Chapter 6: DNA and Biotechnology

Question 1

Nucleosides

Answer 1

Composed of five carbon sugar bonded to a nitrogenous base and are formed by covalently linking the base to C-1’, one of the sugar.

Question 2

DNA

Answer 2

The bulk of DNA is found in chromosomes in the nucleus of aerobic cells, although some is also present in the microconidia and chloroplast.

Question 3

Nucleotides.

Answer 3

Formed when one or more phosphate groups are attached to a C-5’ of a nucleoside. They are the building blocks of DNA.

Question 4

Ribose

Answer 4

The nucleic acid is RNA. The 2’ is an –OH

Question 5

Deoxyribose.

Answer 5

The nucleic acid is DNA. The 2’ is an –H.

Question 6

Sugar phosphate backbone.

Answer 6

Alternating sugar and phosphate groups. It is always read from 5’ end to 3’ end. Nucleotides are joined by 3’ to 5’ phosphodiester bonds. Phosphates carrier negative charge. Thus DNA in RNA strands have an overall negative charge. DNA strands run antiparallel to each other.

Question 7

How to differentiate 5 end and 3 end from each other?

Answer 7

The 5’ end of DNA for instance, will have an OH or phosphate group bonded to C-5 of the sugar. While the 3’end has a free -OH on the C-3 of the sugar.

Question 8

DNA versus RNA.

Answer 8

DNA is generally double stranded (dsDNA) and RNA is generally single stranded (ssRNA).

Question 9

Purines and pyrimidines.

Answer 9

Purines contain two rings in their structure. The two purines found in nucleic acids are adenine and guanine. WATER (AGua) is pure. Pyrimidines contain only one ring in their structure. The three pyrimidines are Cytosine, Thymine, and Uracil.

Question 9

Aromatic

Answer 9

It describes any unusually stable Ring system that adheres to the following four specific rules: The compound is cyclic, the compound is planar, the compound is conjugated, which means it has an alternating single and multiple bonds, creating at least one unhybridized P orbital for each atom in the ring, and the compound has a four 4n + 2 π electrons. This is called Huckel’s rule.

Question 10

Heterocycles.

Answer 10

Are ring structures that contain at least two different elements in the ring. Purines and pyrimidines contain nitrogen in their aromatic rings.

Question 10

Watson-Crick Model

Answer 10

Double helical nature of DNA and propose specific base pairing that would be the basis of copying mechanism. In the double helix, two linear polynucleotide chains of DNA are wound together in a spiral orientation along a common axis. The two strands of DNA are antiparallel, that is, the strands oriented in opposite direction. One strand has polarity 5’ to 3’ down the page. The other strand has 5’ to 3’ polarity up the page. The sugar phosphate backbone is on the outside of the Helix with the nitrogenous bases in the inside. Complementary base pairing, which means the adenine is always base paired with thymine via two hydrogen bonds. A guanine always pairs with cytosine via three hydrogen bronze.

Question 11

What is adenine (A) paired with?

Answer 11

Thyine (T)

Question 12

What is guanine (G) paired with?

Answer 12

Cytosine (C)

Question 13

B-DNA.

Answer 13

The double Helix of most DNA is a right-handed Helix forming what is called B-DNA. Returns every 3.4 nm and contains about 10 bases within that span. Major minor groups can be identified between the interlocking strands and are often the site of protein binding. The other option is Z DNA for its zigzag appearance. It is left-handed Helix that has a turn every 4.6nm and contains 12 base pairs each turn.

Question 13

Histones

Answer 13

The DNA that makes up a chromosome is wound around a group of small basic proteins called histones, forming chromatin. The histone proteins are H2A H2B H3 H4 and H1. All of the DNA’s wrapped around this protein complex, forming a nucleosome. The H1 protein, seals off the DNA and enters and leaves the nucleosome.

Question 13

Chargaff’s rule.

Answer 13

Total purines will equal the total pyrimidines overall.

Question 14

Denaturation and Reannealing

Answer 14

The double helical nature of DNA can be denature by conditions that disrupt hydrogen bonding and base pairing, resulting in the melting of the double Helix into two single strains that have separated from each other. Heat, alkaline pH, and chemicals are commonly used to denature DNA. Denatured DNA can be reannealed (Brought back together) if the denaturing condition is slowly removed.

Question 15

How many chromosomes are found in the nucleus of the cell?

Answer 15

46 chromosomes.

Question 16

Nucleoproteins.

Answer 16

Proteins are associated with DNA, histones are an example.

Question 17

Heterochromatin and Euchromatin.

Answer 17

A small percentage of the chromatin remains compacted during interphase and is referred to as heterochromatin. It appears dark, under a microscope and is transcriptionally silent. Often consists of DNA with highly repetitive sequences. In contrast, the dispersed chromatin is called Euchromatin, which appears light under light microscope. It contains genetically active DNA.

Question 18

Telomeres and centromeres.

Answer 18

Do you need replication? Cannot extend all the way to the end of a chromosome. This will result in losing sequences and information with each round of replication. The solution for our cell is to simple repeat unit at the end of the DNA forming a telomere. Some of the sequences lost in each round of replication and can be replaced by the enzyme telomerase. The telomeres also serve a second function, their high GC content creates exceptionally strong strained action. Centromeres are found in the center of the chromosomes. They are composed of heterochromatin.

Question 19

Strand separation and origins of replication.

Answer 19

The replisome or replication complexes are a set of specialized proteins that assist the DNA polymerases. To begin the process of replication, DNA unwinds at point call Origins of replication. Creating replication forks on both sides of the origin. In eukaryotes cells the there are multiple origins of replication. As the replication forks move towards each other and the sister chromated their created the chromatids. Don’t remain connected at the centromere.

Question 20

In bacterial chromosomes DNA replication.

Answer 20

It is a closed, double stranded circular DNA molecule with a single origin of replication. Eventually, the two replication forks will meet, resulting in the production of two identical circular molecules of DNA.

Question 21

Helicase

Answer 21

Is the enzyme responsible for unwinding the DNA generating two single stranded template strands ahead of the polymerase.

Question 22

Single stranded DNA binding protein.

Answer 22

Will bind to the unraveled strain, preventing both the reassociation of the DNA strand and the degradation of DNA by nucleases.

Question 23

Supercoiling

Answer 23

A wrapping of DNA on itself as its helical structures pushed even farther towards the telomeres during replication. This happens because helicase unwinds the DNA and it causes positive supercoiling that strains the DNA Helix.

Question 24

Parental versus daughter strands.

Answer 24

During replication, these parental strains will serve as templates for the generation of new daughter strains. There were replication processes termed semiconservative because one parental strain is retained in each of the two resulting identical double stranded DNA molecules.

Question 24

DNA topoisomerases.

Answer 24

It is an enzyme that introduced negative supercoil, which alleviates the torsional stress and reduces the risk of strand breakage.

Question 25

DNA polymerases.

Answer 25

Responsible for reading the DNA template or parental strand and synthesizing the new daughter strand. This will result in a new double Helix of DNA that has the required antiparallel orientation. DNA polymerases can read the template strain in a 3’ to 5’ direction while synthesizing the complementary strand in a 5’ to 3’ direction.

Question 26

Lagging strand.

Answer 26

Strain that is copied in a direction opposite the direction of the replicating fork. On this side of the replication fork, the parental strand has a 5 to 3 polarity. Because DNA polymerase can only synthesize in the 5’ to 3’ direction from 3’ to 5’ template, small strains called all Okazaki fragments are produced. As the replication fork continues to move forward, it clears the additional space that DNA polymerase must fill in.

Question 26

Leading strand

Answer 26

In each replication fork is the strain that is copied in a continuous fashion in the same direction as the advancing replication fork. This parental strand will be read 3’ to 5’ and its complement will be synthesized in a 5 to 3 manner.

Question 27

Primase

Answer 27

Synthesizes a short primer, roughly about 10 nucleotides in the five to three direction to start replication on each strand. The first step in the replication of DNA is actually to lay down the RNA primer.

Question 28

DNA polymerase III

Answer 28

Found in prokaryotes. It begins synthesizing the daughter strands of DNA in the 5’ to 3’ manner.

Question 29

DNA polymerase α, δ and ε.

Answer 29

Found in Eukaryotes. It begins synthesizing the daughter strands of DNA in the 5’ to 3’ manner.

Question 29

DNA polymerase I (prokaryotes) and DNA polymerase δ

Answer 29

Add DNA nucleotides where the RNA primer had been.

Question 29

DNA polymerase I (prokaryotes) and RNase H (Eukaryotes)

Answer 29

Our name was eventually be removed to maintaining tegrity of the genomes and these are the enzymes that do that.

Question 29

Pay special attention to the Eukaryotes DNA polymerases:

Answer 29

DNA polymerases: α, β, γ, δ, and ε. DNA polymerases α, δ, and ε and work together to synthesize both the leading and lagging strands; DNA polymerase δ also fills in gaps left behind when RNA primers are removed. DNA polymerase γ replicates mitochondrial DNA. DNA polymerase β and ε are also important to the process of DNA repair. DNA polymerase δ, and ε assisted by the PCNA protein, which assembles into a trimmer to form the sliding clamp. They claim that helps to strengthen the interaction between the DNA polymerase and the template strand.

Question 29

DNA ligase.

Answer 29

Seals the ends of the DNA molecules together, creating one continuous strand of DNA.

Question 29

Oncogenes and tumor suppressor genes.

Answer 29

Certain genes, When mutated, can lead to cancer. Cancer cells proliferate excessively because they are. Able to divide without stimulation from other cells and are no longer subject to normal control on cell proliferation. Cancer cells are able to migrate by local invasion or metastasis. Mutated genes that cause cancer are termed oncogenes. Before these genes are mutated, they can often refer to as proto-oncogenes. Tumor suppressor genes encode proteins that inhibit the cell cycle or participate in DNA repair processes. They normally function to stop tumor progression and are sometimes called antioncogenes.

Question 29

Proofreading.

Answer 29

DNA polymerase enzyme Does proofreading. When the complementary strains have incorrectly paired bases, the hydrogen bonds between the strands can be unstable, and this lack of stability is detected as the DNA passes through the sport of the polymerase. The way it differentiates between the daughter and parent strand it is that it looks at the level of methylation. The template strand has existed in the cell for a longer period of time and therefore is more heavily methylated. The DNA ligase, which closes the gaps between Okazaki fragments, lacks proofreading ability.

Question 30

Nucleotide excision repair.

Answer 30

Thymine dimers are eliminated from DNA by a nucleotide excision repair mechanism, which is a cut and patch process. An excision endonuclease then makes nicks in the phosphodiester backbone of the damage strand on both sides of the thymine dimer and removes the defective oligonucleotide.

Question 30

Nucleotide and base excision repair.

Answer 30

cell machinery recognizes 2 specific types of DNA damage in the G1 and G2 cell cycle phases and fixes them through nucleotide. excision repair or basic excision repair.

Question 30

DNA libraries.

Answer 30

Large collections of known DNA sequences.

Question 30

DNA cloning.

Answer 30

Technique that can produce large amount of a desired sequence. The goal is to produce a large quantity of homogeneous DNA for other applications. Cloning requires that the investigator ligate the DNA of interest into a piece of nucleic acid referred to as a vector, forming a recombinant vector. Vectors are usually bacteria or viral. The bacteria are then growing colonies and a colony containing the recombinant vectors isolated. This can be accomplished by ensuring that the recombinant vector also includes a gene for antibiotic resistance. The resulting colony can then be grown in large quantities. The bacteria can then be made to express the gene of interest or be lysed to reisolate the replicated recombinant vector.

Question 30

Restriction enzymes.

Answer 30

enzymes that recognize specific double stranded DNA sequences. These sequences are palindromic, meaning that the five to three sequence of one strand is identical to five to three sequence on the other strand. Once a specific sequence has been identified, the restriction enzyme can cut through the backbones of the double Helix.

Question 30

Mismatch repair.

Answer 30

Cells also have machinery in G2 phase of the cell cycle for mismatch repair. These enzymes are encoded by genes MSH2 and MLH1. Which detect and remove errors introducing replication that were missed during the S phases of the cell cycle.

Question 30

base excision repair

Answer 30

These are repaired by base excision repair. The affected base is recognized and removed by a glycosylase enzyme. Leaving behind an abasic site (AP). The AP site is recognized by an AP endonuclease that removes the damaged sequence from the DNA. Do you need polymerase in DNA. Ligase can then fill in the gap and seal the strand as described above.

Question 31

Recombinant DNA.

Answer 31

Allows DNA fragment from any source to be multiplied by either gene cloning or polymerase chain reaction. It can provides reagents necessary for genetic testing, such as carrier detection and prenatal diagnosis of genetic diseases.

Question 32

Genomic libraries.

Answer 32

Contain large fragments of DNA including both coding and non coding regions of genome. They cannot be used to make recombinant proteins or for gene therapy.

Question 33

Coding Region.

Answer 33

Exon

Question 34

Non coding region.

Answer 34

Intron.

Question 35

What is another name for cDNA?

Answer 35

Complementary DNA.

Question 36

cDNA libraries (Expression libraries)

Answer 36

contain smaller fragments of DNA and only include the exons of gene expressed by the sample tissue. They can be used to make recombinant proteins or for gene therapy.

Question 36

Polymerase chain reaction (PCR)

Answer 36

automated process by which millions of copies of DNA sequence can be created from a very small sample by hybridization. During PCR, the DNA of interest is teenager replicated and then cooled, repeated several times. It double s the amount of DNA with each cycle until enough copies of DNA sequence are available. ABC or reaction requires primers there are complementary to the DNA that flanks the region of interest.

Question 37

Hybridization

Answer 37

Joining of complementary base pair sequences.

Question 38

Southern blotting

Answer 38

Can be used to detect the presence and quantify your various DNA strands in the sample. After electrophoresis, this sample transferred to woman’s brain that can be probed with single stranded DNA molecules to look for sequence of interest.

Question 38

Gel electrophoresis

Answer 38

Electrophoresis is a technique used to separate macromolecules such as DNA and proteins by size and charge. All molecules of DNA are negatively charged because of the phosphate group in the backbone of the molecule. So all the DNA strands will migrate towards the anode of the electrochemical cell. The preferred gel is at agarose gel.

Question 39

Gene therapy.

Answer 39

Intended for diseases in which a given gene is mutated or inactive, giving rise to pathology. By transferring a normal copy of the gene into the affected tissue, the pathology should be fixed, essentially curing the individual.

Question 39

Transgenic and knockout mice.

Answer 39

Transgenic mice are altered at their germ line by introducing a clone gene into fertilized ova or into embryonic stem cells. The cloned gene that is introduced is referred to as the transgene. If the transgene is a disease producing allele, Then the transgenic mice can be used to study the disease process from early embryonic development. Knockout mice is which a gene has been intentionally deleted. A cloned gene may be microinjected into the nucleus of a newly fertilized ovum.

Question 39

DNA sequencing

Answer 39

Uses dideoxyribonucleotides. Which terminates the DNA chain because they lack a 3’ -OH group. the resulting fragments can be separated by gel electrophoresis and the sequence can be read from the gel.

Question 40

When creating a DNA library, what are some of the advantages of genomic libraries? What about cDNA libraries?

Answer 40

Genomic libraries include all of the DNA now organisms genome including non coding regions. This may be useful for studying DNA in introns, centromeres, or telomeres. cDNA Libraries only include express genes from a given tissue, but can be used to express recombinant proteins or to perform gene therapy.

Question 40

What are chimeras?

Answer 40

Organisms that contain cells from two different lineages.

Question 41

During DNA sequencing, why does the DNA polymer stop growing once a dye deoxyribonucleotide is added?

Answer 41

deoxyribonucleotide lack the 3’end -OH group that is required for DNA strand elongation. Therefore, when it is added to a growing DNA molecule, no more nucleotides can be added.

Question 41

What does PCR accomplish for the researcher? What about South Southern blotting?

Answer 41

PCR including the number of copies of a given DNA sequence and can be used for a sample containing very few copies of the DNA sequence. Southern blotting is useful when you’re searching for a particular DNA sequence because it separates DNA fragments by length and then probes for a sequence of interest.

Question 42

What is the difference between a transgenic and knockout mouse?

Answer 42

Transgenic mice have a gene introduced into their germline or embryonic stem cell to look at the effects of that gene. They are therefore best suited for studying the effects of dominant illus. Knockout minds are those in which a gene of interest has been removed rather than added.

Question 43

How does DNA polymerase recognize which strand in the template strand once the daughter strand is synthesized?

Answer 43

The parents trained as more heavily methylated, whereas the daughter strand is barely methylated at all. This allowed the DNA polymerase to distinguish between them.

Question 44

What are the purine bases?

Answer 44

Adenosine and Guanine.

Question 45

What are the pyrimidine bases?

Answer 45

Cytosine, thymine and uracil.