Download Unit 4

Macias, Rachel
11/28/02
Per. 3
Chapter 16: The Molecular Basis of Inheritance
Objectives
After reading this chapter and attending lecture, a student should be able to:
1. List the three components of nucleotide.
A nitrogenous base, a sugar (deoxyribose), and a phosphate group.
2. Distinguish between deoxyribose and ribose.
Deoxyribose has one less hydroxyl group than ribose and is the sugar of DNA. Ribose is
the sugar of RNA.
3. List the nitrogen bases found in DNA, and distinguish between pyrimidine and purine.
Adenine and guanine are purines because they have two organic rings. Thymine and
cytosine are pyrimidines with only a single ring.
4. Explain the "base-pairing rule" and describe its significance.
Called Chargaff’s rule, adenine pairs with thymine and guanine pairs with cytosine. This
fact makes it a possibility for copying to be accomplished.
5. Describe the structure of DNA, and explain what kind of chemical bond connects the nucleotides of each
strand and what type of bond holds the two strands together.
It is a double helix, like a ladder with the two sides twisting around each other. Hydrogen
bonds hold the two strands together.
6. Explain, in their own words, semiconservative replication.
The two parent cells will separate and serve as templates to create the daughter cells.
7. Describe the process of DNA replication, and explain the role of helicase, single strand binding protein,
DNA polymerase, ligase, and primase.
Part of DNA is untwisted by helicase (enzyme). Single strand binding proteins attach in
chains along the strands holding them straight. DNA polymerases line up new bases to the end of the new
DNA strand. Ligase connects the Okazaki fragments of the second DNA synthesis. In order to begin DNA
synthesis, the end must be “primed” with RNA and primase.
8. Define antiparallel.
The sugar-phosphate strands making up DNA run in opposite directions.
9. Distinguish between the leading strand and the lagging strand.
Leading – first strand to be synthesized. Lagging – synthesized away from replication
fork.
10. Explain how the lagging strand is synthesized when DNA polymerase can add nucleotides only to the 3’
end.
As the replication bubble widens, portions of DNA are able to be synthesized. DNA ligase
joins them together.
11. Explain the role of DNA polymerase, ligase, and repair enzymes in DNA.
Polymerase proof-reads each nucleotide against its template as soon as it’s attached to the
new strand. If there is a problem, the process is backed up to remove and correct it, then it continues.
Repair enzymes – repair accidental damage to DNA. Ligase – attaches fragments of synthesized DNA.
Chapter 17: From Gene to Protein
Objectives
After reading this chapter and attending lecture, a student should be able to:
1. Explain how RNA differs from DNA.
Each has a different sugar and instead of thymine, RNA contains a base called uracil.
2. In your own words, briefly explain how information flows from gene to protein.
DNA synthesizes mRNA which carries the genetic information to tRNA to be translated
into an amino acid chain.
3. Distinguish between transcription and translation.
Transcription – the synthesis of RNA with the aid of DNA. Information copied from one
molecule to the other. Translation – the synthesis of a polypeptide (protein) chain from mRNA.
4. Describe where transcription and translation occur in prokaryotes and in eukaryotes.
In eukaryotes: Transcription – occurs in the nucleus. Translation – occurs on ribosomes
in the cytoplasm. In prokaryotes transcription and translation are coupled and mRNA ribosomes attach to
while transcription is still occurring.
5. Define codon, and explain what relationship exists between the linear sequence of codons on mRNA and
the linear sequence of amino acids in a polypeptide.
Codon – the mRNA triplet code that “codes” for amino acids. The sequence of codons is
what specifies a specific amino acid for a polypeptide chain. The amino acids follow the pattern set out for
their formation by the mRNA codons.
6. Explain the process of transcription including the three major steps of initiation, elongation, and
termination.
RNA polymerases pry strands of DNA apart and hook together RNA nucleotides to be
synthesized. Initiation – RNA polymerases bind to DNA promoters which include the initiation site.
Transcription factors aid polymerases in finding the DNA sites. Elongation – polymerase moves along the
DNA, separating it and exposing about 10 bases at a time for pairing with RNA nucleotides. It adds these to
the end of a growing RNA chain of molecules. In the end it peels away from the DNA template.
Termination – transcription will continue until the polymerase reaches a code for termination.
7. Describe the general role of RNA polymerase in transcription.
It uncoils DNA so RNA nucleotides can be synthesized and form a chain.
8. Distinguish among mRNA, tRNA, and rRNA.
mRNA – the chain synthesized by DNA, contains codons. tRNA – can be used
repeatedly to pick up the specified amino acid. rRNA – one of the proteins making up the subunit of
ribosomes which provide the site for translation.
9. Describe the structure of tRNA and explain how the structure is related to function.
tRNA can be used repeatedly to pick up the specified amino acid and deposit it at the
ribosome. All have four base-paired regions and three loops. At one end is the amino acid attachment site
and within the middle loop is the anticodon triplet unique to each type of tRNA.
10. Given a sequence of bases in DNA, predict the corresponding codons transcribed on mRNA and the
corresponding anticodons of tRNA.
DNA: CCA – mRNA: GGU – tRNA: CCI (codes for glycine).
11. Describe the structure of a ribosome, and explain how this structure relates to function.
Ribosomes provide the sires for translation or protein synthesis. They are made up of
large and small subunits (2) constructed in the nucleus. They join to form a ribosome when attached to an
mRNA molecule.
12. Describe the process of translation including initiation, elongation, and termination.
Initiation – ribosomal subunits bind to mRNA and tRNA. They attach to the sequence
on mRNA and the initiator tRNA (methionine) binds to the initiation codon. Proteins called initiation
factors then bring together the two ribosomal subunits. Elongation – amino acids form a chain based on: 1)
codon recognition, 2) peptide bond formation, and 3) translocation. Termination – a termination codon
reaches the A site of a ribosome. A release factor binds to the site and a water molecule is added to the
chain, hydrolyzing the polypeptide.
13. Describe the difference between prokaryotic and eukaryotic mRNA.
Eukaryotic – synthesis of one polypeptide (one transcription unit). They contain many
types of polymerase. Prokaryotic – one transcription unit contains a few genes that code for proteins of
related function. They have only one type of polymerase.
14. Explain how eukaryotic mRNA is processed before it leaves the nucleus.
During RNA processing, the ends of the RNA molecule are altered. The 5’ end is
“capped” off with a modified version of the guanine nucleotide. The 3’ end gets a poly-A tail which consists
of 30-200 adenine nucleotides. The 5’ cap helps to protect the mRNA from hydrolytic enzymes and once it
reaches the cytoplasm, provides the location for attachment of small ribosomal subunits. The 3’ end may
facilitate the export of mRNA into the cytoplasm and also helps to prevent the breakdown of RNA.
15. Explain why base-pair insertions or deletions usually have a greater effect than base-pair substitutions.
Insertions and deletions alter the reading of the genetic message. The resulting protein
will almost certainly be non-functional. Some base-pair substitutions don’t alter the effect of the protein.
Some proteins then will make sense, but not the “correct” sense, especially if the coding is ended early.
Chapter 18: Microbial Models: The Genetics of Viruses and Bacteria
Objectives
After reading this chapter and attending lecture, a student should be able to:
1. List and describe structural components of viruses.
A virus is essentially a linear or circular molecule of nucleic acid (either DNA or RNA),
enclosed in a protein coat. This protein coat is called a capsid and is made up of smaller subunits called
capsomeres. These capsids can be rod-shaped, polyhedral, or more complex. Some have viral envelopes
cloaking those capsids. These are derived from the membrane of the host cell as well as the membrane of
the virus.
2. Explain why viruses are obligate parasites.
Because they can only reproduce within a host cell.
3. Explain the role of reverse transcriptase in retroviruses.
It can transcribe DNA from and RNA template. The newly-formed viral DNA then flows
into the cell’s nuclei. It replicates and spreads the virus.
4. Describe how viruses recognize host cells.
They identify by a fit/ recognization between proteins on the outside of a virus and
specific receptor molecules on the cell’s surface.
5. Distinguish between lytic and lysogenic reproductive cycles using phage T4 and phage l as examples.
Lytic – Tail fibers of a T4 virus stick to receptor sites of an E. coli bacterial cell. The tail
contracts and thrusts a hollow core through the wall and membrane. ATP powers it. It injects viral DNA
into the cell, leaving an empty capsid. E. coli begins to replicate the viral genes. One of these breaks up the
cell’s own DNA. The cell is under the command of the virus. Recycled parts of the cell’s original DNA are
used to form phage proteins and nucleotides. These come together to form phage heads, tails, and tail
fibers. Then the phage directs the production of an enzyme the digests bacterial walls, releasing 100-200
phage particles.
Lysogenic – The phage binds to the E. coli receptor and injects DNA into it. The DNA forms a circle and a
molecule is incorporated into a specific site of a chromosome. It becomes a prophage. One prophage gene
codes for a protein the suppresses other prophage genes so they pass silently on to other cells as these are
reproduced.
6. Explain how viruses may cause disease symptoms, and describe some medical weapons used to fight viral
infections.
Some viruses damage or kill cells by causing the release of hydrolytic enzymes. Others get
cells to produce toxins that lead to disease symptoms. To prevent these viruses from affecting people,
vaccines – harmless derivatives of the virus – are used to stimulate the immune system to defend itself
against any possible invasion.
7. List some viruses that have been implicated in human cancers, and explain how tumor viruses transform
cells.
Some main categories of viruses that have been found are retroviruses, papoviruses,
adenoviruses, and herpesviruses. Some specific ones from these broader categories are the virus for Hepatitis
B, the Epstein-Barr virus, papilloma viruses, and the HTLV-1.
8. List some characteristics that viruses share with living organisms, and explain why viruses do not fit our
usual definition of life.
It is not considered “alive” because it cannot exist independent of a live host cell. It is
unable to replicate its genes or supply itself with ATP. Some symptoms of these viruses are fever, aches, and
inflammation as a result of the body trying to fight it off.
9. Describe the structure of a bacterial chromosome.
The major component of a bacterial chromosome is one double-stranded DNA molecule
in a circle. It contains 100x more bases than the DNA of a typical virus. This coils into a ball called a
nucleoid.
10. List and describe the three natural processes of genetic recombination in bacteria.
Transformation – the alteration of a bacterial cell’s genotype by the uptake of foreign
DNA from the surrounding environment. Transduction – phages transfer bacterial genes from one host to
another. Conjugation – the direct transfer of genetic material between two bacterial cells temporarily joined.
11. Explain how the F plasmid controls conjugation in bacteria.
The ability of a bacterial cell to form sex pili and donate DNA requires the presence of
this plasmid.
12. Briefly describe two main strategies cells use to control metabolism.
1) Cells can vary the numbers of specific enzyme molecules and regulate gene expression.
2) Feedback inhibition: as the production of one enzyme increases in a cell, the bacteria stops its own
production. 13. Distinguish between structural and regulatory genes.
Structural – genes that code for polypeptides. Regulatory – produces repressor proteins.
Chapter 19: Genome Organization and Expression in Eukaryotes
Objectives
After reading this chapter and attending lecture, a student should be able to:
1. Compare the organization of prokaryotic and eukaryotic genomes.
Prokaryotic DNA – circular, nucleoid form is so small it can only be seen through an
electron microscope. Eukaryotic DNA – precisely complex with large amounts of protein. More loosely
formed.
2. Describe the current model for progressive levels of DNA packing.
A) DNA forms “bead on a string” which are nucleosomes in extended configuration.
Each is made up of DNA wrapped around a protein core that is composed of two molecules of four types of
histone. A fifth type of histone may be added to the outside of the “bead” of four. B) With the aid of the
fifth histone (H 1) the beaded string coils tightly to make a cylinder 30 nm in diameter. C) These coils in
turn for loops with six nucleotides per turn. D) These multiple levels form the chromosome which is visible
at metaphase.
3. Distinguish between heterochromatin and euchromatin.
Heterochromatin – interphase chromain s visible under a light microscope and is not
transcribed. Euchromatin – “true” chromatin is less compact and is available for transcription.
Chapter 20: DNA Technology
Objectives
After reading this chapter and attending lecture, a student should be able to:
1. Explain how advances in recombinant DNA technology have helped scientists study the eukaryotic
genome.
Three general benefits/uses are:
1) to produce a protein product
2) to give a particular organism a metabolic capability it didn’t have before.
3) To create further copies of the gene itself.
2. Describe the natural function of restriction enzymes.
Restriction enzymes protect bacteria from intruding DNA and other organisms. The
foreign DNA is cut up, but its own DNA is protected by adenine and cytosine in the sequence.
3. Describe how restriction enzymes and gel electrophoresis are used to isolate DNA fragments.
Bonds are cut in a staggered way and the resulting fragments have “sticky ends” which are
simply single-stranded ends. These will form complementary bonds with other DNA molecules split by the
same enzyme. These bonds in turn, are sealed by ligase which is recombinant DNA. Electrophoresis
separates either nucleic acids or molecules by size, electrical charge, and other physical traits. The separation
is based on a molecule’s rate of movement through a gel under the influence of an electric field.
4. List and describe the two major sources of genes for cloning.
1) Bacterial plasmids – recombinant plasmids are injected back into the bacteria. These
bacteria replicate (clone) these, producing a colony.
2) Viruses – following infection, it integrates into the chromosome.
5. Describe how "genes of interest" can be identified with the use of a probe.
In a process called hybridization, a probe or complementary molecule to the one being
checked for a gene is synthesized. The segment will hydrogen-bond to the specific “gene of interest” and be
traced by labeling it with a radioactive isotope or something similar.
6. Explain the importance of DNA synthesis and sequencing to modern studies of eukaryotic genomes.
To read a gene’s nucleotide sequence is to be able to more quickly determine the amino
acid sequence of its polypeptide. Also, restriction sites are found and these allow for more manipulation of
the gene. With computers, long sequences can be scanned to search for promoters and enhancers as well as
similarities to the sequences in other genes and organisms. This provides a new tool in the classification of
organisms.