MOLECULAR DIAGNOSTICS EXAM
AND ANSWERS
semiconservative - ANSWER-Semi-conservative replication is the key to maintaining
the sequence of the nucleotides in DNA through new generations.
Every cell in a multicellular organism or in a clonal population of unicellular
organisms carries the same genetic information. It is important that this information,
in the form of the DNA sequence, be transferred faithfully at cell division. The
replication apparatus is designed to copy the DNA strands in an orderly way with
minimal errors before each cell division.
DNA template in DNA replication - ANSWER-The order of nucleotides is maintained
during DNA replication because each strand of the parent double helix is the
template for a newly replicated strand.
In the process of replication, DNA is first unwound from the helical duplex so that
each single strand may serve as a template for the addition of nucleotides to the new
strand.
Fig 1.9 - ANSWER-Simultaneous replication of both strands of the double helix. Both
strands are read in the 3' to 5' direction. The lagging strand is read discontinuously,
with the polymerase skipping ahead and reading back toward the replication fork on
the lagging strand.
lagging strand and Okasaki fragments - ANSWER-Okazaki fragments, were the key
to explaining how both strands were copied at the replication fork.
The replication apparatus jumps ahead a short distance (~1000 bases) on the 5' to 3'
strand and then copies backward toward the replication fork. The 5' to 3' strand
copied in a discontinuous manner is the lagging strand
DNA replication - where does it begin and end? Origin of replication and end of the
molecule - ANSWER-DNA replication begins at origins or replication
Replication fork with RNA primase adding a RNA primer so DNA polymerase can set
out to work.
Fig 1.10 enzymes - ANSWER-DNA polymerase activity involves more than one
protein molecule. Several cofactors and accessory proteins are required to unwind
the template helix (purple), prime synthesis with RNA primers (gray), and protect the
lagging strand (dark gray).
Helicase, Primase, RNA primer
endonucleases, and types 1-4 - ANSWER-Genetic engineering was stimulated by
the discovery of deoxyriboendonucleases, or endonucleases. Endonucleases break
the sugar-phosphate backbone of DNA.
, Restriction enzymes are endonucleases that recognize specific base sequences and
break or restrict the DNA polymer at the sugar-phosphate backbone. Restriction
endonucleases have been classified into four types.
Type I = have both nuclease and methylase activity in a single enzyme. They bind to
host-specific DNA sites of 4-6 bp separated by 6-8 bp and containing methylated
adenines. The site of cleavage of the DNA substrate can be over 1000 bp from this
binding site.
Ex: EcoK from E. coli K 12
Type III = resemble type I enzymes in their ability to both methylate and restrict (cut)
DNA. Like type I, they are complex enzymes with two subunits. Recognition sites for
these enzymes are asymmetrical, and the cleavage of the substrate DNA occurs 24-
26 bp from the site to the 3' side.
Ex: PstIIII from P. stuartii
Type IV = have similar subunit structures and enzyme requirements. Type IV
enzymes have cutting and methyltransferase functions.
Ex: BseMII from Bacillus stearothermophilus
Type II = are those used most frequently in the laboratory. These enzymes do not
have inherent methylation activity. They bind as simple dimers to symmetrical DNA
recognition sites. These sites are palindromic in nature; that is, they read the same 5'
to 3' on both strands of the DNA (Fig. 1-13), referred to as bilateral symmetry. Type II
restriction enzymes cleave the DNA directly at the binding site, producing fragments
of predictable size.
sexual recombination and crossing over - ANSWER-Recombination is the mixture
and assembly of new genetic combinations. Recombination occurs through the
molecular process of crossing over, or physical exchange between molecules.
Sexually reproducing organisms mix genes in three ways. First, at the beginning of
meiosis, duplicated chromosomes line up and recombine by crossing over, or
breakage and reunion of the four DNA duplexes. This generates newly recombined
duplexes with genes from each duplicate.
Then, the recombined duplexes are randomly assorted into gametes so that each
gamete contains one set of each of the recombined parental chromosomes.
Finally, the gamete will merge with a gamete from the other parent carrying its own
set of recombined chromosomes. The resulting offspring will contain a new set or
recombination of genes of both parents.
transduction - ANSWER-In the early 1960s, Francois Jacob and Elie Wollman
studied the transmission of units of heredity carried by viruses from one bacterium to
another (transduction)
RNA base pairs - ANSWER-RNA differs from DNA in the sugar moieties, having
ribose instead of deoxyribose and, in one nitrogen base component, having uracil
instead of thymine
AND ANSWERS
semiconservative - ANSWER-Semi-conservative replication is the key to maintaining
the sequence of the nucleotides in DNA through new generations.
Every cell in a multicellular organism or in a clonal population of unicellular
organisms carries the same genetic information. It is important that this information,
in the form of the DNA sequence, be transferred faithfully at cell division. The
replication apparatus is designed to copy the DNA strands in an orderly way with
minimal errors before each cell division.
DNA template in DNA replication - ANSWER-The order of nucleotides is maintained
during DNA replication because each strand of the parent double helix is the
template for a newly replicated strand.
In the process of replication, DNA is first unwound from the helical duplex so that
each single strand may serve as a template for the addition of nucleotides to the new
strand.
Fig 1.9 - ANSWER-Simultaneous replication of both strands of the double helix. Both
strands are read in the 3' to 5' direction. The lagging strand is read discontinuously,
with the polymerase skipping ahead and reading back toward the replication fork on
the lagging strand.
lagging strand and Okasaki fragments - ANSWER-Okazaki fragments, were the key
to explaining how both strands were copied at the replication fork.
The replication apparatus jumps ahead a short distance (~1000 bases) on the 5' to 3'
strand and then copies backward toward the replication fork. The 5' to 3' strand
copied in a discontinuous manner is the lagging strand
DNA replication - where does it begin and end? Origin of replication and end of the
molecule - ANSWER-DNA replication begins at origins or replication
Replication fork with RNA primase adding a RNA primer so DNA polymerase can set
out to work.
Fig 1.10 enzymes - ANSWER-DNA polymerase activity involves more than one
protein molecule. Several cofactors and accessory proteins are required to unwind
the template helix (purple), prime synthesis with RNA primers (gray), and protect the
lagging strand (dark gray).
Helicase, Primase, RNA primer
endonucleases, and types 1-4 - ANSWER-Genetic engineering was stimulated by
the discovery of deoxyriboendonucleases, or endonucleases. Endonucleases break
the sugar-phosphate backbone of DNA.
, Restriction enzymes are endonucleases that recognize specific base sequences and
break or restrict the DNA polymer at the sugar-phosphate backbone. Restriction
endonucleases have been classified into four types.
Type I = have both nuclease and methylase activity in a single enzyme. They bind to
host-specific DNA sites of 4-6 bp separated by 6-8 bp and containing methylated
adenines. The site of cleavage of the DNA substrate can be over 1000 bp from this
binding site.
Ex: EcoK from E. coli K 12
Type III = resemble type I enzymes in their ability to both methylate and restrict (cut)
DNA. Like type I, they are complex enzymes with two subunits. Recognition sites for
these enzymes are asymmetrical, and the cleavage of the substrate DNA occurs 24-
26 bp from the site to the 3' side.
Ex: PstIIII from P. stuartii
Type IV = have similar subunit structures and enzyme requirements. Type IV
enzymes have cutting and methyltransferase functions.
Ex: BseMII from Bacillus stearothermophilus
Type II = are those used most frequently in the laboratory. These enzymes do not
have inherent methylation activity. They bind as simple dimers to symmetrical DNA
recognition sites. These sites are palindromic in nature; that is, they read the same 5'
to 3' on both strands of the DNA (Fig. 1-13), referred to as bilateral symmetry. Type II
restriction enzymes cleave the DNA directly at the binding site, producing fragments
of predictable size.
sexual recombination and crossing over - ANSWER-Recombination is the mixture
and assembly of new genetic combinations. Recombination occurs through the
molecular process of crossing over, or physical exchange between molecules.
Sexually reproducing organisms mix genes in three ways. First, at the beginning of
meiosis, duplicated chromosomes line up and recombine by crossing over, or
breakage and reunion of the four DNA duplexes. This generates newly recombined
duplexes with genes from each duplicate.
Then, the recombined duplexes are randomly assorted into gametes so that each
gamete contains one set of each of the recombined parental chromosomes.
Finally, the gamete will merge with a gamete from the other parent carrying its own
set of recombined chromosomes. The resulting offspring will contain a new set or
recombination of genes of both parents.
transduction - ANSWER-In the early 1960s, Francois Jacob and Elie Wollman
studied the transmission of units of heredity carried by viruses from one bacterium to
another (transduction)
RNA base pairs - ANSWER-RNA differs from DNA in the sugar moieties, having
ribose instead of deoxyribose and, in one nitrogen base component, having uracil
instead of thymine
