A. Mendel's first experiment.
1. The two alleles (forms of a gene) for a trait segregate into separate gametes.
2. Start
with true-breeding parental pea plants.
a. homozygous dominant and homozygous
recessive
3. Cross
the parentals, obtain F1 (first filial) generation.
a. all heterozygous (one dominant and one recessive
allele)
4. Allow
the F1 to self-fertilize.
a. Obtain a 3:1 ratio of dominant to recessive phenotypes
(what you see).
b. Obtain 1 homozygous dominant, 2 heterozygous, and
1 homozygous recessive
for every four possible genotypes (the actual alleles inherited).
B. Mendel's dihybrid cross.
1. Two genes assort independently during meiosis.
2. Start with true-breeding parental plants for both characteristics.
3. Cross the parentals, obtain F1 heterozygotes for both genes.
4. After self-fertilization, F2 showed 50% parental and 50% recombinant genotypes.
5. The
phenotypes were 9 (both dominant parentals): 3 (1 dominant and 1 recessive
recombinant): 3 (1 recessive and 1 dominant recombinant): 1 (both recessive
parentals).
6. Can
determine inheritance patterns by Punnett Square or mathematically
by multiplying
individual probabilities together.
C. Test Cross
1. Can
determine the genotype of an unknown dominant phenotype by crossing the
unknown with a homozygous recessive and observing phenotype of offspring.
a. If offspring are 50% dominant and 50% recessive, the
unknown was heterozygous.
b. If offspring have 100% dominant, the unknown was homozygous
dominant.
D. Non-Mendelian Inheritance
1. incomplete
dominance
a. Heterozygote shows intermediate phenotype.
b. example: pink snapdragon color
2. co-dominance
a. Heterozygote equally expresses both alleles.
b. examples: AB blood type and sickle-cell
allele carriers.
3. Some
phenotypes are determined by polygenetic inheritance.
a. One phenotype caused by more than one gene.
b. example: human eye color
4. Some
genes have more than one phenotype that they effect. This is pleitropy.
a. example: sickle-cell allele helps malaria resistance
in heterozygotes, causes the
disease in homozygotes.
5. In a
population, more than 2 alleles are possible. (Only 2 from this pool will
enter one
individual). Called a multiple-allelic series
a. example: human blood type alleles (A, B, i)
6. Genetic
linkage
a. When genes are carried on the same chromosome they
display linkage.
b. They can only be separated from one another by crossing-over
and recombination.
c. The greater the percentage of recombination (recombination
frequency) the
farther apart on a chromosome the genes lie.
d. If recombination frequency = 0, genes are completely
linked. Usually means
they are the same gene.
e. If recombination frequency = 50% or greater, genes
are unlinked. Usually
means they are on different chromosomes.
f. If recombintion freqencey is between 0 and 50%, genes
are partially linked.
7. Sex-linkage
a. All genes carried on the X and Y chromosomes are hemizygous
in the male.
- means there is no corresponding homologous chromosome with that gene
b. Females must be homozygous recessive for a mutated
gene carried on the X
chromosome for it to have an effect.
c. Females can be carriers for mutant genes on the X
chromosome.
- have one normal (wild-type) allele and one mutant allele.
d. Males always show the effect of a mutated gene carried
on the X chromosome.
e. There can be sex-linked genes carried on the Y chromosome
but they are rare.
-these could never be seen in a female, because she has no Y chromosome.
8. Other things to remember.
1. Not all mutations are recessive,
there are many examples of mutated alleles that are
dominant. We did not study any specific examples of these.
2. The terms dominant and recessive
actually refer to the function of the protein produced by
the gene. Recessive alleles, therefore, usually make a non-functional protein.
II. Experimental evidence for DNA as the genetic material:
A. Bacterial transformation of R strain cells by
killed S strain cells produced live S strain cells. (Frederick
Griffiths).
B. Hershey-Chase experiment using
35 S (into proteins) or 32 P (into DNA) radioactively labeled
bacteriophage. Radioactivity found in the infected cell pellet (DNA) not
in the supernatant (protein).
III. DNA structure
A. DNA is made up of nucleotides.
1. Nucleotides consist of a nitrogenous organic base,
either pyrimidine or purine,
a 5 carbon sugar called a ribose, and one, two, or three phosphates.
2. The pyrimidines in DNA are either thymine or cytosine.
The purines are either
adenine or guanine. In RNA, the pyrimidine uracil is found, not thymine.
3. The ribose in DNA is deoxy-ribose, meaning the number
2 carbon has only H,
not OH attached. In RNA, the nucleotides have OH at carbon 2 of the
ribose.
B. The 5' Phosphate of the incoming nucleotide
is linked to the 3'OH of the nucleotide
that has
just been added to the growing polymer.
C. The first nucleotide in the
polymer has a free 5' Phosphate and defines the 5'end of
the DNA polymer.
D. The last nucleotide in the polymer has a free 3' OH and defines the 3' end of the polymer.
E. The most common overall structure of DNA is:
1.
Two right-handed alpha helices wind around each other in antiparallel orientation.
a. 5'-3' on one DNA polymer strand; 3'-5' on the other.
b. The winding creates alternate major and minor grooves.
2.
The polymers are held together by hydrogen bonding between bases of one
polymer
and bases of the other polymer.
a. Adenine (A) bonds to Thymine (T); Guanine (G) bonds
to Cytosine (C).
3. The bases lie in the interior of the double-helix, perpendicular to its axis.
4. The sugar-phosphate backbone is at the exterior of the double-helix.
5.
Evidence for this structure came from
a. Chargaff's Rules: concentration of A = T and of C=G
in all species of DNA
b. Viscous (thick) DNA becomes fluid when heated.
(Suggested H bonds)
c. x-ray crystallography of DNA showed repeated patterns.
6.
A nucleosome is 146 base pairs of DNA wound twice around a histone octamer.
These are
separated by DNA linker regions without nucleosomes. This is the structure
of chromatin
in eukaryotic nuclei.
a. this can be further folded up into a 30-nanometer
fiber
IV. DNA replication
A. DNA replicates by a semi-conservative mechanism.
B. One
parental DNA strand serves as the template for a newly synthesized strand
that is complementary and antiparallel.
1.
The bases are complementary, an A in the parental strand will cause a T
to be put
at that location in the daughter strand, etc.
2. The new strand grows in a 5'-3- direction.
3. The enzyme, DNA polymerase, catalyzes the polymerization reaction.
4.
Each new strand must be "primed" using a short RNA primer, put there
by the
enzyme primase.
5. A replication fork forms where the new strands are growing.
6.
The parental strand that is 3'-5' relative to the replication fork can
be replicated
in a continuous fashion.
a. The daugher strand is called the leading strand.
7.
The parental strand that is 5'-3' relative to the replication fork can
only be
replicated discontinuously.
a. Short strands grow 5'-3' from the fork. They are called
Okazaki fragments.
b. The Okazaki fragments are later connected to
one another using DNA ligase.
8. To melt the original double stranded DNA molecule requires DNA helicase.
9.
To keep the unreplicated DNA from tangling up as the replication fork progresses
requires DNA topoisomerase.
10.
DNA replication begins at an origin of replication (ori) where an
initiator protein
binds and gets things started.
a. Two replication forks move from the ori in opposite
directions to speed up DNA
synthesis. This produces a replication bubble.
V. Genes code for the amino acid sequence of a polypeptide.
1. The code is a triplet code where the three sequential
bases in a DNA molecule
determine one amino acid that should be put into a growing polypeptide.
a. These are called codons.
2. There are three codons that mean STOP. One codon,
AUG means START.
a. The START codon sets the reading frame.
b. The START codon always codes for methionine.
c. If the reading frame is not read accurately, a frameshift
mutation results.
VI. The DNA code is copied into a single-stranded
RNA molecule called
messenger RNA (mRNA).
1. This process is called transcription.
2. RNA polymerase is the enzyme that catalyzes
the polymerization reaction that
makes the mRNA polymer (pre-mRNA or the primary RNA transcript
in eukaryotic cells).
3. Ribonucleotide triphosphates are attached using the
energy from removal
of a PP from the incoming nucleotide.
4. The 5' Phosphate of the incoming nucleotide is linked
to the 3'OH of the
nucleotide just added to the growing polymer, similar to DNA replication.
5. Only one DNA strand is used as the template.
6. The mRNA strand is made in the 5'-3' direction.
7. The mRNA is antiparallel and complementary to the template DNA strand.
8. The base uracil in the mRNA is specified by adenine
in the template
strand. All other pairings are as they are in DNA-DNA pairings.
9. A promoter is a regulatory region in the DNA,
usually preceeding the gene,
that is the site where RNA polymerase binds to the DNA and is directed
to
where it should begin making the mRNA molecule.
10. The pre-mRNA molecule in eukaryotic cells is processed
after it has been made.
a. A Cap is attached to the 5'end. It is an unusual
G containing nucleotide.
- it helps set the reading frame and in mRNA transport to the cytoplasm.
b. A poly-A tail is attached to the 3' end.
- it helps in transport out of the nucleus to the cytoplasm and during
protein synthesis. It increases mRNA stability.
c. Non-coding intervening sequences called introns
in the mRNA are
removed by a process called splicing. The remaining coding
sequences, called exons, are attached to one another during the
splicing reaction.
11. Once this is finished, the mRNA is sent into the
cytoplasm to be used for
protein synthesis.
12. Evidence for introns was found by observing electron
microscope pictures of hybrids between
template DNA strand for globing gene with its mRNA. Looped out region was
the intron.
VII. Protein synthesis is done by a process called translation.
1. Translation requires a particle called the ribosome.
a. Ribosomes consist of one large subunit and one small
subunit.
b. Each subunit contains RNA molecules called ribosomal
RNAs.
(rRNA) and many ribosomal proteins.
c. Ribosomes contain the enzymes, both in protein and
RNA form, that
catalyze the events of translation that lead to peptide bond formation
and
a completed, accurate polypeptide.
2. The amino acids that will be attached to the polypeptide
are first loaded onto a
small RNA molecule called transfer RNA (tRNA).
a. This reaction is called "charging".
b. It is catalyzed by an enzyme called amino acyl tRNA
synthetase.
c. The enzyme specifically attaches the correct amino
acid to the 3'end of the
tRNA molecule.
d. The tRNA is in a cloverleaf structure, stabilized
by hydrogen bonds between
complementary bases within the tRNA molecule.
e. One of the loops so formed is the anticodon loop which
has the three
nucleotide sequence called the anticodon that is complementary and
antiparallel to the codon of the mRNA molecule that specifies that amino
acid carried by the tRNA.
3. Translation begins with Initiation.
a. The small ribosomal subunit interacts with the tRNA
carrying methionine.
b. It finds the Cap on the 5' end of the mRNA and moves
over to the first AUG
that serves as the START codon. The anticodon and codon bind.
c. The large subunit joins, making the functional ribosome.
d. The ribosome now has a P site where the tRNA
carrying methionine is
located and an A site covering the next codon on the mRNA.
4. The next phase of translation is called elongation.
a. The next tRNA moves into the A site. The anticodon
will interact with the
codon.
b. The enzyme peptidyl transferase catalyzes the breaking
of the bond holding
methionine to the tRNA at the AUG and attaches the methionine to the
amino acid carried by tRNA at the A site, creating a peptide bond.
c. The first tRNA leaves the P site, exiting through
the E site.
d. The ribosome moves over one codon on the mRNA. This
is called
translocation.
e. Now, the P site is at the tRNA previously at the A
site and the A site
is now empty and situated over the third codon.
f. The third tRNA carrying the amino acid coded for by
the third codon moves
into the A site. Anticodon binds to codon.
g. Peptidyl transferase breaks the bond between
the second tRNA at the P site
and the two amino acids and attaches them to the amino acid on the tRNA
at
the A site by a peptide bond.
h. tRNA two leaves the P site, translocation occurs,
and the A site is ready to
receive tRNA number 4.
i. This process proceeds again and again sequentially
until a stop codon is
located at the A site.
5. Protein synthesis ends by termination.
a. There is no tRNA with an anticodon for a stop codon.
b. A termination factor binds to the A site instead.
c. Peptidyl transferase breaks the bond between the final
tRNA at
the P site, adds water to the end of the polypeptide chain creating the
COOH terminus, and releases the polypeptide.
6. Many ribosomes (polyribosomes) are translating a single
mRNA at the same
time, with the ones closer to the 3'end the farthest along.
7. In bacteria (prokaryotes) coupling of transcription
and translation occurs since they
happen in the same place, the cytoplasm. As soon as transcription has begun,
translation will soon follow, even while transcription is still continuing.
Also, bacterial mRNAs can be polycistronic. More than one protein can be
made from
different AUG start codons, preceeded by Shine-Dalgarno sequences. These
bind to
the ribosomal RNA in complementary, antiparallel fashion, and direct the
ribosome to the next AUG
to use as a start codon.
8. Cell membrane proteins, secreted proteins, lysosomal
proteins, and secretory vesicle proteins
are synthesized on the rough ER membrane.
a. they begin to be translated in the cytoplasm.
b. a signal peptide is made first at the amino terminal
end of the protein and this binds to the
signal recognition particle (SRP), halting translation
c. the SRP binds to a receptor in the ER membrane, depositing
the signal peptide in the memberane.
d. the SRP leaves and translation continues, with the
elongating polypeptide being pumped into the
ER lumen.
e. the signal peptide is usually cleaved off, releasing
the protein completely. Alternatively, it
can remain, anchoring the protein in the membrane and eventually it will
become a cell membrane
protein.
9. Sickle cell anemia is caused by a base-pair
substitution point mutation in the protein, substituting a
non-polar amino acid for a negatively charged amino acid.
a. Changes the 3D shape of the polypeptide.
b. Caused by a single nucleotide change in the DNA causing
an incorrect mRNA codon.
c. Other point mutations include pase pair insertions
or deletions (can cause nonsense, frameshift,
or loss of a codon mutations).
d. Also can have base-pair substitutions that introduce
stop codons. Some are silent and do not
cause a change in the protein.