Download Chapter 9, 10, and 11

Chapter 9, 10, 11, and 12
Gregor Mendel
A. The Blending Concept of Inheritance
1. This theory stated that offspring would have traits intermediate between those
of the parents.
2. Red and white flowers produce pink flowers; any return to red or white
offspring was considered instability in the genetic material.
3. Charles Darwin wanted to develop a theory of evolution based on hereditary
principles; blending theory was of no help.
a. A blending theory did not account for variation (differences) and could not
explain species diversity.
B. Mendel’s Particulate Theory of Inheritance
1. Mendel was an Austrian monk.
2. Mendel formulated two fundamental laws of heredity in the early 1860s.
3. He had previously studied science and mathematics at the University of
Vienna.
4. At time of his research, he was a substitute science teacher at a local technical
high school.
5. Because Mendel had a mathematical background, he used a statistical basis
for his breeding experiments.
6. Mendel’s particulate theory is based on the existence of minute particles—
now called genes.
C. Mendel Worked with the Garden Pea
1. Mendel prepared his experiments carefully and conducted preliminary studies.
a. He chose the garden pea, Pisum sativum, because peas were easy to
cultivate, had a short generation time, and could be cross-pollinated by
hand.
b. From many varieties, Mendel chose 22 true-breeding varieties for his
experiments.
c. True-breeding varieties had all offspring like the parents and like each
other.
d. Mendel studied simple traits (e.g., seed shape and color, flower color,
etc.).
2. He used his understanding of mathematical principles of probability to
interpret results.
Mendel’s Law
A. Law of Segregation
1. Mendel confirmed that his tall plants always had tall offspring, i.e., were truebreeding, before crossing two different strains that differed in only one trait—
this is called a monohybrid cross.
2. A monohybrid cross is between two parent organisms true-breeding for two
distinct forms of one trait.
3. Mendel tracked each trait through two generations.
a. P generation is the parental generation in a breeding experiment.
b. F1 generation is the first-generation offspring in a breeding experiment.
c. F2 generation is the second-generation offspring in a breeding
experiment.
4. He performed reciprocal crosses, i.e. pollen of tall plant to stigma of short
plant and vice versa.
5. His results were contrary to those predicted by a blending theory of
inheritance.
6. He found that the F1 plants resembled only one of the parents.
7. Characteristics of other parent reappeared in about 1/4 of F2 plants; 3/4 of
offspring resembled the F1 plants.
8. Mendel saw that these 3:1 results were possible if:
a. F1 hybrids contained two factors for each trait, one being dominant and the
other recessive;
b. factors separated when gametes were formed; a gamete carried one copy
of each factor;
c. and random fusion of all possible gametes occurred upon fertilization.
9. Results of his experiments led Mendel to develop his first law of
inheritance—the law of segregation:
a. Each organism contains two factors for each trait.
b. Factors segregate in the formation of gametes.
c. Each gamete contains one factor for each trait.
d. Fertilization gives each new individual two factors for each trait.
B. Mendel’s Cross as Viewed by Classical Genetics
1. The gene locus is the specific location of alleles on homologous
chromosomes.
2. Alternate versions of a genes are called alleles.
3. A dominant allele masks or hides expression of a recessive allele; it is
represented by an uppercase letter.
4. A recessive allele is an allele that exerts its effect only in the homozygous
state; its expression is masked by a dominant allele; it is represented by a
lowercase letter.
5. The process of meiosis explains Mendel’s law of segregation.
6. In Mendel’s cross, the parents were true-breeding; each parent had two
identical alleles for a trait–they were homozygous, indicating they possess
two identical alleles for a trait.
7. Homozygous dominant genotypes possess two dominant alleles for a trait.
8. Homozygous recessive genotypes possess two recessive alleles for a trait.
9. After cross-pollination, all individuals of the F1 generation had one of each
type of allele.
10. Heterozygous genotypes possess one of each allele for a particular trait.
11. The allele not expressed in a heterozygote is a recessive allele.
C. Genotype Versus Phenotype
1. Two organisms with different allele combinations can have the same outward
appearance (e.g., TT and Tt pea plants are both tall; therefore, it is necessary to
distinguish between alleles present and the appearance of the organism).
2. Genotype refers to the alleles an individual receives at fertilization (dominant,
recessive).
3. Phenotype refers to the physical appearance of the individual (tall, short,
etc.).
D. Mendel’s Law of Independent Assortment
1. This two-trait (dihybrid) cross is between two parent organisms that are truebreeding for different forms of two traits; it produces offspring heterozygous
for both traits.
2. Mendel observed that the F1 individuals were dominant in both traits.
3.. He further noted four phenotypes among F2 offspring; he deduced second law
of heredity.
4. Mendel’s law of independent assortment states that members of one pair of
factors assort independently of members of another pair, and that all
combinations of factors occur in gametes.
5. The law of independent assortment only applies to alleles on different
chromosomes.
6. A phenotypic ratio of 9:3:3:1 is expected when heterozygotes for two traits are
crossed and simple dominance is present for both genes.
7. Independent assortment during meiosis explains these results.
E. Mendel’s Laws and Meiosis (Science Focus box)
1. Scientists now know that Mendel’s laws hold true because of meiosis.
2. For example, a parent cell will have two pairs of homologous chromosomes.
3. During metaphase I, all alignments of homologous chromosomes can occur,
following the lines of Mendel’s law of independent assortment.
4. During metaphase II, there is only one member of each homologous pair,
following the lines of Mendel’s law of segregation.
5. All possible gametes result since one daughter cell has both dominant alleles
(AB), one daughter cell has both recessive alleles (ab), and two daughter cells
have one dominant and one recessive allele (Ab and aB), following Mendel’s
two laws.
F. Mendel’s Law of Probability
1. A Punnett square is used for two-trait crosses.
2. Probability is the likely outcome a given event will occur from random
chance.
a. For example, with every coin flip there is a 50% chance of heads and 50%
chance of tails.
3. The product rule of probability states that the chance of two or more
independent events occurring together is the product of the probability of the
events occurring separately.
a. The chance of inheriting a specific allele from one parent and a specific
allele from another is ½ x ½ or 1/4.
b. Possible combinations for the alleles Ee of heterozygous parents are the
following:
EE = ½ x ½ = 1/4
eE = ½ x ½ = 1/4
Ee = ½ x ½ = 1/4
ee = ½
x½=¼
4. The sum law of probability calculates the probability of an event that occurs
in two or more independent ways; it is the sum of individual probabilities of
each way an event can occur; in the above example where unattached earlobes
are dominant (EE, Ee, and eE), the chance for unattached earlobes is 1/4 + 1/4
+ 1/4 = 3/4.
G. Testcrosses
1. A testcross is used to determine if an individual with the dominant phenotype
is homozygous dominant or heterozygous for a particular trait.
2. By Mendel performing a testcross, the law of segregation was supported.
3. A one-trait testcross is used between an individual with dominant phenotype and
an individual with a recessive phenotype to see if the individual with
dominant phenotype is homozygous or heterozygous.
4. A two-trait testcross tests if individuals showing two dominant characteristics are
homozygous for both or for one trait only, or heterozygous for both.
a. If an organism heterozygous for two traits is crossed with another recessive
for both traits, the expected phenotypic ratio is 1:1:1:1.
b. In dihybrid genetics problems, the individual has four alleles, two for each
trait.
H. Mendel’s Laws and Human Genetic Disorders
1. Genetic disorders are medical conditions caused by alleles inherited from parents.
2. An autosome is any chromosome other than a sex (X or Y) chromosome.
3. In a pedigree chart, males are designated by squares, females by circles; shaded
circles and squares are affected individuals; line between square and circle
represents a union; vertical line leads to offspring.
4. A carrier is a heterozygous individual with no apparent abnormality but able to
pass on an allele for a recessively-inherited genetic disorder.
5. Autosomal dominant and autosomal recessive alleles have different patterns of
inheritance.
a. Characteristics of autosomal dominant disorders
1)
Affected children usually have an affected parent.
2)
Heterozygotes are affected.: two affected parents can
produce unaffected child; two unaffected parents will not have affected
children.
b. Characteristics of autosomal recessive disorders
1)
Most affected children have normal parents since
heterozygotes have a normal phenotype.
2)
Two affected parents always produce an affected child.
3)
Close relatives who reproduce together are more likely to
have affected children.
I. Autosomal Recessive Disorders
1. Methemoglobinemia
a. Relatively harmless disorder resulting from an accumulation of
methemoglobin in the blood.
b. Cause and genetic link still remain a mystery.
c. Symptoms include bluish-purple skin due to inability to clear abmornal
blue protein from blood.
d. People with methemoglobinemia lack the enzyme diaphorase, which is
coded for by a gene on chromosome 22.
2. Cystic Fibrosis
a. This is the most common lethal genetic disease in Caucasians in the U.S.
b. About 1 in 20 Caucasians is a carrier, and about 1 in 3,000 newborns has
this disorder.
c. An increased production of a viscous form of mucus in the lungs and
pancreatic ducts is seen.
1) The resultant accumulation of mucus in the respiratory tract interferes
with gas exchange.
2) Digestive enzymes must be mixed with food to supplant the pancreatic
juices.
d. New treatments have raised the average life expectancy to up to 35 years.
e. Chloride ions (Cl–) fail to pass plasma membrane proteins.
f. Since water normally follows Cl–, lack of water in the lungs causes thick
mucus.
g. The cause is a gene on chromosome 7; attempts to insert the gene into
nasal epithelium has had little success.
h. Genetic testing for adult carriers and fetuses is possible.
3. Niemann-Pick Disease
a. Infant symptoms include jaundice, difficulty feeding, enlarged abdomen,
and pronounced mental retardation.
b. Type A and B forms of Niemann-Pick disease are caused by defective
versions of the same gene located on chromosome 11.
c. This disease is marked by the inability to break down lipids. Lipid droplets
accumulate in liver, lymph nodes, and spleen, and in severe cases, the
brain.
J. Autosomal Dominant Disorders
1. Osteogenesis Imperfecta
a. This is an autosomal dominant disorder that affects one in 5,000 newborns
and is distributed equally around the world.
b. Affected individuals have weakened, brittle bones. Additional symptoms
include unusual blue tine in the sclerea of the eye, reduced skin elasticity,
weakened teeth, and sometimes heart valve abnormalities.
c. The disease may be treated by long-term medicine.
2. Hereditary Spherocytosis
a. This genetic blood disorder results from a defective copy of a gene found
on chromosome 8.
b. Symptoms include: spherical shape of red blood cells, and enlarged
spleen.
c. Hereditary spherocytosis affects 1 in 5,000 people and is one of the most
common hereditary blood disorders.
H. Testing for Genetic Disorders (Science Focus box)
1. Two genetic disorders resulting from faulty genes are Huntington disease and
cystic fibrosis.
2 Researchers are tests that can detect particular DNA base sequencing that may be
able to identify individuals who may either have a genetic disease or if they are
carriers to a particular genetic disease.
a.
A carrier is a person who does not exhibit traits of the disease, but who has
the potential of passing the recessive allele of a genetic disorder.
3. In order to develop a test for a particular genetic disorder, scientists must first
obtain family pedigrees.
a. Family pedigrees trace particular genes through many family generations.
b. In the example of Huntington disease, the family pedigree illustrated that the
offspring of an affected individual has a 50% of having the disease.
c. When blood testing can be conducted, DNA base sequencing is determined
and compared to see if there are similarities in base sequencing with people
who have the disease.
d. However, this gene is only linked to the disease and not the disease itself.
e. More than one allele can occure on the same chromosome, meaning the alleles
are linked.
f. Linked alleles are found together on the same gamete. However, even though
they are considered to be linked, crossing over and unlinking can occur.
4. Assocation studies are another method to discover potential base sequencing to
identify if an individual has a genetic disorder.
a. DNA of the general population is tested to identify similar base sequences.
b. The exploration of the human genome project has made it possible to identify
genes that may be linked to particular genetic disorders.
5. Base sequencing identification can be used for prenatal testing and carrier testing.
Extending the Range of Mendelian Genetics
A. Multiple Allelic Traits
1. This occurs when a gene has many allelic forms or alternative expressions.
2. ABO Blood Types
a. The ABO system of human blood types is a multiple allele system.
b. Two dominant alleles (IA and IB) code for presence of A and B
glycoproteins on red blood cells.
c. This also includes a recessive allele (iO) coding for no A or B
glycoproteins on red blood cells.
d. As a result, there are four possible phenotypes (blood types): A, B, AB,
and O
e. This is a case of codominance, where both alleles are fully expressed.
3. The Rh factor is inherited independently from the ABO system; the Rh+ allele
is dominant.
B. Incomplete Dominance and Incomplete Penetrance
1. Incomplete dominance: offspring show traits intermediate between two
parental phenotypes.
a. True-breeding red and white-flowered four-o’clocks produce pinkflowered offspring.
b. Incomplete dominance has a biochemical basis; the level of gene-directed
protein production may be between that of the two homozygotes.
c. One allele of a heterozygous pair only partially dominates expression of
its partner.
d. This does not support a blending theory; parental phenotypes reappear in
F2 generation.
2. Human Examples of Incomplete Dominance
a. Curly versus Straight Hair
1) A curly-haired Caucasian and a straight-haired Caucasian will have
wavy-haired offspring.
2) Two wavy-haired parents will produce a 1:2:1 ratio of curly-wavystraight hair children.
b. Cystic fibrosis is considered an example of incomplete dominance.
3. Incomplete penetrance: offspring does not always show dominant allele.
a. Polydactyly is an example of incomplete penetrance.
C. Pleioptropic Effects
1. Pleiotropy describesva gene that affects more than one characteristic of an
individual.
2. Examples include Marfan syndrome, porphyria, and sickle-cell anemia.
3. Marfan syndrome
a. The gene is on chromosome 15
b. symptoms include disproportionately long arms, legs, hands, and feet;
weakened aorta; poor eyesight.
4. Porphyria
a. This disease is caused by a chemical insufficiency in the production of
hemoglobin.
b. Symptoms include photosensitivity, strong abdominal pain, port-winecolored urine, paralysis in arms and legs.
5. Sickle-cell anemia
a. This disease is the most common inherited disorder in blacks, affecting
about 1 in 500 African Americans.
b. The gene is on chromosome 11.
c. In affected individuals, the red blood cells are shaped like sickles—an
abnormal hemoglobin molecule, Hbs, causes the defect.
1) Normal hemoglobin, HbA, differs from Hbs by one amino acid in the
protein globin.
d. Sickling of the red blood cells occurs when the oxygen content of the
person’s blood is low, thereby slowing down blood flow and clogging
small vessels.
e. Signs and symptoms include anemia, weakness, fever, pain, rheumatism,
low resistance to disease, kidney and heart failure.
f. Treatment includes pain management, blood transfusions, and bone
marrow transplants.
g. The disease can be diagnosed prenatally.
h. Individuals with the sickle cell trait (carriers), who normally do not have
any sickle-shaped cells unless they experience dehydration or mild oxygen
deprivation, are resistant to the disease malaria.
D. Polygenic Inheritance
1. Polygenic inheritance occurs when one trait is governed by two or more sets
of alleles.
2. Dominant alleles have a quantitative effect on the phenotype: each adds to the
effect.
3. The more genes involved, the more continuous is the variation in phenotypes,
resulting in a bell-shaped curve.
4. Multifactorial traits are controlled by polygenes subject to environmental
influences.
a. The coats of Siamese cats and Himalayan rabbits have darker tipped ears,
nose, paws, etc. due to the enzyme encoded by an allele which is only active
at the extremities at low temperatures.
5. Human Examples of Polygenic Inheritance
a. A hybrid cross for skin color provides a range of intermediates.
b. Parents with intermediate skin color can produce children with the full
range of skin colors.
c. Other examples include cleft lip, clubfoot, congenital dislocations of the
hip, hypertension, diabetes, schizophrenia, allergies and cancers.
d. Behavioral traits including suicide, phobias, alcoholism, and
homosexuality may be associated with particular genes but are not likely
completely predetermined.
e. All behavioral traits are partly heritable, and genes work together and are
susceptible to environmental influences.
E. X-Linked Inheritance
1. Sex chromosomes in the human female are XX; those of the male are XY.
2. Males produce X-containing and Y-containing gametes; therefore males
determine the sex of offspring.
3. Besides genes that determine sex, sex chromosomes carry many genes for traits
unrelated to sex.
4. An X-linked gene is any gene located on X chromosome; used to describe genes
on X chromosome that are missing on the Y chromosome.
5. Work with fruit flies (Drosophila) by Thomas Hunt Morgan (early 1900s)
confirmed genes were on chromosomes.
a. Fruit flies are easily and inexpensively raised in common laboratory
glassware.
b. Females only mate once and lay hundreds of eggs.
c. The fruit fly generation time is short, allowing rapid experiments.
6. Fruit flies have an XY sex chromosome system similar to the human system;
experiments can be correlated to the human situation.
a. Newly discovered mutant male fruit flies had white eyes.
b. Cross of the hybrids from the white-eyed male crossed with a dominant
red-eyed female yielded the expected 3:1 red-to-white ratio; however, all
of the white-eyed flies were males.
c. An allele for eye color on the X but not on the Y chromosome supports the
results of this cross.
d. Behavior of this allele corresponds to the behavior of the chromosome;
this confirmed the chromosomal theory of inheritance.
7. Solving X-Linked Genetic Problems
a. X-linked alleles are designated as superscripts to the X chromosome.
b. Heterozygous females are carriers; they do not show the trait but can
transmit it.
c. Males are never carriers but express the one allele on the X chromosome;
the allele could be dominant or recessive.
d. One form of color-blindness is X-linked recessive.
F. Human X-Linked Disorders
1. More males have X-linked traits because recessive alleles on the X
chromosome in males are expressed in males.
2. Color Blindness
a. Color blindness can be an X-linked recessive disorder involving mutations
of genes coding for green or red sensitive cone cells, resulting in the
inability to perceive green or red, respectively; the pigment for bluesensitive protein is autosomal.
b. About 8% of Caucasian men have red-green color blindness.
3. Menkes Syndrome (kinky hair syndrome)
a. Caused by a defective allele on the X chromosome.
b. Symptoms include: poor muscle tone, seizures, low body temperature,
skeletal abnormalities, and brittle, steely hair.
c. Treatments include copper injections, but prognosis is poor and many
individuals die within the first few years of life.
4. Muscular Dystrophy
a. Duchenne muscular dystrophy is the most common form and is
characterized by wasting away of muscles, eventually leading to death; it
affects one out of every 3,600 male births.
b. This X-linked recessive disease involves a mutant gene that fails to
produce the protein dystrophin.
c. Signs and symptoms (e.g., waddling gait, toe walking, frequent falls,
difficulty in rising) soon appear.
d. Muscles weaken until the individual is confined to a wheelchair; death
usually occurs by age 20.
e. Affected males are rarely fathers; the gene passes from carrier mother to
carrier daughter.
f. Lack of dystrophin protein causes calcium ions to leak into muscle cells;
this promotes action of an enzyme that dissolves muscle fibers.
g. As the body attempts to repair tissue, fibrous tissue forms and cuts off
blood supply to the affected muscles.
h. A test now detects carriers of Duchenne muscular dystrophy; treatments
are being attempted.
7. Adrenoleukodystrophy (ADL)
a. This disease is an X-linked recessive disorder.
b. Fatty acids are not broken down and severe nervous system damage
occurs.
c. Symptoms include failure to properly develop after the age of five, loss of
adrenal gland function, exhibit poor coordination, show progressive loss
of hearing, speech, vision.
6. Hemophilia
a. About one in 10,000 males is a hemophiliac with impaired ability of blood
to clot.
b. The two common types: Hemophilia A, due to the absence of clotting
factor IX; Hemophilia B, due to the absence of clotting factor VIII.
c. Hemophiliacs bleed externally after an injury and also suffer internal
bleeding around joints.
d. Hemorrhages stop with transfusions of blood (or plasma) or concentrates
of clotting protein.
e. Factor VIII is now available as a genetically-engineered product.
f. Of Queen Victoria’s 26 offspring, five grandsons had hemophilia and four
granddaughters were carriers.
The Genetic Material
1. Early researchers knew that the genetic material must be:
a. able to store information used to control both the development and the
metabolic activities of cells;
b. stable so it can be replicated accurately during cell division and be transmitted
for generations; and,
c. able to undergo mutations providing the genetic variability required for
evolution.
A. Transformation of Bacteria
1. Bacteriologist Frederick Griffith (1931) experimented with Streptococcus
pneumoniae (a pneumococcus that causes pneumonia in mammals).
2. Mice were injected with two strains of pneumococcus: an encapsulated (S)
strain and a non-encapsulated (R) strain.
a. The S strain is virulent (the mice died); it has a mucous capsule and forms
“shiny” colonies.
b. The R strain is not virulent (the mice lived); it has no capsule and forms
“dull” colonies.
3. In an effort to determine if the capsule alone was responsible for the virulence
of the S strain, he injected mice with heat-killed S strain bacteria; the mice
lived.
4. Finally, he injected mice with a mixture of heat-killed S strain and live R
strain bacteria.
a. The mice died; living S strain pneumococcus were recovered from their
bodies.
b. Griffith concluded that some substance necessary for synthesis of the
capsule--and therefore for virulence--must pass from dead S strain bacteria
to living R strain bacteria so the R strain were transformed.
c. This change in phenotype of the R strain must be due to a change in the
bacterial genotype, suggesting that the transforming substance passed
from S strain to R strain.
B. DNA: The Transforming Substance
1. Oswald Avery et al. (1944) reported that the transforming substance was
DNA.
2.
In the early twentieth century, it was shown that nucleic acids contain four
types of nucleotides.
a. DNA was composed of repeating units, each of which always had just one
of each of four different nucleotides (a nitrogenous base, a phosphate, and
a pentose).
b. In this model, DNA could not vary between species and therefore could
not be the genetic material; therefore some other protein component was
expected to be the genetic material.
3. Purified DNA is capable of bringing about the transformation. Evidence:
a. DNA from S strain pneumococcus causes R strain bacteria to be
transformed.
b. Digestion of the transforming substance with enzyme that digests DNA
prevents transformation.
c. The molecular weight of the transforming substance is great enough for
some genetic variability.
d. Enzymes that degrade proteins cannot prevent transformation, nor can
enzymes that digest RNA.
4. Avery’s experimental results demonstrated DNA is genetic material and DNA
controls biosynthetic properties of a cell.
C. Transformation of Organisms Today
1. Transformation experiments today are important especially in biotechnology
labs.
2. Transformation of organisms are being used in commercial products.
3. In order to illustrate that transferring genes was possible from one organism to
another, scientists used a green fluorescent protein from jellyfish and
transferred it to other organisms. The result was that these organisms glowed
in the dark.
4. Mammalian genes have the ability to function in other species: bacteria,
invertebrates, plants.
D. The Structure of DNA
1. Erwin Chargaff (1940s) analyzed the base content of DNA.
2. It was known DNA contained four different nucleotides:
a. two with purine bases, adenine (A) and guanine (G); a purine is a type of
nitrogen-containing base having a double-ring structure.
b. two with pyrimidine bases, thymine (T) and cytosine (C); a pyrimidine is a
type of nitrogen-containing base having a single-ring structure.
3. Results: DNA does have the variability necessary for the genetic material, and,
4. For a species, DNA has the constancy required of genetic material.
5. This constancy is given in Chargaff’s rules:
a. The amount of A, T, G, and C in DNA varies from species to species.
b. In each species, the amount of A = T and the amount of G = C (A +G = T
+C).
6. The tetranucleotide hypothesis (proposing DNA was repeating units of one of
four bases) was disproved: each species has its own constant base composition.
7. The variability is in base sequences is staggering; a human chromosome contains
about 140 million base pairs.
8. Since any of the four possible nucleotides can be present at each nucleotide
6
position, the total number of possible nucleotide sequences is 4140 x 10 =
4140,000,000.
9. Rosalind Franklin produced X-ray diffraction photographs.
10. Franklin’s work provided evidence that DNA had the following features:
a. DNA is a helix.
b. Some portion of the helix is repeated.
11. American James Watson joined with Francis H. C. Crick in England to work on
the structure of DNA.
12. Watson and Crick received the Nobel Prize in 1962 for their model of DNA.
13. Using information generated by Chargaff and Franklin, Watson and Crick
constructed a model of DNA as a double helix with sugar-phosphate groups
on the outside, and paired bases on the inside.
14. Their model was consistent with both Chargaff’s rules and Franklin’s X-ray
diffraction studies.
15.
Complementary base pairing is the paired relationship between purines and
pyrimidines in DNA: A is hydrogen-bonded to T and G is hydrogen-bonded
to C.
Replication of DNA
1. DNA replication is the process of copying a DNA molecule. Replication is
semiconservative, with each strand of the original double helix (parental
molecule) serving as a template (mold or model) for a new strand in a daughter
molecule. This process consists of:
a. Unwinding: old strands of the parent DNA molecule are unwound as weak
hydrogen bonds between the paired bases are “unzipped” and broken by the
enzyme helicase.
b. Complementary base pairing: free nucleotides present in the nucleus bind
with complementary bases on unzipped portions of the two strands of DNA;
this process is catalyzed by DNA polymerase.
c. Joining: complementary nucleotides bond to each other to form new strands;
each daughter DNA molecule contains an old strand and a new strand; this
process is also catalyzed by DNA polymerase.
A. Aspects of DNA Replication (Science Focus box)
1. For complementary base pairing to occur, the DNA strands need to be
antiparallel, as discovered by Watson and Crick
2. One strand of DNA is 5’ at the top and the other strand is 3’ at the top of the
strand.
3. During replication the DNA polymerase can ony join to the free 3’ end of the
previous nucleotide.
4. DNA polymerase cannot start the sythesis of a DNA chain, so an RNA
polymerase lays out an RNA primer that is complementary to the replicated
strand.
5. Now the DNA polymerase can join the DNA nucleotides to the 3’ end of the new
strand.
6. The helicase enzyme unwinds the DNA and one strand (called the leading new
strand) can be copied in the direction of the replication fork.
7. The other strand of DNA is copied in the direction away from the fork, and
replication begins again.
a. This new lagging strand is discontinuous and each segment is called an
Okazaki fragment, after the scientist who discovered them.
8. Replication is only complete when RNA primers are removed.
9. During replication, DNA molecules gets smaller and smaller.
10. The end of eukaryotic DNA molecules have nucleotide sequences called
teleromers.
a. Teleromeres on’t code for proteins. They are repeats of short nucleotide
sequences (i.e. TTAGGG).
11. In normal mammalian cells divide approximately 50 times and then stop.
However in cancer cells the telomerase can be turned on and cancer cells then
divide without limit.
C. Prokaryotic Versus Eukaryotic Replication
1. Prokaryotic DNA Replication
a. Bacteria have a single loop of DNA that must replicate before the cell
divides.
b. Replication in prokaryotes may be bidirectional from one point of origin
or in only one direction.
c. Replication only proceeds in one direction, from 5' to 3'.
d. Replication begins at a special site on a bacterial chromosome, called the
origin of replication.
e. Bacterial cells can complete DNA replication in 40 minutes; eukaryotes
take hours.
2. Eukaryotic DNA Replication
a. Replication in eukaryotes starts at many points of origin and spreads with
many replication bubbles—places where the DNA strands are separating
and replication is occurring.
b. Replication forks are the V-shape ends of the replication bubbles; the
sites of DNA replication.
c. Eukaryotes replicate their DNA at a slower rate – 500 to 5,000 base pairs
per minute.
d. Eukaryotes take hours to complete DNA replication.
D. Accuracy of Replication
1. A mismatched nucleotide may occur once per 100,000 base pairs, causing a
pause in replication.
2. Proofreading is the removal of a mismatched nucleotide; DNA repair
enzymes perform this proofreading function and reduce the error rate to one
per billion base pairs.
The Genetic Code of Life
1. Sir Archibald Garrod (early 1900s) introduced the phrase inborn error of
metabolism.
a. Garrod proposed that inherited defects could be caused by the lack of a
particular enzyme.
b. Knowing that enzymes are proteins, Garrod suggested a link between
genes and proteins.
2. Linus Pauling and Harvey Itano (1949) compared hemoglobin in red blood
cells of persons with sickle-cell disease and normal individuals.
a. They discovered that the chemical properties of a protein chain of sicklecell hemoglobin differed from that of normal hemoglobin.
b. Pauling and Itano formulated the one gene–one polypeptide hypothesis:
each gene specifies one polypeptide of a protein, a molecule that may
contain one or more different polypeptides.
A. RNA Carries the Information
1. Like DNA, RNA is a polymer of nucleotides.
2. Unlike DNA, RNA is single-stranded, contains the sugar ribose, and the base
uracil instead of thymine (in addition to cytosine, guanine, and adenine).
3. There are three major classes of RNA.
a. Messenger RNA (mRNA) takes a message from DNA in the nucleus to
ribosomes in the cytoplasm.
b. Ribosomal RNA (rRNA) and proteins make up ribosomes where proteins
are synthesized.
c. Transfer RNA (tRNA) transfers a particular amino acid to a ribosome.
B. The Genetic Code
1. DNA undergoes transcription to mRNA, which is translated to a protein.
2. DNA is a template for RNA formation during transcription.
3. Transcription is the first step in gene expression; it is the process whereby a
DNA strand serves as a template for the formation of mRNA.
4. During translation, an mRNA transcript directs the sequence of amino acids in
a polypeptide.
5. The genetic code is a triplet code, comprised of three-base code words (e.g.,
AUG).
6. A codon consists of 3 nucleotide bases of DNA.
7. Four nucleotides based on 3-unit codons allows up to 64 different amino acids
to the specified.
8. Finding the Genetic Code
a. Marshall Nirenberg and J. Heinrich Matthei (1961) found that an enzyme
that could be used to construct synthetic RNA in a cell-free system; they
showed the codon UUU coded for phenylalanine.
b. By translating just three nucleotides at a time, they assigned an amino acid
to each of the RNA codons, and discovered important properties of the
genetic code.
c. The code is degenerate: there are 64 triplets to code for 20 naturally
occurring amino acids; this protects against potentially harmful mutations.
d. The genetic code is unambiguous; each triplet codon specifies one and
only one amino acid.
e. The code has start and stop signals: there are one start codon and three
stop codons.
9. The Code Is Universal
a. The few exceptions to universality of the genetic code suggests the code
dates back to the very first organisms and that all organisms are related.
b. Once the code was established, changes would be disruptive.
First Step: Transcription
A. Messenger RNA is Formed
1. A segment of the DNA helix unwinds and unzips.
2. Transcription begins when RNA polymerase attaches to a promoter on
DNA. A promoter is a region of DNA which defines the start of the gene, the
direction of transcription, and the strand to be transcribed.
3. As RNA polymerase (an enzyme that speeds formation of RNA from a DNA
template) moves along the template strand of the DNA, complementary RNA
nucleotides are paired with DNA nucleotides of the coding strand. The strand
of DNA not being transcribed is called the noncoding strand.
4. RNA polymerase adds nucleotides to the 3'-end of the polymer under
construction. Thus, RNA synthesis is in the 5’-to-3’ direction.
5. The RNA/DNA association is not as stable as the DNA double helix;
therefore, only the newest portion of the RNA molecule associated with RNA
polymerase is bound to DNA; the rest dangles off to the side.
6. Elongation of mRNA continues until RNA polymerase comes to a stop
sequence.
7. The stop sequence causes RNA polymerase to stop transcribing DNA and to
release the mRNA transcript.
8. Many RNA polymerase molecules work to produce mRNA from the same
DNA region at the same time.
9. Cells produce thousands of copies of the same mRNA molecule and many
copies of the same protein in a shorter period of time than if a single copy of
RNA were used to direct protein synthesis.
B. RNA Molecules Are Processed
1. Newly formed pre-mRNA transcript is processed before leaving the nucleus.
2. Pre-mRNA transcript is the immediate product of transcription; it contains
exons and introns.
3. The ends of the mRNA molecule are altered: a cap is put on the 5' end and a
poly-A tail is put on the 3' end.
a. The “cap” is a modified guanine (G) where a ribosome attaches to begin
translation.
b. The “poly-A tail” consists of a 150–200 adenine (A) nucleotide chain that
facilitates transport of mRNA out of the nucleus and inhibits enzymatic
degradation of mRNA.
4. Portions of the primary mRNA transcript, called introns, are removed.
a. An exon is a portion of the DNA code in the primary mRNA transcript
eventually expressed in the final polypeptide product.
b. An intron is a non-coding segment of DNA removed by spliceosomes
before the mRNA leaves the nucleus.
5. Ribozymes are RNAs with an enzymatic function restricted to removing
introns from themselves.
a. RNA could have served as both genetic material and as the first enzymes
in early life forms.
6. Spliceosomes are complexes that contains several kinds of
ribonucleoproteins.
a. Spliceosomes cut the primary mRNA transcript and then rejoin adjacent
exons.
7. Smaller nucleolar RNA (snoRNA) is present in the nucleolus, to assist in the
processing of rRNA and tRNA molecules.
C. Function of Introns
1. Introns give a cell the ability to decide which exons will go in a particular
mRNA
2. mRNA do not have all of the possible exons available from a DNA sequence.
In one mRNA what is an
exon could be an intron in another mRNA.
This process is termed alternative mRNA splicing.
3. Some introns give rise to microRNAs (miRNA). miRNA regulate mRNA
translation by bonding with mRNA through complementary base pairing and
preventing translation from occurring.
4. Exon shuffling occurs when introns encourage crossing over during meiosis.
Second Step: Translation
1. Translation takes place in the cytoplasm of eukaryotic cells.
2. Translation is the second step by which gene expression leads to protein
synthesis.
3. One language (nucleic acids) is translated into another language (protein).
A. The Role of Transfer RNA
1. Transfer RNA (tRNA) molecules transfer amino acids to the ribosomes.
2. The tRNA is a single-stranded ribonucleic acid that doubles back on itself to
create regions where complementary bases are hydrogen-bonded to one
another.
3. The amino acid binds to the 3’ end; the opposite end of the molecule contains
an anticodon that binds to the mRNA codon in a complementary fashion.
4. There is at least one tRNA molecule for each of the 20 amino acids found in
proteins.
5. There are fewer tRNAs than codons because some tRNAs pair with more than
one codon; if an anticodon contains a U in the third position, it will pair with
either an A or G–this is called the wobble hypothesis.
6. The tRNA synthetases are amino acid-activating enzymes that recognize
which amino acid should join which tRNA molecule, and covalently joins
them. This requires ATP.
7. An amino acid–tRNA complex forms, which then travels to a ribosome to
“transfer” its amino acid during protein synthesis.
B. The Role of Ribosomal RNA
1. Ribosomal RNA (rRNA) is produced from a DNA template in the nucleolus
of the nucleus.
2. The rRNA is packaged with a variety of proteins into ribosomal subunits, one
larger than the other.
3. Subunits move separately through nuclear envelope pores into the cytoplasm
where they combine when translation begins.
4. Ribosomes can float free in cytosol or attach to endoplasmic reticulum.
5. Prokaryotic cells contain about 10,000 ribosomes; eukaryotic cells contain
many times more.
6. Ribosomes have a binding site for mRNA and binding sites for two tRNA
molecules.
7. They facilitate complementary base pairing between tRNA anticodons and
mRNA codons; rRNA acts as an enzyme (ribozyme) that joins amino acids
together by means of a peptide bond.
8. A ribosome moves down the mRNA molecule, new tRNAs arrive, the amino
acids join, and a polypeptide forms.
9. Translation terminates once the polypeptide is formed; the ribosome then
dissociates into its two subunits.
10.
Polyribosomes are clusters of several ribosomes synthesizing the same
protein.
11.
To get from a polypeptide to a function protein requires correct bending
and twisting; chaperone molecules assure that the final protein develops the
correct shape.
12.
Some proteins contain more than one polypeptide; they must be joined to
achieve the final three-dimensional shape.
C. Translation Requires Three Steps
1. During translation, mRNA codons base-pair with tRNA anticodons carrying
specific amino acids.
2. Codon order determines the order of tRNA molecules and the sequence of
amino acids in polypeptides.
3. Protein synthesis involves initiation, elongation, and termination.
4. Enzymes are required for all three steps; energy (ATP) is needed for the first
two steps.
5. Chain Initiation
a. A small ribosomal subunit attaches to mRNA in the vicinity of the start
codon (AUG).
b. First or initiator tRNA pairs with this codon; then the large ribosomal
subunit joins to the small subunit.
c. Each ribosome contains three binding sites: the P (for peptide) site, the A
(for amino acid) site, and the E (for exit) site.
d. The initiator tRNA binds to the P site although it carries one amino acid,
methionine.
e. The A site is for the next tRNA carrying the next amino acid.
f. The E site is to discharge tRNAs from the ribosome.
g. Initiation factor proteins are required to bring together the necessary
translation components: the small ribosomal subunit, mRNA, initiator
tRNA, and the large ribosomal subunit.
6. Chain Elongation
a. The tRNA with attached polypeptide is at the P site; a tRNA-amino acid
complex arrives at the A site.
b. Proteins called elongation factors facilitate complementary base pairing
between the tRNA anticodon and the mRNA codon.
c. The polypeptide is transferred and attached by a peptide bond to the newly
arrived amino acid in the A site.
d. This reaction is catalyzed by a ribozyme, which is part of the larger
subunit.
e. The tRNA molecule in the P site is now empty.
f. Translocation occurs with mRNA, along with peptide-bearing tRNA,
moving to the P site and the spent tRNA moves from the P site to the E
site and exits the ribosome.
g. As the ribosome moves forward three nucleotides, there is a new codon
now located at the empty A site.
h. The complete cycle is rapidly repeated, about 15 times per second in
Escherichia coli.
i. The ribosomes will reach a stop codon, termation will occur, and the
peptide will be released.
7. Chain Termination
a. Termination of polypeptide synthesis occurs at a stop codon that does not
code for amino acid.
b. The polypeptide is enzymatically cleaved from the last tRNA by a release
factor.
c. The tRNA and polypeptide leave the ribosome, which dissociates into its
two subunits.
d. Proteomics is a new field ob biology that aims to understand protein
structures, and the functions of metabolic pathways.
D. Protein Synthesis and the Eukaryotic Cell
1. The first few amino acids of a polypeptide act as a signal peptide that
indicates where the polypeptide belongs in the cell or if it is to be secreted by
the cell.
2. After the polypeptide enters the lumen of the ER, it is folded and further
processed by addition of sugars, phosphates, or lipids.
3. Transport vesicles carry the proteins between organelles and to the plasma
membrane.
Structure of Eukaryotic Chromosomes and Genes
1. The DNA is wound around a core of eight protein molecules (“beads on a
string”); the proteins are called histones and each “bead” is called a
nucleosome.
a. Histones play a structural role in chromosome structure and package the
large DNA into the small nucleus.
b. The nucleosomes also contribute to the shortening of DNA by folding it
into a “zigzag” structure.
2. During interphase, some chromatin is highly compact, darkly stained, and
genetically inactive heterochromatin.
3. The rest is diffuse, lightly colored euchromatin thought to be genetically
active.
a. Euchromatin activity is related to the extent nucleosomes are coiled and
condensed.
b. A nucleosome is a bead-like unit made of a segment of DNA wound
around a complex of histone proteins.
Prokaryotic Regulation
1. Bacteria do not require the same enzymes all the time; they produce just those
needed at the moment.
2. Francois Jacob and Jacques Monod (1961) proposed the operon model to
explain regulation of gene expression in prokaryotes.
a. In the operon model, several genes code for an enzyme in the same
metabolic pathway and are located in a sequence on a chromosome;
expression of structural genes is controlled by the same regulatory genes.
b. An operon is a group of structural and regulatory genes that function as a
single unit; it includes the following:
1) A regulator gene, located outside the operon, codes for a repressor
protein molecule that controls whether the operon is active or not.
2) A promotor is the sequence of DNA where RNA polymerase attaches
when a gene is to be transcribed.
3) An operator is a short sequence of DNA where an active repressor
binds, preventing RNA polymerase from attaching to the promotor-transcription therefore does not occur.
4) Structural genes are one to several genes coding for enzymes of a
metabolic pathway that are transcribed as a unit.
A. The trp Operon
1. Some operons in E. coli usually exist in the “on” rather than the “off”
condition.
2. E. coli produces five enzymes as part of the anabolic pathway to synthesize
the amino acid tryptophan.
3. If tryptophan is already present in medium, these enzymes are not needed and
the operon is turned off .
a. The regulator codes for a repressor that usually is unable to attach to the
operator.
b. The repressor has a binding site for tryptophan (if tryptophan is present, it
binds to the repressor).
c. This changes the shape of the repressor that now binds to the operator.
4. The entire unit is called a repressible operon; tryptophan is the corepressor.
5. Repressible operons are involved in anabolic pathways that synthesize
substances needed by cells.
B. The lac Operon
1. If E. coli is denied glucose and given lactose instead, it makes three enzymes
to metabolize the lactose.
2. These three enzymes are encoded by three genes.
a. One gene codes for beta-galactosidase that breaks lactose to glucose and
galactose.
b. A second gene codes for a permease that facilitates entry of lactose into
the cell.
c. A third gene codes for enzyme transacetylase, which is an accessory in
lactose metabolism.
3. The three genes are adjacent on a chromosome and under control of one
promoter and one operator.
4. The regulator gene codes for a lac operon repressor protein that binds to the
operator and prevents transcription of the three genes.
5. When E. coli is switched to medium containing an allolactose, this lactose
binds to the repressor and the repressor undergoes a change in shape that
prevents it from binding to the operator.
6. Because the repressor is unable to bind to the operator, the promoter is able to
bind to RNA polymerase, which carries out transcription and produces the
three enzymes.
7. An inducer is any substance (lactose in the case of the lac operon) that can
bind to a particular repressor protein, preventing the repressor from binding to
a particular operator; consequently, RNA polymerase can bind to the promoter
and transcribe the structural genes.
C. Further Control of the lac Operon
1. Since E. coli prefers to break down glucose, how does E. coli know how to
turn on when glucose is absent?
2. When glucose is absent, cyclic AMP (cAMP) accumulates; cAMP has only
one phosphate group and attaches to ribose at two locations.
a. CAP is a catabolite activator protein (CAP) in the cytoplasm.
b. When cAMP binds to CAP, the complex attaches to a CAP binding site
next to the lac promoter.
c. When CAP binds to DNA, DNA bends, exposing the promoter to RNA
polymerase.
d. Only then does RNA polymerase bind to the promoter; this allows
expression of the lac operon structural genes.
3. When glucose is present, there is little cAMP in the cell.
a. CAP is inactive and the lactose operon does not function maximally.
b. CAP affects other operons when glucose is absent.
c. This encourages metabolism of lactose and provides a backup system for
when glucose is absent.
4. Active repressors shut down the activity of an operon—this is negative
control..
5. CAP is an example of positive control; when the molecule is active, it
promotes the activity of the operon.
6. Use of both positive and negative controls allows the cell to fine-tune control
of its metabolism.
7. If both glucose and lactose are present, the cell preferentially metabolizes
glucose.
Eukaryotic Regulation
1. Different cells in the human body turn on different genes that code for
different protein products.
2. Eukaryotes have four levels of regulatory mechanisms to control gene
expression; two in the nucleus and two in the cytoplasm.
3. There are several levels of control that can modify the amount of gene
product.
a. Chromatin structure: if genes are not accessible to RNA polymerase,
they cannot be transcribed.
1) Chromatin structure is part of epigenetic inheritance, the transmission
of genetic information outside the coding sequences of a gene.
b. Transcriptional control in the nucleus determines which structural genes
are transcribed and the rate of transcription; it includes transcription
factors initiating transcription and transposons (DNA sequences that move
between chromosomes and shut down genes).
c. Posttranscriptional control occurs in the nucleus after DNA is
transcribed and preliminary mRNA forms.
1) This may involve differential processing of mRNA before it leaves the
nucleus.
2) The speed that mature mRNA leaves nucleus affects the ultimate
amount of gene product.
d. Translational control occurs in cytoplasm after mRNA leaves the
nucleus but before there is a protein product.
1) The life expectancy of mRNA molecules can vary, as well as their
ability to bind ribosomes.
2) Some mRNAs may need additional changes before they are translated
at all.
e. Posttranslational control occurs in the cytoplasm after protein synthesis.
1) Polypeptide products may undergo additional changes before they are
biologically functional.
2) A functional enzyme is subject to feedback control; binding of an end
product can change the shape of an enzyme so it no longer carries out
its reaction.
A. Chromatin Structure
1. Eukaryotic DNA is in the form of chromatin, a stringy material associated
with proteins.
2. During interphase, some chromatin is highly compact, darkly stained, and
genetically inactive heterochromatin.
3. Barr bodies are an example of heterochromatin.
a. Since human males have only one X chromosome, it might be supposed
that they produce half the gene product of a female with two X
chromosomes.
b. However, females have in each nucleated cell a darkly staining Barr
body, a condensed, inactive X chromosome.
c. Which X chromosome is condensed in each cell is determined by chance.
d. Thus, the body of heterozygous females is “mosaic”; half her cells express
alleles on one X chromosome and half of her cells express the alleles on
the other X chromosome.
e. Female gonads do not show Barr bodies; both X chromosomes are needed
in development.
f. Only one active X chromosome in the female zygote means that lower Xcoded gene products are normal.
g. Other examples of this mosaic effect include: ocular albinism, Duchenne
muscular dystrophy, and female calico cat coat color.
4.
Euchromatin activity is related to the extent nucleosomes are coiled and
condensed.
c. A nucleosome is a bead-like unit made of a segment of DNA wound
around a complex of histone proteins.
d. When DNA is transcribed, activators called remodeling proteins are able
to push aside the histone proteins so transcription can begin.
5.
Epigenetic inheritance is the term used to describe inheritance patterns
that do not depend on the genes themselves.
a. Histones proteins have different chemical modifications in
heterochromatin and euchromatin.
b. Methylation of DNA accounts for a phenomenon called genomic
imprinting: gene expression is dependent on whether the chromosome
carrying the gene is inherited from the mother or the father.
c. Epigenetic inheritance explains unusual inheritance patterns, and may
have implications in growth, aging and cancer.
B. Transcriptional Control
1. Transcriptional control is the most critical level of genetic control.
a. Transcription is controlled by DNA-binding proteins called transcription
factors.
b. Each cell contains many different types of transcription factors.
c. A group of transcription factors binds to a promoter adjacent to a gene; the
complex attracts and binds RNA polymerase, but transcription may still
not begin.
d. Transcription activators are often involved in controlling transcription in
eukaryotes.
1) Different combinations may regulate different genes.
2) Transcription activators bind to DNA regions called enhancers.
3) Enhancers can be quite a distance from the promoter, but a hairpin
loop in the DNA brings the activator attached to an enhancer into
contact with the promotor.
4) Mediator proteins act as a bridge between transcription factors and
transcription factors at the promotor.
e. Transcription factors are always present in the cell and most likely they
have to be activated in some way (e.g., regulatory pathways involving
kinases or phosphatases) before they bind to DNA.
C. Posttranscriptional Control
1. Posttranscriptional control includes mRNA processing and the speed at
which mRNA leaves the nucleus.
2. Messenger RNA molecules are processed before they leave the nucleus and
enter the cytoplasm.
3. Differential excision of introns and splicing of mRNA can vary the type of
mRNA that leaves nucleus.
a. The hypothalamus and thyroid glands produce calcitonin but the mRNA
that leaves the nucleus is not the same in both types of cells.
b. Evidence of different patterns of mRNA splicing is found in cells that
produce neurotransmitters, muscle regulatory proteins, and antibodies.
4. Speed of transport of mRNA from nucleus into cytoplasm affects the amount
of gene product realized per unit time following transcription; there is a
difference in the length of time it takes various mRNA molecules to pass
through nuclear pores.
D. Translational Control
1. Translational control begins when the processed mRNA molecule reaches
the cytoplasm and before there is a protein product.
a. The longer an active mRNA molecule remains in the cytoplasm, the more
product is produced.
b. Mature red blood cells eject their nucleus but synthesize hemoglobin for
several months; the mRNAs must persist during this time.
c. Ribonucleases are enzymes associated with ribosomes that degrade
mRNA.
d. Mature mRNA has non-coding segments at 3' cap and 5' poly-A tail ends;
differences in these segments influence how long the mRNA avoids being
degraded.
e. MicroRNAs are small, processed pieces of intron; after microRNAs are
degraded, they combine with protein and the complex binds to mRNAs,
destroying them.
E. Posttranslational Control
1. Posttranslational control begins once a protein has been synthesized and has
become active.
a. Some proteins are not active after synthesis; the polypeptide product has
to undergo additional changes before it is biologically functional.
b. Bovine proinsulin, for example, is inactive when first produced; a single
long polypeptide folds into a three-dimensional structure, a sequence of 30
amino acids is removed from the middle, and the two polypeptide chains
are bonded together by disulfide bonds resulting in an active protein.
c. Many proteins are short-lived in cells and degraded or destroyed so they
are no longer active.
d. Giant protein complexes called proteasomes carry out this task.
e. One example is the cyclins that control the cell cycle; they are only
temporarily present.
F. Alternative mRNA Splicing in Disease (Science Focus box)
1. Gorlin syndrome is an autosomal dominant disease that causes an aggressive
type of skin cancer.
a. It is linked to the tumor suppressor genet, patched, which is found on
chromosome 9.
b. It was previously thought that patched mutations occurred within the gene’s
exons, but current research demonstrated, these mutations occur within the
gene’s introns.
c. The mutations caused mRNA to be incorrectly spliced and then making the
proteins nonfunctional.
2. Spinal muscular dystrophy (SMD) is also caused by defective pre-mRNA
splicing.
a. Exon 7 of the SMN2 gene is omitted from the mature mRNA in SMA patients,
making the protein nonfunctional.
b. SMA results in loss of spinal cord motor neurons, paralysis and skeletal
muscle atrophy.
c. Scientists are working with a new technique, antisense oligonucleotide
technology, to reverse the effects of the disorder.
d. The results support that tarketing pre-mRNA splicing may be a tangeable
treatment for many of these disorders.
3. Scientist are using pre-mRNA splicing to design more powerful mecidine with
less side effects, as well as new therapy that may be more effective in battling
many disorders.
Regulation Through Gene Mutations
1. A gene mutation is a permanent change in the sequence of bases in DNA;
mutations range from having no effect to total inactivity.
A. Causes of Mutations
1. Spontaneous mutations result in abnormalities in normal biological processes.
a. Mutations due to replication errors are very rare.
b. DNA polymerase constantly proofreads new DNA against the old, and repairs
any irregularities, thereby reducing mistakes to one out of every one
billion nucleotide pairs replicated.
2. Induced mutations are changes in the DNA base sequencing from exposure to
toxic chemicals or radtation.
a. Carcinogens are mutagens that increase the chances of cancer.
1) The Ames test is commonly-used to determine if a chemical is
carcinogenic.
2) A histidine-requiring strain of bacteria is exposed to a chemical.
3) If the chemical is mutagenic, the bacteria regain the ability to grow
without histidine.
c. Tobacco smoke contains a number of known carcinogenic chemicals.
d. X rays and gamma rays are ionizing radiation that creates free radicals,
ionized atoms with unpaired electrons.
e. Ultraviolet (UV) radiation is easily absorbed by pyrimidines in DNA.
1) Where two thymine molecules are near each other, UV may bond them
together as thymine dimers.
2) Usually dimers are removed from damaged DNA by special enzymes
called DNA repair enzymes.
B. Effect of Mutations on Protein Activity
1. Point mutations change a single nucleotide and therefore change a single
specific codon.
a. The effect of the point mutation depends on the specific base change in the
codon.
b. Changes to codons that code for the same amino acid have no effect; e.g.,
UAU to UAC both code for tyrosine.
c. A change from UAC to UAG (a stop codon) results in a shorter protein,
and a change from UAC to CAC incorporates histidine instead of tyrosine.
d. Sickle cell disease results from a single base change in DNA where the
beta-chain of hemoglobin contains valine instead of glutamate at one
location and the resulting distorted hemoglobin causes red blood cells to
clog vessels and die off sooner.
2. Frameshift Mutations
a. The reading frame depends on the sequence of codons from the starting
point: THE CAT ATE THE RAT.
b. If, for example, C is deleted, the reading frame is shifted: THE ATA TET
HER AT.
c. Frameshift mutations occur when one or more nucleotides are inserted or
deleted from DNA.
d. The result of a frameshift mutation is a new sequence of codons and
nonfunctional proteins.
C. Nonfunctional Proteins
1. A single nonfunctioning protein can cause dramatic effects.
2. The human transposon Alu is responsible for hemophilia when it places a
premature stop codon in the gene for clotting factor IX.
3. PKU results when a person cannot convert phenylalanine to tyrosine;
phenylalanine builds up in the system, leading to mental retardation.
4. A faulty code for an enzyme in the same pathway results in an albino
individual.
5. Cystic fibrosis is due to inheriting a faulty code for a chloride transport
protein in the plasma membrane.
6. Androgen insensitivity is due to a faulty receptor for male sex hormones; body
cells cannot respond to testosterone and the individual develops as a female
(even though all of the body cells are XY).
D. Mutations Can Cause Cancer
1. The development of cancer involves a series of various types of mutations.
2. Tumor-suppressor genes normally act as brakes on cell division when it
begins to occur abnormally.
3. When proto-oncogenes mutate, they become oncogenes.
4. Tumor-suppressor genes and proto-oncogenes often code for transcription
factors or proteins that control transcription factors.
5. P53, a major tumor-suppressor gene, is more frequently mutated in human
cancers than any other known gene.
a. The p53 protein acts as a transcription factor to turn on the expression of
genes whose products are cell cycle inhibitors.
b. The p53 can also stimulate apoptosis (programmed cell death).
6. Other proto-oncogenes code for Ras proteins, which are needed for normal
cell growth and DNA synthesis.
DNA Cloning
1. Cloning is the production of identical copies of DNA through some asexual
means.
a. An underground stem or root sends up new shoots that are clones of the
parent plant.
b. Members of a bacterial colony on a petri dish are clones because they all
came from division of the same cell.
c. Human identical twins are clones; the original single embryo separate to
become two individuals.
2. Gene cloning is production of many identical copies of the same gene.
a. If the inserted gene is replicated and expressed, we can recover the cloned
gene or protein product.
b. Cloned genes have many research purposes: determining the base
sequence between normal and mutated genes, altering the phenotype,
obtaining the protein coded by a specific gene, etc.
c. Humans can be treated with gene therapy: alteration of the phenotype in a
beneficial way.
d. Otherwise transgenetic organisms are used to produce products desired
by humans.
e. To clone DNA, scientists use recombinant DNA (rDNA) and
polymerase chain reaction (PCR).
A. Recombinant DNA Technology
1. Recombinant DNA (rDNA) contains DNA from two or more different
sources.
2. To make rDNA, technician selects a vector.
3. A vector is a plasmid or a virus used to transfer foreign genetic material into a
cell.
4. A plasmid is a small accessory ring of DNA in the cytoplasm of some
bacteria.
5. Plasmids were discovered in research on reproduction of intestinal bacteria
Escherichia coli.
6. Introduction of foreign DNA into vector DNA to produce rDNA requires two
enzymes.
a. Restriction enzyme is a bacterial enzyme that stops viral reproduction by
cleaving viral DNA.
1) The restriction enzyme is used to cut DNA at specific points during
production of rDNA.
2) It is called a restriction enzyme because it restricts growth of viruses
but it acts as a molecular scissors to cleave any piece of DNA at a
specific site.
3) Restriction enzymes cleave vector (plasmid) and foreign (human)
DNA.
4) Cleaving DNA makes DNA fragments ending in short single-stranded
segments with “sticky ends.”
5) The “sticky ends” allow insertion of foreign DNA into vector DNA.
b.
DNA ligase seals the foreign gene into the vector DNA
1) Treated cells take up plasmids, and then bacteria and plasmids
reproduce.
2) Eventually, there are many copies of the plasmid and many copies of
the foreign gene.
3) When DNA splicing is complete, an rDNA (recombinant DNA)
molecule is formed.
7. If the human gene is to express itself in a bacterium, the gene must be
accompanied by the regulatory regions unique to bacteria and meet other
requirements.
a. The gene cannot contain introns because bacteria do not have introns.
b. An enzyme called reverse transcriptase can be used to make a DNA copy
of mRNA.
c. This DNA molecule is called complementary DNA (cDNA) and does not
contain introns.
d. A laboratory DNA synthesizer can produce small pieces of DNA without
introns.
B. The Polymerase Chain Reaction (PCR)
1. PCR can create millions of copies of a single gene or a specific piece of DNA
in a test tube.
2. PCR is very specific—the targeted DNA sequence can be less than one part in
a million of the total dxcDNA sample; therefore a single gene can be
amplified using PCR.
3. PCR uses the enzyme DNA polymerase to carry out multiple replications (a
chain reaction) of target DNA.
4. PCR automation is possible because heat-resistant DNA polymerase from
Thermus aquaticus, which grows in hot springs, is an enzyme that withstands
the temperature necessary to separate double-stranded DNA.
5.
Analyzing DNA
a. Mitochondria DNA sequences in modern living populations can decipher
the evolutionary history of human populations.
b. Short tandem repeat (STR) profiling is a technique used to analyze
DNA.
1) Advantages include not being limited to using restriction enzymes.
2) Since the chromosomal location of STRs are known, these are the only
locations that are subjected to PCR and gel electrophoresis.
3) The band patterns are different in each person because each individual
has their own number of STR repeats at different locations.
c. DNA fingerprinting is a technique of using DNA fragment lengths,
resulting from restriction enzyme cleavage and amplified by PCR, to
identify particular individuals.
d. DNA is treated with restriction enzymes to cut it into different sized
fragments.
e. During gel electrophoresis, fragments separate according to length,
resulting in a pattern of bands.
f. DNA fingerprinting can identify deceased individuals from skeletal
remains, perpetrators of crimes from blood or semen samples, and genetic
makeup of long-dead individuals or extinct organisms.
6. PCR amplification and DNA analysis is used to:
a. detect viral infections, genetic disorders, and cancer;
b. determine the nucleotide sequence of human genes, and, because it is
inherited,
c. associate samples with DNA of parents.
Biotechnology Products
1. Genetically engineered organisms can produce biotechnology products.
2. Organisms that have had a foreign gene inserted into them are transgenic.
A. Genetically Modified Bacteria
1. Bacteria are grown in large vats called bioreactors.
a. Foreign genes are inserted and the product is harvested.
b. Products on the market include insulin, hepatitis B vaccine, t-PA, and
human growth hormone.
2. Transgenic bacteria have been produced to protect and improve the health of
plants.
a. Frost-minus bacteria protect the vegetative parts of plants from frost
damage.
b. Root-colonizing bacteria receive genes from bacteria for insect toxin,
protecting the roots.
c. Bacteria that colonize corn roots can be endowed with genes for insect
toxin.
3. Transgenic bacteria can degrade substances.
a. Bacteria selected for ability to degrade oil can be improved by
bioengineering.
b. Bacteria can be bio-filters to prevent airborne chemical pollutants from
being vented into the air.
c. Bacteria can also remove sulfur from coal before it is burned and help
clean up toxic dumps.
d. Bacteria can also be given”suicide genes” that caused them to die after
they have done their job.
4. Transgenic bacteria can produce chemical products.
a. Genes coding for enzymes can be manipulated to catalyze synthesis of
valuable chemicals.
b. Phenylalanine used in artificial sweetener can be grown by engineered
bacteria.
5. Transgenic bacteria process minerals.
a. Many major mining companies already use bacteria to obtain various
metals.
b. Genetically engineered “bio-leaching” bacteria extract copper, uranium,
and gold from low-grade ore.
B. Genetically Modified Plants
1. Plant cells that have had the cell wall removed are called protoplasts.
2. Electric current makes tiny holes in the plasma membrane through which
genetic material enters.
3. The protoplasts then develop into mature plants.
4. Foreign genes now give cotton, corn, and potato strains the ability to produce
an insect toxin and soybeans are now resistant to a common herbicide.
5. Plants are being engineered to produce human proteins including hormones,
clotting factors, and antibodies in their seeds; antibodies made by corn, deliver
radioisotopes to tumor cells and a soybean engineered antibody can treat
genital herpes.
C. Genetically Modified Animals
1. Animal use requires methods to insert genes into eggs of animals.
a. It is possible to microinject foreign genes into eggs by hand.
b. Vortex mixing places eggs in an agitator with DNA and silicon-carbide
needles that make tiny holes through which the DNA can enter.
c. Using this technique, many types of animal eggs have been injected with
bovine growth hormone (bGH) to produce larger fishes, cows, pigs,
rabbits, and sheep.
2. Gene pharming is the use of transgenic farm animals to produce
pharmaceuticals; the product is obtainable from the milk of females.
a. Genes for therapeutic proteins are inserted into animal’s DNA; animal’s
milk produces proteins.
b. Drugs obtained through gene pharming are planned for the treatment of
cystic fibrosis, cancer, blood diseases, and other disorders.
D. Cloning Transgenic Animals
1. For many years, it was believed that adult vertebrate animals could not be
cloned; the cloning of Dolly in 1997 demonstrated this can be done.
2. Cloning of an adult vertebrate would require that all genes of an adult cell be
turned on again.
3. Cloning of mammals involves injecting a 2n nucleus adult cell into an
enucleated egg.
4. The cloned eggs begin development in vitro and are then returned to host
mothers until the clones are born.
E. Applications of Transgenic Animals
1. Scientists are using transgenic animals to illustrate that maleness is due to a
section of DNA called SRY (the sex determining region of the Y
chromosome).
2. Transgenic animals are also being used to investigate various treatments for
diseases by eliminating genes from these animals.
3. Organ transplants from transgenic animal donors to human recipients is called
xenltransplantation.
Gene Therapy
1. Gene therapy involves procedures to give patients healthy genes to make up
for a faulty gene.
2. Gene therapy also includes the use of genes to treat genetic disorders and
various human illnesses.
3. There are ex vivo (outside body) and in vivo (inside body) methods of gene
therapy.
A. Ex Vivo Gene Therapy
1. Children with severe combined immunodeficiency (SCID) underwent ex vivo
gene therapy.
a. Lacking the enzyme ADA involved in maturation of T and B cells, they
faced life-threatening infections.
b. Bone marrow stem cells are removed, infected with a retrovirus that
carries a normal gene for the enzyme ADA, and returned.
c. Use of bone marrow stem cells allows them to divide and produce more
cells with the same genes.
d. Patients who undergo this procedure show significant improvement.
2. Gene therapy include treatment of familial hypercholesterolemia where liver
cells lack a receptor for removing cholesterol from blood.
a. High levels of blood cholesterol make the patient subject to fatal heart
attacks when young.
b. A small portion of the liver is surgically removed and infected with
retrovirus with normal gene for receptor.
c. This has lowered cholesterol levels following the procedure.
B. In Vivo Gene Therapy
1. Cystic fibrosis patients lack a gene for trans-membrane chloride ion carriers;
patients die from respiratory tract infections.
a. Liposomes, microscopic vesicles that form when lipoproteins are in
solution, are coated with healthy cystic fibrosis genes and sprayed into a
patient’s nostrils.
b. Various methods of delivery are being tested for effectiveness.
2. A gene for vascular endothelial growth factor (VEGF) can be injected alone or
within a virus into the heart to stimulate branching of coronary blood vessels.
3. Another strategy is to make cancer cells more vulnerable, and normal cells
more resistant, to chemotherapy.
4. Injecting a retrovirus containing a normal p53 gene–that promotes apoptosis–
into tumors may stop the growth of tumors.
Genomics
1. Genetics in the 21st century concerns genomics: the study of genomes of humans
and other organisms.
A. Sequencing the Genome
1. The Human Genome Project has produced a record of all the base pairs in
all our chromosomes.
2. The task took 13 years to learn the sequence of the three billion base pairs
along the length of our chromosomes.
3. Most of the DNA regions that differed among individuals are called single
nucleotide polymorphisms (SNPs)
a. Many SNPs have no effect and others could contribute to enzymatic
differences affecting the phenotype.
4. Following functional genomics, structural genomics investigates base
sequences and the number of genes in humans.
B. Genome Comparisons
1. There is little difference between the sequence of our bases and other
organisms whose DNA sequences are known.
C.
D.
E.
F.
2. We share a large number of genes with simpler organisms (e.g., bacteria,
yeast, mice); perhaps our uniqueness is due to regulation of these genes.
3. Researchers found that certain genes on chromosome 22 differed in humans
and chimpanzees: those for speech development, hearing, and smell.
4. Many genes found were responsible for human diseases.
Eukaryotic Gene Structure
1. Eukaryotic chromosome structure is much more complex than prokaryotic
chromosome structure.
2. Generally speaking, more complex organisms have more complex genes with
more and larger introns.
3. Introns are currently regarded as gene expression regulators and determine
which genes are expressed and how they are spliced.
Intergenic Sequences
1. Intergenic sequences are DNA sequences that occur between genes.
2. Generally speaking, as the complexity of an organism increases, so does the
proportion of its non-protein-coding DNA sequences.
3. Most of the human chromosomes are comprised of intergeneic sequences,
genes represent 1.5% of human’s total DNA, and the remainder represents
“junk DNA,” which scientists believe serve many important functions.
4. Three types of intergenic sequences found in the human genome are:
repetitive elements, tranposons, and unknown sequences.
Repetative Elements
1. Repetative DNA elements occur when the same sequence of two or more
nucleotides are repeated many times along the length of one or more
chromosomes.
2. They make up nearly half of the human genome and scientists believe that
their true significance have yet to be discovered.
3. Repetative DNA elements occur as tandem repeats – meaning the repeated
sequences are next to each other on the chromosome.
4. One type of tandem repeat sequence, short tandem repeats (STRs), are a
standard method in forensic science for identifying one individual from
another and also determining family relationships.
5. Another type of repetitive DNA element is called an interspersed repeat –
meaning the repetitions may be placed intermittently along a single
chromosome, or across multiple chromosomes.
6. Interspersed repeat are common and thought to play a role in the evolution of
new genes.
Transposons
2. Transposons are specific DNA sequences that move with and between
chromosomes.
a. Their movement may increase or decrease the expression of neighboring
genes.
b. They are among the 40% of the human genome consisting of the same
short sequence of DNA continuously repeated.
c. They are noncoding sequences that play regulatory functions, and could
thus be cansidered part of epigenetic inheritance.
G. Unknown Sequences
1. Unknown DNA sequences were once dismissed as junk DNA since scientists
could not identify any function they have.
2. Recently scientists believe observed that many of the unknown sequences is
transcribed into RNA.
3. This RNA may carry out regulatory functions more easily than proteins at
times.
H. What Is a Gene?
1. A modern definition of a gene, suggested by Mark Gerstein and associates,
“…is a genomic sequence (either DNA or RNA) directly encoding functional
products, either RNA or protein.”
2. This definition recognizes that the genetic material involved does not have to
be DNA, and that coding does not only mean DNA sequencing, but the gene
product could be RNA or protein as well.
I. Functional and Comparative Genomics
1. The emphasis on genome structure is both on functional and comparative
genomics.
2. Functional genomics aims to understand the exact role of the genome in cells
or organisms.
a. To meet the aim of functional genomics, DNA microarrays can be used
(see Science Focus box).
b. DNA microarrays rapidly identify a person’s complete genotype; this is
called the genetic profile.
3. Comparative genomics aims to compare the human genome to the genome
of other organisms.
a. This type of genomics has revealed little difference between DNA
sequence of our bases and those of many other organisms.
J. Proteomics
1. Proteomics is the study of the structure, function, and interaction of cellular
proteins.
2. The information obtained from proteomic studies can be used in designing
better drugs, and to correlate drug treatment to the particular genome of the
individual.
K. Bioinformatics
1. Bioinformatics is the application of computer technologics to the study of
the genome.
2. Information obtained from computer analysis of the genome can show
relationships between genetic profiles and genetic disorders.
H. DNA Microarray Technology (Science Focus box)
1. DNA microarray technology is being used to identify genes associated with gene
tissues, and identify links between disease and chromosomal variation.
2. DNA microarray technology techniques begin with labeling the test mRNA with a
fluorescent dye and added to the chip.
3. The tagged mRNA can then be followed by identifying the genes that exhibit the
fluorescent dye.
I. Copy Number Variations (Science Focus box)
1. Scientists have recently discovered small duplications and deletions in
chromosomes, referred to as copy number variations (CNVs)
2. The change in genes may arise from fork stalling and template switching.
3. Some CNVs may be linked to diseases such as HIV, Lupus, and autism.
4. Evolutionarily speaking, it may be advantageous for species to have multiple
copies of genes because if one copy of an allele does not function properly,
the duplicate allele may be able to restore normal function.
5. In addition, having two normal alleles may free the extra gene copy to accumulate
mutations, which could ultimately lead to the formation of a new gene.