Exam 2 biochem CH.5

PCR Steps for one cycle

1 strand separation by heating to 95°C 2 hybridization of the primers by cooling to 54°C 3 synthesis of new DNA

Suppose that you have isolated an enzyme that digest paper pulp and have obtained its cDNA. The goal is to produce a mutant that is effective at high temperature. You have engineered a pair of unique restriction sites in the cDNA that flank a 30 dp coding region. Propose a rapid technique for generating many different mutations in this region.

A simple strategy for generating many mutants is to synthesize a degenerate set of cassettes by using a mixture of activated nucleosides in particular rounds of oligonucleotide synthesis. Suppose that the 30-bp coding region begins with GTT, which encodes valine. IF a mixture of all four nucleotides is used in the first and second rounds of synthesis, the resulting oligonucleotides will begin with the sequence XYT (where X and Y denote, A,C,G, or T). These 16 different versions of the cassette will encode portions containing either, Phe, Leu, Ole, Val, Sir, Pro, Thr, Ala, Try, His, Ask, asp, Cyc, Arg or Fly at the first position. Likewise, degenerate cassettes can be made in which two or more codons are simultaneously varied.

The restriction of enzymes Kpnl and Acc65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from Kpnl cleavage cannot be ligated directly to the sticky end formed from Acc65I cleavage. Explain why.

Although the two enzymes cleave the same recognition site, they each break different bonds within the 6-bp sequence. Cleavage by KpnI yields an overhand on the 3' strand, wheres cleavage by Acc65I produces an overhand on the 5' strand. These sticky ends do not overlap.

The stringency of PCR amplification can be controlled by altering the temperature at which the primers and the target DNA undergo hybridization. How would altering the temperature of hybridization affect the amplification? Suppose that you have a particular yeast hens A and that you wish to see if it has a counterpart in humans. How would controlling the stringency of the hybridization help you?

At high temperatures of hybridization, only very close matches between primer and target would be stable because all of the bases would need to find partners to stabilize the primer-target helix. As the temperature is lowered, more mismatches would be tolerated; so the amplification is likely to yield genes with less sequence similarity. In regard to the yeast gene, synthesize primers corresponding to the ends of the gene, and then use these primers and human DNA as the target. If nothing is amplified at 54°C, the human gene differs from the yeast gene, but a counterpart may still be present. Repeat the experiment at a lower temperature of hybridization

The power of PCR can also create problems. Suppose someone claims to have isolated dinosaur DNA by using PCR. What questions might you ask to determine if it is indeed dinosaur DNA?

Because PCR can amplify as little as one molecule of DNA, statements claiming the isolation of ancient DNA need to be greeted with some skepticism. The DNA would need to be sequenced. It is similar to human, bacterial, or fungal DNA? If so, contamination is the likely source of the amplified DNA. It is similar to that of birds or crocodiles? this sequence similarity would strengthen the case that it is dinosaur DNA because these species are evolutionarily close to dinosaurs.

Site Directed Mutagenesis (oligonucleotide directed)

Change one amino acid. Mutation of one to three bases

Plasmids

Circular duplex DNA occurring naturally in some bacteria. modified to enhance the delivery of recombinant DNA into bacteria and to facilitate the selection of bacteria containing them. LacZ gene Bacteria lack the enzymes for posttraslation modifications such as peptide cleavage and glycosylation

Southern Blotting

DNA fragment containing a specific sequence can be identified by separating a mixture of fragments by electrophoresis, transferring them to nitrocellulose and hybridizing with P-labled probe

Deletions

Deleting amino acid from the amino acid sequence of a protein

PCR is typically used to amplify DNA that lies between two known sequences. Suppose that you want to explore DNA on both sides of a single known sequence. Devise a variation of the usual PCR protocol that would enable you to amplify entirely new genomic terrain.

Digest genomic DNA with a restriction enzyme, and select fragment that contains the known sequence. Circularize this fragment. Then carry our PCR with the use of a pair of primers that serve as templates for the synthesis of DNA away from the known sequence.

Insertions(Cassette Mutagenesis)

Inserting new amino acids to an amino acid sequence of a protein

Suppose that a human genome library is prepared by exhaustive digestion of human DNA with the EcoRi restriction enzyme. Fragments averaging about 4kb in length would be generated. Is this procedure suitable for cloning large genes? Why or why not?

No, because most human genes are much longer than 4kb. A fragment would contain only a small part of a complete gene.

You have identified a gene that is located on human chromosome 20 and wish to identify its location within the mouse genome. On which chromosome would you be most likely to find the mouse counterpart of this gene?

On the basis of the comparative genome map, the region of greatest overlap with human chromosome 20 can be found on mouse chromosome 2

Ovalbumin is the major protein of egg white. The chicken ovalbumin gene contains eight exons separated by seven introns. Should ovalbumin cDNA or ovalbumin genomic DNA be used to form the protein in E. coli why?

Ovalbumin cDNA should be used. E. coli lacks the machinery to splice the primary transcript arising from genomic DNA

DNA sequences that are highly enriched in G-C base pairs typically have high melting temperatures. Moreover, once separated, single strands containing these regions can form rigid secondary structures. How might the presence of G-C-rich regions in a DNA template affect PCR amplification?

PCR amplification is greatly hindered by the presence of G-C-rich regions within the template. Owing to their high melting temperatures, these templates do not denature easily, thus preventing the ignition of an amplification cycle. In addition, rigid secondary structures prevent the progress of DNA polymerase along the template strand during elongation.

Mutants of S. pyogenes Cas9 have been developed which recognize different PAM sequences, such as 5'-NGAN-3'. Why might these mutant be useful for genome editing studies?

Since PAM sequences are required downstream of a particular target site, variants of Cas9 that recognize different PAMs would increase the likelihood that a sgRNA could be designed against a particular target sequence.

Targeting a DNA sequence for editing by the CRISPER-Cas9 System requires that a PAM sequence be present just downstream of the target sequence. For S. progenies Cas9 this PAM is 5'-NGG-3', where N represents any nucleotide. Assuming a fragment of DNA includes equal amounts of A, C,G, and T bases, how far apart would one expect to find these PAMs?

Since the N could be one one of the four bases, the presence of the NGG trinuclotide would occur with a frequency of 4/4*1/4*1/4, or once every 16 bases, on average

Sickle[cell anemia arises from a mutation in the gene for beta chain of human hemoglobin. The change from GAG to GTG in mutant eliminates a cleavage site for the restriction enzyme Mstll, which recognized the target sequence CCTGAGG. These finding form the basis of a diagnostic test for the sickle-cell gene. Propose a rapid procedure for distinguishing between the normal and the mutant gene. Would a positive result prove that the mutant contains GTG in place of GAG?

Southern blotting of an MstII digest would be distinguish between the normal and the mutant genes. The loss of a restriction site would lead to the replacement of two fragments on the Southern blot by a single longer fragment. Such a finding would not prove the GTG replaced GAG; other sequence changes at the restriction site could yield the same result.

Why is Taq polymerase especially useful for PCR

Taw polymerase is the DNA polymerase from the thermophilic bacterium that lives in hot springs. Consequently, it is heat stable and can withstand the high temperatures required for PCR without denaturing. Thus, it isn't necessary to add it to the reaction before each new cycle.

Which of the following amino acid sequences would yield the most optimal oligonucleotide probe. Ala-MEt-Ser-Leu-Pro-Trp Fly-Trp-Asp-Met-His-Lys Cys-Cal-Trp-Asn-Lys-Ile Arg-ser-Met-Leu-Gln-Asn

The codons for each amino acid can be used to determine the number of possible nucleotide sequences that encode each peptide sequence Ala-MEt-Ser-Leu-Pro-Trp 4x1x6x6x4x1=576 Fly-Trp-Asp-Met-His-Lys 4x1x2x1x2x2=32 Cys-Cal-Trp-Asn-Lys-Ile 2x4x1x2x2x3=96 Arg-ser-Met-Leu-Gln-Asn 6x6x1x6x2x2=864 The set of DNA sequences encoding the peptide Gly-Trp-Asp-Met-His-Lys would be most ideal for probe design because it encompasses only 32 oligonucleotides

A gel pattern displaying PCR products show four strong bands. The four pieces of DNA have lengths that are approximately in the rations of 1:2:3:4. The largest band is cut out of the gen, and PCR is repeated with the same primers. Again, a ladder of four bands is evident in the gel. What does this result reveal about the structure of encoded protein?

The encoded protein contains four repeats of a specific sequence

The restriction enzyme Alul cleaves at the sequence 5'-AGCT-3' and Note cleaves at 5'-GCGGCCGC-3'. What would be the average distance between cleavage sites for each enzyme on digestion of double-stranded DNA? Assume that the DNA contains equal proportions of A, G, C and T

The presence of the AluI sequence would, on average, be (1/4)^4, or 1/256 because the likelihood of any base being at any position is 1/4 and there are 4 positions. By the same reasoning, the presence of the NotI sequence would be (1/4)^8, or 1/65,536/ Thus, the average product of digestion by AluI would be 250 base pairs in length, whereas that by NotI would be 66,000 base pairs in length

Propose a method for isolating a DNA fragment that is adjacent in the genome to a previously isolated DNA fragment. Assume that you have access to a complete library of DNA fragments in a BAC vector but that the sequence of the genome under study has not yet been determined.

Use chemcical synthesis or the polymerase chain reaction to prepare hybridization probes that are complementary to both ends of known DNA fragment. Challenge clones representing the library of DNA fragments with both of the hybridization probes. Select clones that hybridize to one of the probes but not the other; such clones are likely to represent DNA fragment that contain one ens of the known fragment along with the adjacent region of particular chromosome.

Why might the genomic analysis of dogs be particularly useful for investigating the genes responsible for body size and other physical characteristics?

Within a single species, individual dogs show enormous variation in body size and substantial diversity in other physical characteristics. Therefore, genomic analysis of individual dogs would provide valuable clues concerning the genes responsible for the diversity within the species

DNA sequencing

base pair to get the DNA sequenced (template) --->Change T to U in synthesized sequence to get the mRNA that can be synthesized from template --->Convert RNA code to amino acids from NH3- COO- terminal (5'-3')

cDNA

inserted into a plasmid contains only exons

Mutagenesis

making mutations in proteins.

Restriction enzymes

recognize specific base sequences in double -delical DNA and cleave both strands of duplex at a specific site. twofold symmetry

Northern Blotting

same as southern but for RNA

Polymerase Chain Reaction (PCR)

used to amplify specific DNA sequences or rapid making copies of a strand of DNA Components: 1 target dna 2 a pair of primers that hybridize with the flanking sequence of template 3 dNTPs 4 Heat stable DNA polymerase