Name: _______________________________
Section Time: _________________________
Bio 93 : Final Study Guide
This study guide covers material from Lecture 18 – 25. Its purpose is to provide a framework in which
you can input your knowledge from this class. It is not all-inclusive, but covers topics focused on in
lectures and rocketmix modules. I will not be collecting this study guide, but please bring it (and
questions about it) to section on Wednesday 12/10.
Directions: Please fill this out without looking at your notes, then go back and fill in more information (in
a different color pen) using your notes, book and rocketmix modules.
“It’s mainly about working hard and proving to people you’re serious about it, and stretching yourself
and learning.” – Daniel Radcliffe (Harry Potter)
Lecture 18: Alterations of Chromosomes + Fruit Fly Genetics Rocketmix
Reading: 286-302/292-309
1) Who was Thomas Hunt Morgan? Please summarize his experiment concerning red and white
eyed flies and the contributions of the experiment to scientific knowledge.
Thomas Hunt Morgan – 1933 Nobel Prize for his discoveries concerning the role of
chromosomes in heredity
Experiment:
- Bred flies:
o Wild type female (red eyes)
o Mutant Male (white eyes)
o F1  all offspring had red eyes
o F2  2 females with red eyes, one male with red eyes, one male with white
eyes
- Concluded that eye color is sex-linked
o Gene located on X chromosome
o Males have only one X chromosome, XY (females have 2, XX)
2) What are recombination frequencies?
a. What is the recombination frequency for genes on different chromosomes?
Zero, because genes on two different chromosomes will never cross over.
b. Genes that are located more closely on a given gene have a ___________________
recombination frequency, and those located more far apart have a ________________
recombination frequency.
(more close = higher; far apart = lower)
3) Why are sex-linked disorders (such as colorblindness) more common in males?
Sex-linked disorders are more common in males because females inherit two x
chromosomes, only one of which is active. Males only inherit one X chromosome, so
their chances are higher.
4) Please define genomic imprinting and describe one example.
Genomic imprinting is the silencing of the gene from one parent by a gene from the
other parent through the process of DNA methylation. Examples are provided in lecture
and in the book.
Lecture 19: DNA  the Molecular Basis of Inheritance
Reading: 305 - 314
5) Please define the following terms and their importance to DNA structure and replication.
a) DNA – heritable genetic material
b) Transformation – the process by which bacteria take up genetic information (DNA) .This was
important for the experiments that determined that DNA is the genetic information in cells
c) Bacteriophages (phages) – viruses that infect Bacteria; used in experiments do discover that
DNA is the genetic information in cells.
d) Double Helix – the structure of DNA, where two twisting strands are held together by
Hydrogen bonding. Watson and Crick received the Nobel prize for their model, however
Rosalind Franklin was the first to capture the structure with x-ray crystallography
e) DNA Strand – a single strand of DNA consisting of nucleotides: A,C,T,G
f) Anti-Parallel Strands – each parent strand of DNA runs is the reverse compliment of the
other
g) Base Pairing – in DNA, A pairs with T, and C with G
o Purines pair with pyrimidines
h) DNA Replication – the semi-conservative process by which DNA is split and then daughter
strands formed to create two new pairs of strands of DNA
i) Semi-Conservative Replication – A result of the nature of DNA replication. Since a parent
strand is “conserved” in each of the offspring, DNA replication is said to be semiconservative. This means that each “new” double-stranded DNA formed has one parent
strand and one daughter strand.
j) Origin of Replication – sequence consisting of mostly A’s and T’s where DNA replication
begins
k) Template Strand – the strand that is read by DNA polymerase during DNA replication, so
that a complimentary daughter strand can be synthesized
6) Please describe the experimental process by which Griffith and others determined that DNA is
the genetic information in bacterial cells.
Griffith and others experimented with live and dead strains of a pneumonia bacteria,
and discovered that information could be passed from dead infectious bacteria to live
non-infectious bacteria that would then make the live bacteria infectious. This
happened via transformation, where bacteria took up information from dead cells in
order to acquire the coating that was needed in order to become infectious.
Once this principle was established, another 20 years of experimentation proved that it
was DNA that was taken up by the live, non-infectious cells that caused them to become
infectious. The experiments that resulted in this finding were ones where proteins, fats,
RNA and DNA were removed in a factorial manner, so that the transferred material
could be singled out.
7) Please describe the procedures and findings of the Hershey-Chase experiments with viruses that
concluded that DNA was the genetic material.
a. Background:
These experiments used bacteriophages (viruses that infect bacteria) to determine the
genetic material that was passed from bacteriophage to bacteria that causes infection.
Using knowledge that Sulfur is found in proteins, but not DNA, and that Phosphorous is
found in DNA and not Sulfur, researchers were able to distinguish between protein and
DNA as a possible genetic material.
b. Methods:
Half of the phages were grown in media with radioactive Sulfur that were tagged with
fluorescent pink. This labeled the protein shell of all of the phages pink. The other half
were grown in media with radioactive Phosphorous that was tagged with fluorescent
blue. This labeled the DNA of this group pink. The phages were allowed to infect two
groups of bacteria. Afterward, the phages were separated from the bacteria using a
blender. The resulting fluid was then centrifuged.
c. Results:
The results showed that the bacteria were labeled with the blue fluorescent tag, where
the fluid was labeled with the pink fluorescent tag.
d. Conclusion:
The bacteriophages had transferred the blue-labeled DNA into the bacteria upon
infection, and the protein shell (pink) was not the agent that was transferred. This study
provided evidence that genetic material was DNA.
8) What are the 4 bases in DNA? How do they differ from the 4 bases in RNA?
The 4 bases in DNA are: A-T, G-C. In RNA, they are: A-U, G-C
9) Please explain Chargaff’s Rule and answer the following question. You are examining the DNA of
a newt whose genome contains 34% thymine. What percentage of the bases in its genome are
cytosine?
Chargaff’s Rule states that the percentage of A = T and the percentage of C = G, and that
A%+G%+C%+T% = 100%.
Using this rule, you know that A = T = 34 %, and 100 – (34 + 34) = C + G.
This means C + G = 100 – 68% = 32%
If there are equal numbers of C and G, then the % of C = ½ x 32 %, or 16%.
C = 16%
10) Name the three important people in the discovery of DNA structure and base pairing and their
contributions.
Rosalind Franklin – used x ray crystallography to take a picture of the structure of DNA
Watson + Crick – used the x ray picture to make a model, and determined base pairing by width
of purines and pyrimidines
Lecture 20: DNA Replication and Repair
Reading: 311-319
Concepts Needed:
-
Base pairing happens between a purine and a pyrimidine
DNA replication is semi-conservative
DNA replication occurs in the 5’ to 3’ direction
DNA replication is carried out by DNA polymerases, and many other enzymes
DNA has a leading and lagging strand
Replicated DNA is proofread and repaired
Telomeres are sequences of DNA at the end of chromosomes
11) Please define the following terms and their importance to DNA structure and replication
a. DNA Replication
b. Semi-Conservative Replication
c. Origin of Replication
d. Helicase – enzyme that “un-zips” the two parental strands of DNA, breaking H bonds
between the bases
e. RNA Primase – enzyme that creates RNA primer so that DNA Polymerase can begin
adding nucleotides to the daughter strand during DNA replication (works in 5’ to 3’
direction)
f. Topoisomerase – enzyme that relaxes the parental strands ahead of Helicase so that
they do not break due to the force of unwinding
g. Single-Strand Binding Protein – after parental strands are separated, these proteins bind
to the unpaired bases, making them more stable in their single-stranded state
h. Replication Forks – corners of the replication bubble where DNA replication occurs
i. Template Strand – parental strand from which the daughter strand is synthesized
j.
DNA polymerase – enzyme that adds nucleotides to a forming strand of DNA. Needs a
primer and a template strand in order to operate. Works in the 5’ to 3’ direction.
k. Leading Strand – strand that is synthesized continuously in the overall 5’ to 3’ direction
l. Lagging Strand –strand that is synthesized in fragments in the 5’ to 3’ direction, but
overall in the 3’ to 5’ direction
m. Okazaki Fragments – the fragments in which the lagging strand is synthesized
n. Telomeres – end sequences of chromosomes that shorten with replications
12) Please illustrate and provide explanations of the following steps of DNA replication:
a. Parental DNA Strands Parted
The Parental DNA strands are first parted at the origin of replication. This is a sequence
that contains mostly A’s and T’s, as they have less H bonds than C’s bonded to G’s, and
are more easily parted. Topoisomerase first comes through and relaxes the parental
strands by cutting and pasting bonds in the backbone that promote unwinding. Next,
Helicase comes through and unwinds the DNA in either direction, parting the Hydrogen
bonds between the bases of the two parental strands, creating 2 unwound and “flat”
parental strands that are separate. Lastly, single-strand binding proteins stabilize the
hydrophobic exposed bases so that the strands do not come back together immediately.
Illustrations for the portions of this guide can be found in the “DNA Walkthrough”
posted on the section web site.
b. Extension of the Daughter Strands : LEADING vs LAGGING
DNA polymerase is the enzyme that adds nucleotides in the 5’ to 3’ direction,
synthesizing the daughter strands of DNA. It needs two things to function: a primer, and
a template strand. At this point in replication, the parental (template) strands have been
exposed allowing DNA polymerase in. Primase is an enzyme that adds a short RNA
primer to a site of replication. The DNA polymerase then attaches to the 3’ end of the
primer, elongating the daughter DNA in the 5’ to 3’ direction.
For the leading strand, DNA synthesis occurs continuously, whereas on the lagging
strand, it occurs in fragments called Okazaki Fragments.
c. Finishing of replication
Since RNA doesn’t last as long as DNA, the RNA primers must be replaced by DNA. DNA
polymerase 1 is responsible for this process. Then, there are some holes left in the
backbone by the DNA polymerase process, which are patched by DNA ligase. The end
result is two double-stranded DNA molecules, each with one parental and one daughter
strand.
Please do not hesitate to email me or come to OH (office hours) if you have any
questions about this concept or any others.
13) Please label the parts of the following diagram including: enzymes, 3’ and 5’ ends of each strand
and primer, leading and lagging strands, RNA primers, Parental DNA strands, and Daughter DNA
strands
14) Please complete the following chart concerning the function and roles of important players in
DNA replication.
15) Different Types of Proofreading and Repair:
a. What is the key proofreading mechanism?
DNA polymerases proofread newly made DNA, replacing any incorrect nucleotides
b. What is mismatch repair?
When bases are not matched up, repair enzymes correct the errors.
c. What is nucleotide excision repair?
When a nuclease cuts out a stretch of damaged DNA, which is replaced by DNA
polymerase and then sealed by DNA ligase.
16) Why is mutation important?
Without mutation, natural selection would not occur.
17) What are telomeres and where are they located? Why are they important? Why are they of
interest for aging?
Telomeres are repeated sequences of TTAGGG at the ends of eukaryotic chromosomal DNA.
They are important because they provide a buffer for the eventual shortening of the lagging
strand during DNA replication. Our telomeres shorten as we age (or as our cells replicate more
and more) in our somatic cells. In our germ cells (gametes), an enzyme called telomerase
elongates the telomeres. Telomeres are a popular target in aging research because, perhaps, if
they are lengthened, we can extend youth.
Lecture 21: Transcription
Reading: 325-336
Concepts Needed:
-
Genes produce proteins via transcription, then translation
Transcription is the production of mRNA from DNA
Translation is the formation of a protein sequence based off the mRNA and occurs
at ribosomes
3 base pairs code for an amino acid, and make up a codon
RNA is spliced (cut and rejoined) at introns and exons
18) Please define the following terms and their importance to DNA transcription, translation and
gene expression.
a. Gene expression – the process by which DNA directs protein synthesis. It includes two
stages: transcription and translation
b. Genes – specific areas of the chromosomes that contain sequences of bases that give
specific information to give a specific protein (protein makes up phenotype)
c. Transcription – production of mRNA from DNA
d. Translation – production of protein from mRNA
e. mRNA – messenger RNA , created in transcription and read in translation
f. tRNA – transfer RNA, involved in translation
g. rRNA – ribosomal RNA, makes up ribosomes, which perform translation
h. Ribosomes – bind mRNA and tRNA to conduct translation
i. Polyribosomes – groups of ribosomes transcribing the same mRNA transcript at once,
enabling many protein copies to be made simultaneously
j. Primary Transcript – pre-mRNA that is the direct product of transcription before splicing
and the cap and tail are added
k. Codon – group of 3 bases that code for a single amino acid
l.
m.
n.
o.
p.
q.
r.
s.
t.
u.
v.
w.
Template Strand – strand which the mRNA is based off of (can be either parental
strand, however the same parental strand is transcribed for each gene). This strand is
also called the non-coding strand and is read in the 3’ to 5’ direction.
Reading Frame – set by the start codon (AUG) which codes for methionine. The reading
frame establishes which 3 bases will make up a given codon
RNA Polymerase – enzyme that performs transcription, using DNA as a template, adding
RNA nucleotides to create the primary transcript
Promoter – sequence of DNA where RNA polymerase attaches
Terminator – sequence of nucleotides that signals RNA polymerase to stop transcribing
the DNA into the primary transcript
Transcription Unit – stretch of DNA transcribed into pre-mRNA
Transcription Factor – specific proteins found in specific nuclei that can bind to
promoters and recruit RNA polymerase, controlling which genes will be transcribed
TATA Box – sequence at the beginning of a promoter consisting of T’s and A’s, where the
strands are easily separated
Poly-A Tail – series of A nucleotides added to the end of a pre-mRNA transcript that will
protect from degradation
RNA Splicing – the removal of introns and gluing together of exons of a primary mRNA
transcript as it becomes mRNA. This is done by spliceosomes.
Exons – coding portion of the gene; the DNA RNA sequence that will eventually be
expressed and used to make protein
Introns – intervening sequences between the exons that are not coded and will not be
expressed as a protein (get cut out during splicing)
19) Please fill in the central dogma below and the processes it entails:
DNA --------------------- RNA ------------------------- protein
Transcription
Translation
20) Please describe the following steps in transcription and draw a figure for each step containing
the strands of DNA and RNA involved and the enzymes and proteins involved. Where does each
process occur?
a. Initiation
The promoter region of a gene (TATA box) recruits transcription factors which bind to
the promoter, and recruit RNA polymerase to the scene. These three elements are
called the initiation complex.
b. Elongation
After RNA polymerase has been recruited, the transcription unit is transcribed into pre
mRNA. The gene is identified and the parental strands are parted by RNA polymerase
(no helicase here). The template strand is read by RNA polymerase in the 3’ to 5’
direction as the pre mRNA is created in the 5’ to 3’ direction. The pre mRNA is the
reverse compliment of the template strand. The primary transcript is made of RNA and
contains both introns and exons and does not yet have a cap and tail. The entirety of
transcription occurs in the nucleus.
c. Termination
Termination occurs once a termination sequence has been reached by RNA polymerase.
This causes RNA polymerase to stop transcribing and frees the primary transcript from
the enzymes. This happens in the nucleus.
d. RNA Modification
The primary transcript must undergo three principal changes in order to become mature
RNA (mRNA).
1) A 5’ cap must be added containing a modified Guanine nucleotide
2) A poly A tail must be added (bunch of A’s on the 3’ end)
3) An enzyme called spliceosome cuts out the introns and glue the exons together in a
process called RNA splicing
This process happens in the nucleus.
21) Why is it important that a codon consists of 3 amino acids?
3 bases give us a total of 64 possibilities for amino acids, and we need to have 20 distinct
possibilities present. If codons were made up of 2 bases, there would only be 16 combinations,
which would not be enough for specificity.
22) Why are the three principal functions of proteins?
They function as cell structure, enzymes, and act as signaling molecules.
23) Please explain template, non-template, coding and non-coding strands. Which is transcribed?
Will two genes on the same chromosome always be transcribed from the same template
strand?
The template strand is another name for the non-coding strand, and the coding strand is
another name for the non-template strand. The template strand is the strand that is
transcribed. For each gene, the same strand is transcribed, however, for different genes on the
same chromosome, each may use a different template strand.
24) How did scientists discover the genetic code? Please describe both the experiment and the
findings.
Scientists deciphered the genetic code by synthesizing 3-letter RNA molecules and putting them
in a test tube with ribosomes, free nucleotides, tRNA, amino acids, ATP, and other enzymes.
After letting it sit for a few hours, the RNA molecules were all transcribed into amino acids. By
analyzing which amino acid was produced and knowing the code that it came from, the codons
were understood.
25) What is the start codon and which amino acid does it code for? What is the anti-codon from the
tRNA ? Why is the start codon important?
The start codon is: AUG, and it codes for the amino acid methionine (MET). The anti-codon from
the tRNA is: UAC
The start codon is important because it sets the reading frame for the rest of the amino acids.
The reading frame determines the following codons, for example:
AUG GCC AUU UGG etc.
Vs.
MAU GGC CAU UUG G, which would end up being read as AUG GCC AUU UGG because the AUG
codon always starts the amino acid
26) Please draw the difference between pre mRNA (the primary transcript) and mRNA. Include the
5’ UTR, Start Codon, 5’ Cap, 3’ Poly A Tail, 3’ UTR, and stop codon as well as introns and exons.
See lecture slides for figures.
Lecture 22: Translation
Reading: 345-357
Concepts:
-
Genes are produced via transcription then translation
Translation is the formation of a protein sequence based off of the mRNA, and
occurs at ribosomes
- The proteins produced in translation are targeted to specific cellular locations
27) Please define the following terms:
a. Translation – the creation of an amino acid sequence (protein) from the mRNA, occurs
in the cytoplasm and is facilitated by the ribosome
b. RNA c. Ribosomes d. Polyribosomes e. Codon – a set of 3 RNA bases in mRNA that code for a single amino acid
f. Anti-Codon – a set of 3 RNA bases in tRNA that are reverse complimentary to those of
the codon
g. Amennoacyl-tRNA synthetase - creates correct matches between tRNAs and the correct
amino acid so that translation can occur accurately
h. Mutations – changes in the DNA sequence that can alter gene expression of a cell or
virus
28) Please describe the following processes that occur during translation and draw a figure to go
along with it.
a. mRNA leaves the nucleus and travels into the cytoplasm
b. Initiation:
During initiation, the small subunit of the ribosome binds to the 5’ end of the mRNA.
Matching tRNA binds to the start codon and recruits the large ribosomal subunit. The
three described components make up the initiation complex, which requires GTP to be
formed. The ribosome has 3 sites: E, P and A. Once the initiation complex is formed, the
MET tRNA sits in the P site.
c. Elongation:
During the process of elongation, amino acids are added one-by-one to the C terminus
of the growing amino acid chain. The following describes this process;
-
A site is empty
Codon recognition occurs  the correct tRNA is paired with the correct codon from
the exposed A site, so that tRNA “parks” at the A site
The ribosome catalyzes the formation of a peptide bond, so that the amino acid
from the P site is now attached to the one in the A site
The chain of amino acids is now held by the tRNA in the A site
Translocation occurs –> the tRNAs are now shifted over, so that the tRNA that was
once in the A site is now in the P site and holding onto the chain of the amino acids
The tRNA that was once in the P site is now in the E (exit) site and will leave
The A site is now empty and ready to find a matching tRNA
d. Termination :
- Occurs when the stop codon reaches the A site of the ribosome
- Instead of the polypeptide chain being added to the A site, a protein called a release
factor binds instead
- This release factor stimulates hydrolysis of the amino acid chain so that it can be
released
- The complex breaks apart and is recycled (ribosome, tRNAs etc.)
29) After the amino acid sequence is produced, it is not always ready to become a functioning
protein. What are some ways in which it is processed in order to become a functional protein?
-
Sometimes enzymes need bits of protein to be cleaved off in order to be activated
Sometimes phosphorylation or glycosylation needed for activation
-
Sometimes larger proteins need multiple polypeptide chains to be brought together
to be functional
30) How do proteins get transported to the part of the cell where they are needed?
Some proteins have a 20 amino acid long sequence that tells the cell where the protein needs to
go. This sequence is recognized by a signal recognition protein, which binds to the amino acid
and tows it to the correct location. In some cases, they are transported to the ER membrane,
where they are then secreted out of the cell.
31) Please provide a summary of the following mutations that can occur and the effects they have
on proteins formed:
a. Point Mutations: nucleotide pair substitutions and sickle cell anemia
Nucleotide pair substitutions occur when one nucleotide and its partner are
replaced with another pair of nucleotides
A single base change from T-->A changes a single base in the mRNA, which results in a
different amino acid being coded for by the codon. This single amino acid change results
in abnormal folding of the resulting protein that makes up red blood cells
b. Silent Mutations – have no effect on the amino acid produced by the codon because of
redundancy in the genetic code
c. Missense mutations – still code for an amino acid, but not the correct amino acid (as in
sickle cell anemia)
d. Nonsense Mutations – change an amino acid codon into a stop codon, nearly always
leading to a non-functional protein
e. Insertions + Deletions – are additions or losses of nucleotide pairs in a gene. They create
a disaster in the resulting protein, often producing a frame-shift mutation (MOST
HARMFUL)
f. Frame-shift mutation – mutations that alter the reading frame
g. What are mutagens? Agents that cause mutations
DNA Rocket mix: please review the DNA rocketmix module and the questions it contains. They are good
practice for questions on the Central Dogma that may be on the exam. You should have no problem
converting DNARNA protein, and also backwards from protein to DNA, keeping in mind the 3’ and
5’ ends, as well as the directions in which the enzymes that perform such processes work.
Lecture 23: Regulation of Gene Expression
Concepts:
-
Genes have promoters (regions where transcription is initiated)
o Include the TATA box
Transcription Factors (TFs) bind to the promoter and control elements, and may be
activators or repressors
mRNA degradation is controlled via regulation
protein creation is also regulated via proteasomes (eat up proteins)
32) Please define the following terms:
a. Transcription Factors – activating or repressing elements that bind to the promoter and
control elements, effectively controlling transcription, and therefore gene expression
b. TATA Box – region in promoter with T’s and A’s that are easily separated
c. Promoters
d. Enhancers
e. RNA Polymerase II
f. Alternative RNA Splicing
g. Proteasome
33) There are many places where gene expression can be regulated. Please describe the following
mechanisms of gene expression regulation and provide an illustration. Which is the most
important/powerful?
a. Chromatin
Epigenetic Mechanism
Chromatin is a complex of DNA and proteins that make up the chromosomes. They are
necessary to pack a large amount of DNA into a small volume. Nucleosomes consist of 8
histone proteins with DNA wrapped around it. The tails of the histone protein are
positively charged, and attract negatively charged DNA. Therefore, when the attraction
of the DNA to the histone protein is strong, it is difficult for transcription to occur. We
call this state heterochromatin. When enzymes in the nucleus attach acetyl groups to
the positively charged Glycine amino acids on the tails of the histone proteins, the
charge is nullified, leaving the histone protein less positively charged. This causes a
weaker attractive force between the DNA and the histones, allowing for transcription to
occur. We call this state euchromatin.
b. DNA Methylation
Epigenetic Mechanism
Gene expression can also be controlled through DNA methylation. This occurs when
enzymes attach methyl groups onto the DNA, blocking RNA polymerase from being able
to transcribe the DNA. This is a LONG TERM mechanism that underlies the process of
genomic imprinting. An example of DNA methylation occurs in the XX chromosome pairs
of girls. One of the X chromosomes is methylated so that it is inactivated for life.
c. Transcription
Most IMPORTANT
Transcription factors influence transcription initiation, and gene expression by
controlling which genes are transcribed in a given cell type. They turn genes “on” and
“off” by binding to control elements of a targeted gene. Different groups of
transcription factors are present in different cell types so that different groups of genes
can be transcribed in different cell types, creating differences among (for example) a
liver and muscle cell.
d. RNA Processing and Regulation
Once the primary transcript is created, RNA processing and regulation determine how
much of RNA actually becomes protein. An example of this is Alternative RNA splicing,
where different combinations of exons in a primary transcript can be spliced into the
mRNA. This process allows two or more different proteins (with different amino acid
sequences) to be made from a single gene.
e. RNA Degradation:
The longevity of RNA determines how many times it will be translated into protein.
When RNA is degraded, the poly A tail is first removed, then the 5’ cap. Nucleases then
“chew up” the rest of the RNA molecule for recycling.
f.
Protein Processing:
Some proteins must be modified in order to become functional. This means that
whether a protein is processed or not can determine whether it turns into a functional
product. Like RNA, protein can also be degraded by first being tagged with ubiquitin,
then transported to the proteasome where it is “eaten up” and then recycled. Proteins
can also be recycled by lysosomes, which digest them in a separate vacuole within the
cell compartment.
34) How do the two types of transcription factors differ? How do activators and repressors differ?
There are two types of transcription factors, ones that help with the function of RNA polymerase
II, and ones that bind to control elements. Activators and repressors are ones that bind to the
control elements. Activators activate transcription of a gene, and repressors prevent
transcription of the gene.
Lecture 24: Biotechnology
Reading: 408- 433
35) Please define:
a. Recombinant DNA – nucleotide sequences from two different sources that are
combined in vitro into the same DNA molecule
b. Biotechnology – manipulation of organisms or their components to make useful
products
c. Clone a Gene – make multiple copies of a gene
d. Clone - 3 different definitions
i. A lineage of genetically identical individuals or cells
ii. A single individual organism that is identical to another
iii. To make one or more genetic replicas of an individual or cell
36) How do you clone a gene? Please describe the steps and product.
In order to clone a gene, you must insert a gene of interest into a bacterial plasmid and
transform the bacteria so that the plasmid is inside. The bacteria will make many copies of the
gene, which can later be harvested. The gene or protein product can be isolated and studied or
used for other applications.
37) What are restriction enzymes and how can they be used to make recombinant DNA?
Restriction enzymes are bacterial enzymes that evolved to cut up phages (viruses that infect
bacteria) to protect the bacteria. They recognize specific DNA sequences and chop them up into
pieces called restriction fragments. The sequences they recognize are palindromes (same in
both directions…think RACECAR, which is RACECAR spelled backwards). The most useful ones
create “sticky ends”, which overhang from the DNA molecule. Blunt ends do not.
The restriction enzymes are incubated with both bacteria and plasmid separately. This yields
linear pieces of DNA and a linear piece of plasmid that can then be annealed by DNA ligase.
38) What is the process of PCR and what is it useful for?
PCR (polymerase chain reaction) is useful because it produces many copies of a specific target
segment of DNA. IT allows for the study of small DNA sequences, and the identification of
perpetrators in court.
Requires: DNA, dNTPs, water, DNA polymerase, Primers
It involves a 3-Step Cycle:
Heating: 90*C : double stranded DNA is separated
Cooling: 70*C : allows primers to anneal to single strands, which provide a 3’ end for DNA
polymerase to add nucleotides
Replication: DNA polymerase adds nucleotides, creating daughter strands of interest
39) What is gel electrophoresis and how does it work?
Gel electrophoresis occurs when you apply current to samples of DNA strands in a gel plate.
Since DNA is negatively charged, it moves toward the positively charged end of the plate.
Smaller pieces of DNA migrate faster than large ones, separating the resulting pieces into bands.
This can be used in criminology to identify criminals.
40) How was Dolly the sheep cloned? Why did she die so young?
The technique used to create Dolly is called nuclear transplantation. Mammary cells are taken
from sheep 1 (donor) and the nucleus from that cell is placed into the egg of sheep 2. The
resulting cell is jolted with electricity to start the creation of the embryo, which is implanted into
a third sheep. The third sheep is treated with hormones to fake its pregnancy.
Dolly didn’t live long because she had telomeres from an adult sheep, and aged prematurely.
41) What is the difference between an embryonic stem cell, an adult hematopoetic stem cell and
an induced pleuripotent stem cell?
Embryonic stem cell – comes from blastocycst, can be turned into any cell type
Adult hematopoetic stem cell – lives in bone marrow of an adult, can only become blood cells
Induced pleuripotent stem cell – (iPS) when four genes are inducted in an adult (skin) cell, it can
return to an embryonic-like state and become any other cell type.
Lecture 25: The Genetic Base of Development
Reading:
Concepts:
-
Cells go through differentiation, then morphogenesis
The fate of cells is determined by cytoplasmic determinants
Cleavage generates small undifferentiated cells, and is followed by gastrulation and
the formation of germ layers
- The single-celled zygote gives rise to cells of many types, each with a different
structure and corresponding function
42) Please define the following terms:
a. Differentiation
b. Morphogenesis
c. Cytoplasmic Determinants
d. Induction
e. Differentiation
f. Pattern Formation
g. Positional Information
h. Bicoid
i. Cleavage
j. Blastocyst
k. Organogensis
l. Fertilization
m. Morphogenesis
n. Gastrulation
o.
p.
q.
r.
Germ Layers
Ectoderm
Endoderm
Mesoderm
43) How are the axes defined in an embryo (what gives rise to pattern formation)? How does this
relate to bicoid?
44) How does cell determination occur?
45) Please draw a diagram of the stages of embryonic development of a tadpole.