly indicate the 5' and 3' ends of each.

Complimentary strand sequence: 3' AGGACTTTGACAT 5' Template strand sequence: 3' ATGTCAAAGTCCT 5'

Referring again to pBR322, explain the significance of the amp and tet genes in the context of creating recombinant DNA.

First, it is important to know that amp stands for ampicillin and tet stands for tetracycline antibiotics. The amp and tet are the selection markers that are tagged into the plasmid vector so that we can verify after inserting these plasmid vectors into cells that the surviving cells are resistant to amp and tet. Essentially, the newly derived cells survived compared to control, non-transformed cells. Furthermore, these genes can be used for screening of transformants carrying recombinant inserts; if insert is in ampR gene then they will no longer be resistant to ampicillin, and colonies will not show(died bc not resistant). Vice versa for TetR meaning that if the insert is in tetR gene then they will be resistant to tetracycline and colonies will grow in the presence of tetracycline

Reverse transcriptase

Following mRNA binding to oligo-dT, reverse transcriptase is a vital enzyme used in cDNA synthesis. Its role is to work together with the dNTPs to generate DNA from a strand of mRNA that is why it is called 'reverse.'

The DNA base composition for two different bacterial species is indicated in the table below. Based on this information, what can you say about the G/C & A/T content of these species? What would you predict about the relative DNA melting points of these two species? Explain.

For species 1 I can see that A/T(total = 60) are present in higher quantities compared to the GC(total = 40). A/T have only two H-bonds between, compared to G/G who have 3 H-bonds between. For species 2, I can see that G/C(total = 70) are present at higher quantities than A/T(total = 30). I know that G/C has 3H-bonds between thus making species 2 the more stable (more G/C content = H-binds more stable) species than species 1. Species 2 has a higher melting point(higher heat required) for DNA due to more G/C bonds.

Recount the significance of the famous photograph 51 generated by Rosalind Franklin. What particular details were deduced from this image?

Franklin's experiments using an x-ray illustrated the diffraction pattern of DNA when shot with an x-ray machine. Double helix; sugar-phosphate backbone; nitrogenous bases stacked in the center; purine/pyrimidine pairing; major groove/minor groove; one turn of helix = ~10 bp

Purines

G and A

Compare and contrast genomic libraries vs. cDNA libraries; indicate similarities and differences in how they are created; and highlight the general kinds of information that can be gained from each

Genomic libraries contain a lot of the genetic information, like a database of genes, for specific organisms. It is useful when aiming to study a chromosomal DNA fragment from an organism's genomic library. You would go through the experimental steps of incorporating restriction enzymes(RE) into a mixture with chromosomal DNA known as a "digest." This would break up the strands depending on how the RE cut allowing these fragments to join inside a vector, and subsequently be transformed into a host cell. cDNA libraries are different because they work with DNA that has been derived from "copy/complimentary DNA." The cDNA library is created in multiple steps. From an organisms, we isolate the mRNA that is hen mized with the enzyme, reverse transcriptase. What this enzyme does is attach itself to the mRNA strand and generates a complimentary strand called cDNA that is why it is called 'reverse.' The next step involves inserting the newly formed DNA into bacterial plasmid. The plasmid will take up the new DNA which we can incorporate into our bacteria in a series of steps of transformation/plating. The colonies that grow will be the key ones to place in tubes and isolate the plasmid from. A few more steps are taken to clean the DNA from any other residues to form the official cDNA sequence.

Describe the chemical basis of base pairing that occurs in the DNA double helix and that occurs in DNA-RNA interactions. Be sure to mention the types of chemical bonds and the specific pairing interactions.

In a DNA double helix, base-pairing involves bringing together two single strands in an anti-parallel fashion, one strand is 5 to 3 direction and other is 3 to 5 direction, that way the Hydrogen(H) bonds can form appropriately between the two strands. The H bonds form between the A&T and G&C which are the pyrimidine/purines located in the inner side of the double helix. Between A&T nucleotides there are 2 H bonds; however, between the G&C nucleotides there are 3 H bonds.

PCR

PCR is a process for which we can make multiple copies of a gene that we are interested in experimenting with. The following ingredients go into a PCR mixture tube: Oligonucleotide primer, Thermostable DNA polymerase, buffer, DNA template, and dNTPs. Once combined, this mixture goes into a special tube that is inserted in the PCR machine, in which it will run the steps of denaturation, annealing, and extension.

Explain why DNA & RNA are described as exhibiting polarity. Be sure to mention the particular molecular components that account for this polarity.

Polarity is another way of saying directionality of a strand or in other words, in which direction to read a strand. DNA and RNA are described as having polarity because their structure runs from 5' to 3' meaning that distinct 5' end / 3'end; 5' end = free phosphate group bound to 5' carbon of sugar; 3' end = free OH group of 3' carbon of the sugar

Which of the following is not true about RNA in comparison to DNA:

Presence of 2' hydroxyl group increases stability of RNA

Components of a nucleotide:

Purine(A,G) or Pyrimidine base (T,C, U), 5-carbon sugar, phosphate group

Explain why the relative instability and structural diversity characteristics of RNA are important.

RNA instability important so that levels of RNA can fluctuate, important component of gene regulation; structural diversity reflective of function diversity: e.g. tRNA; hairpin loops for regulation

Which of the following is not suggested by the Watson-Crick DNA Double Helix model:

RNA polymerase binding to promoter sequence

Which of the following methods can be used to amplify mRNA sequences?

RT-PCR (uses mRNA and converts it to cDNA using RT)

Southern Blotting

Research method for analysis of DNA sequences; 1) extract and purify DNA from cells 2) DNA is restricted with enzymes 3) DNA fragments separated by agarose gel; 4) Denature DNA 5) Transfer to nitrocellulose paper (blotting); 6) Add labeled probe for hybridization to happen 7) Wash off unbound probe 8) Autoradiograph

Northern Blotting

Research method for analysis of RNA sequences; 1) extract and purify RNA from cells 2) RNA is restricted with enzymes 3) RNA fragments separated by agarose gel; 4) Denature RNA 5) Transfer to nitrocellulose paper (blotting); 6) Add labeled probe for hybridization to happen 7) Wash off unbound probe 8) Autoradiograph

Hybridization

Research method of annealing between complementary nucleotide sequences of DNA/DNA or DNA/RNA - occurs in like PCR, blotting techniques, cloning ,etc

Creates "sticky ends" in DNA:

Restriction endonuclease

Restriction mapping

Restriction mapping is the method for identification of restriction enzyme cutting sites within a particular segment of DNA(any span of DNA); involves restriction enzyme digestion followed by gel electrophoresis to pinpoint the specific cut sites by the size of the fragments created

A method for analysis of DNA fragments transferred from an electrophoresis gel:

Southern blot

Central dogma

The Central Dogma is a proposed and now proved molecular concept that explains formation of a protein starts with a replicated amount of DNA undergoing transcription to make an RNA strand, then RNA undergoes translation to produce the protein strand.

Protein secondary structure

The secondary structure for a protein can take on two different structures. One structure is the alpha helix, which is characteristic of a helical/coiled shape. The second structure is the Beta sheet which looks more flat in shape like a sheet and come in 2 or more strands bound together, never alone. Structure is due to: hydrogen bonds between carbonyl & hydroxyl groups of every 4th amino acid within a polypeptide chain

tertiary protein structure

The tertiary structure of a protein is more complex than the secondary because it requires the bridging of both types of secondary structures in a 3D figure, rather than straight 2D as seen in protein secondary structures. These tertiary structures have a greater capacity for flexibility and bond interactions. Structure due to R-group interactions: ionic bonds, H bonds, hydrophobic interactions, covalent disulfide bonds between cysteine amino acids

Which of the following is not necessary of RT-PCR analysis

Thermoresistant DNA ligase

Oligonucleotide primer

These primers are small fragment of DNA less than ~50 nucleotides designed in an external polynucleotide synthesis machine, and the person designing them has a general understanding of the gene they are targeting thus they can pull a primer from the sequence database desired. Usually this oligonucleotide primer is within a specific length according to its stable melting point(Tm). As will be learned in the next term, these primers are crucial for binding to another PCR ingredient to allow for amplification of the targeted sequence.

Protein quaternary structure

These structures are a combination of protein tertiary structures that each contains its specialized compartment. In order to maintain the structure very intact, this higher-level structure is due to R-group interactions: ionic bonds, H bonds, hydrophobic interactions, covalent disulfide bonds between cysteine amino acids

Dideoxynucleotide

This nucleotide is composed of the regular three parts of a nucleotide(sugar, nit base, P); however, it is missing a 3'OH on the sugar which is necessary for synthesis in Sanger sequencing. Therefore, when synthesis reaches this dideoxynucleotide, the process terminates synthesis since there is no phosphodiester bond to allow for continuation.

follow up questions from above.... How would you be able to confirm that HindIII, EcoRV and BamHI are located between EcoRI and SalI, but not between EcoR1 and PstI? Be sure to explain by indicating restriction fragment sizes.

To confirm where the restriction sites are, we would have to perform restriction digests. This molecular technique involves mixing the plasmid with restriction enzymes, which cut at their specific sites to generate bands through gel electrophoresis analysis. These bands show because they have been tagged with a dye and therefore, can be analyzed for their bp size by comparing to a general reference ladder of dye measurements. Additionally, we could use 1 RE or 2 RE depending on what we are trying to confirm. For example, if we wanted to know confirm that HindIII, EcoRV and BamHI are located between EcoRI and SalI, but not between EcoR1 and PstI, we would make a digest with all the following RE: EcoRI, HindIII , EcoRV, BamH1, Sal I, and PstI. The result would show multiple fragments with one being an uncut piece between EcoRI (4359) and PStI (3607) that has a size of 4359-3607 = 752. This goes to show that there the other RE are not located between PstI and EcoRI. We can clearly see that the other RE are between EcoR1 and SalI because it led to many fragments between.

Transcription

Transcription is the process in which the DNA double helix uncoils itself partly to allow RNA polymerase, an enzyme that extends along the uncoiled strands of the open helix. RNA polymerase essentially creates a copy strand side by side the DNA open strands, and these new strands are called the mRNA templates or RNA.

You are working in a lab focused on lung cancer research. A protein called LUNC1 is detected in abnormally high levels in some types of lung cancer. In some cancers, gene amplifications are known to occur, and in other types, some cancer-associated genes are transcribed at very high levels. To explore this further, you decide to use Southern and Northern blotting techniques to compare LUNC1 gene levels and expression levels from normal brain, heart, and lung samples as well as from lung cancer cells. The results of these blots are shown below. Based on these data, what can you conclude about the LUNC1 gene? Explain.

We can see that the southern blot result shows DNA sequences that are fairly similar if not equal. However, in the Northern blot result, we can identify that the expression of RNA varies. The brain shows a decreased expression of RNA. The heart has greater expression than the brain, the lung has greater expression than both brain and heart, and the lung cancer cells show the greatest expression over the other three samples. Therefore, we can conclude that the protein LUNC1 has a higher level of expression of RNA in lung cancer cells. We can also say that there is more hybridization with the probe (single-stranded fragment) for Northern blot since there is more complimentary sequences.

Photograph 51 technique:

X-ray diffraction

In deoxyribonucleotides, which of the following correctly describes the bonding between the pentose sugar and a pyrimidine base?

a. 1' carbon of the pentose forms glycosidic bond with N-1 of the base

End product is representative of genes that are transcriptionally active in a particular cell or tissue sample:

cDNA library (using cell and tissue sample) northern blot analyzes the gene expression for RNA so it could also work but not the correct answer

Pyrimidines:

cytosine, thymine, uracil

This occurs when DNA is heated above its melting temperature:

denaturation

Used in Sanger sequencing, this component interrupts the continued synthesis of a new DNA strand:

dideoxynucleotide

1. Examine the structure of pBR322 as depicted in the image provided on the next page to help you answer the following questions: a. Indicate how many fragments would be generated in each of the following four different restriction digestion of pBR322: 1) EcoR1 2) EcoR1 + BamH1 3) EcoR1 + Pst1 4) EcoR1 + SalI

fragment # 1) EcoR1 = there is only one EcoR1 restriction enzyme (RE) cut site so it would cut and make the circular DNA into linear so 1 fragment 2) EcoR1 + BamH1 = there is one RE for each so that's a total of two cuts made to the circular DNA making it two fragments. 3) EcoR1 + Pst1 = again, there is only one RE site for each of these for a total of 2 cut sites and therefore 2 fragments. 4) EcoR1 + SalI = again, there is only one RE site for each of these for a total of 2 cut sites and therefore 2 fragments.

Chromosomal DNA fragments inserted into cloning vectors and subsequently introduced into bacterial cells creates a:

genomic library

High G/C content is associated with:

high DNA melting temperature

Probe

hybridization A probe is typically a single-stranded fragment that can be either DNA or RNA. This probe tends to be used in hybridization. The probe is tagged with usually labeled with florescent or radioactive tag which binds to the gene fragments and can help with the visualization step that is viewed under autoradiogram.

follow up questions from above.... The entire plasmid is 4,361 bp in size, and the numerical position of each restriction site is indicated in the blue numbers. Using this information, the sizes of the DNA fragments generated in each of the restriction digest reactions described in your answer above. 1) EcoR1 2) EcoR1 + BamH1 3) EcoR1 + Pst1 4) EcoR1 + SalI

i. EcoR1 alone: Since the EcoR1 only makes one cut and simply opens up the circular DNA there is one fragment that maintains the same bp number as the entire plasmid = 4361 ii. EcoR1 + BamH1: We are shown in the image that EcoR1 cuts at 4359bp and BamH1 cuts at 375bp therefore we would start by finding the fragment size between EcoR1 (4359) and Bam H1(375), which is 375 + 2bp to the left of zero= 377. Then, we would take the entire plasmid value 4361 -377 = 3984, which is the other fragment size. iii. EcoR1 + Pst1: This time with start with finding the fragment size between the sites EcoR1(4359) and PSt1(3607) by simply subtracting the two 4359-3607= 752. Then, we would find the outer fragment by taking the entire plasmid value 4361 - frag 1(752) = 3609bp iv. EcoR1 + SalI: we would start by finding the size of the fragment between Sal1(651) and EcoR1(4359) which is 651 + 2 to the left of zero = 653. Now, we just subtract thus value from the entire plasmid value 4361-653 = 3708.

nucleoside vs. nucleotide

nucleoside is composed of only a nitrogenous base and a 5Csugar while a nucleotide is composed of the following three parts: nitrogenous base, 5C sugar, phosphate

Which of the following levels of protein structure does not involve hydrogen bonding?

primary

purine vs pyrimidine

purine = double ringed nitrogenous base (adenine & guanine) pyrimidine = single ring nitrogenous base (cytosine and thymine and uracil)

Briefly describe at least three types of studies enabled by recombinant DNA, and indicate the utility of this technology to the pharmaceutical industry.

question is asking about the kinds of studies these techniques enable & utility for pharmaceutical companies? e.g. DNA library can enable study of: 1. Gene structure/regulation 2. Expression vectors can be used for large scale production of proteins that can be purified 3. used by pharmaceutical companies to produce biological drugs (e.g. antibody therapies & vaccines) 4. Can use these to study protein function

Enzyme that cleaves phosphodiester bonds at palindrome sequences

restriction enzyme

ribose vs. deoxyribose

ribose is the sugar used by RNA while deoxyribose is the sugar used by DNA; both 5 carbon sugars; difference is ribose has 2' OH while DNA has 2'H.

Key requirements for cloning vectors include all except: - origin of replication for propagation into host cell - suitable restriction sites for insertion of DNA fragments - sequences homologous to genes of host cells - selectable markers such as antibiotic resistance genes

sequences homologous to genes of host cells

LUCA

stands for the "last universal common ancestor"; hypothetical ancestral microorganisms existing ~3.5 + billion years ago on earth

Genetic code

the genetic code encompasses all the codon patterns available involving the A,U,G,C as are specifically relevant for RNA strands. Codons are formed by three nucleotides in a row. Therefore, every codon signals for a different action which can be formation of an amino acid, start codon to begin a chain of polypeptide, or stop codon to terminate the formation of a strand.

Restriction site

this is the specific site(usually palindrome sequence 4-6bp) on any DNA sequence (not limited to plasmid vectors) where an enzyme, protein used for cutting gene sequences, comes and cuts the site leaving an opening. The gap is a result of breaking the phosphodiester bonds and two spots. Although there can be more than 2, usually two restriction sites, in order, to remove a fragment of base pairs from the circular plasmid vector

Which of the following is not typically a component of a ribonucleotide?

thymine(we use uracil instead)

Procedure for delivery of exogenous DNA into a host cell:

transformation

Which of the following is NOT a characteristic of DNA structure as proposed by Watson & Crick:

two grooves of equal size (major and minor grooves)