Personalities and Discoveries

Garrod's 1903 Book on the "Inborn Errors of Metabolism "outlined a theory
of the genetic nature of the diseases of PKU, albinism and alkaptouria. Today,
we know that defective enzymes in a metabolic pathway of amino acid
metabolism are responsible.

Muller discovered in the 1920s that radiation, X-rays ultra violet rays, can
cause lethal mutations in fruit flies, Drosophila melanogaster. He concluded
that most mutations (structural changes of DNA) of genes were harmful.
Today we think that most mutations of human DNA are neutral in effect.
Most human DNA surprisingly does not seem to make proteins through
transcription and translation.

Chargaff in 1950 gave Watson and Crick an important clue about the
structure of DNA when he showed that the important DNA building block
Adenine was approximately equal in amount to the Thymine, and that the
Guanine was approximately equal to the amount of the Cytosine.

Watson and Crick in 1953 interpreted Rosalind Franklin's pictures of
X-rays bounced off the atoms of DNA in order to determine it's structure.
They created the first model of DNA and proposed the Central DOGMA
of its action as a carrier of inherited messages (see replication and transcription
below).

In 1941, Beadle and Tatum constructed the "One Gene-One Enzyme"
theory, now called the "One Gene-One Polypeptide" theory of gene action.
Using X-rays, they mutated the DNA of a strain of the red bread mold,
Neurospora: it became deficient in one or more metabolic "vitamins."
They determined that certain inactive enzymes were responsible for the
mutant forms that required the missing nutrients to grow and thrive.
Remember this: genes make proteins (polypeptides).

DNA Replication Events - See p. 98, 7th ed.; p. 97, 8th ed.

1. DNA strands separate assisted by two enzymes, gyrase and helicase:
a replication 'bubble' is formed with two replication forks proceeding in opposite
directions. One strand serves as a Template or mold for the new replicate strand.

2. The enzymes DNA Polymerase I-III insert the appropriate complementary
DNA triphosphate nucleotides into the exposed DNA bases on the original strands.
DNA Polymerase also 'patrols' the DNA molecule, correcting errors in base pairing.

3. The enzyme DNA Ligase ties the discontinuous fragments together by making
the ester bonds mentioned in the handout. Because replication ends with one original
DNA strand and one new strand, it is termed semiconservative
(sorta like democrats running for high office).

DNA Structure
 

Name the three building blocks for a nucleotide.

Xeroderma Pigmentosum

When ultraviolet radiation in sunlight strikes DNA in skin cells, two
vertically-adjacent thymine bases can form an abnormal bond between their
methyl (CH₃-) groups, creating a thymine dimer. Repair chemicals called
photoreactivating enzymes are normally present to collect the damage.
A recessive gene mutation inactivates this enzyme in those who have the most
common form of Xeroderma pigmentosum. Any exposure to sunlight will
blister the skin and eventually lead to cancer, usually the generally fatal melanoma
in which pigmented melanocytes multiply rapidly and invade other organs.
A mutation inactivating the DNA Polymerase repair enzyme will cause a
similar disease.
 

Protein Synthesis Events or "How to make a Protein" - See p. 102, 7th ed.; p. 101, 8th ed.

1. Transcription: RNA polymerase enzyme attaches to a region called the
Promoter. The downstream Operator gene will then open the DNA strands as the
RNA polymerase proceeds downstream from the Promoter site. When RNA
polymerase gets to the Structural Gene, the formation of a single-stranded
messenger RNA will occur with RNA triphosphate nucleotides inserted one a
t a time into the free base codes of one (sense strand) of the two strands of the DNA.

The messenger RNA formed is different from DNA in that: (1) the RNA
nitrogenous base uracil substitutes for the Thymine of DNA, (2) that m-RNA
is single stranded, and (3) that the pentose sugar in Ribose has one more
Oxygen atom than the DNA sugar, deoxyribose.

The start codon or polypeptide chain initiator is AUG. The stop codons include UGA,
UAA and UAG. AUG codes for Methionine and the stop codons translate for no
amino acids. Therefore, the Start codon will be at the beginning of the m-RNA
strand and the stop code at the end. Transcription can be blocked if an antibiotic
such as rifampicin binds to RNA polymerase preventing it from forming m-RNA.

2. Translation: Transcribed m-RNA leaves the nucleus and attaches its CAP
end to the small subunit of the ribosome. Transfer RNA (mostly double stranded)
which has the anticodon UAC (complimentary to the start codon AUG brings in the
amino acid Methionine to the small subunit. Then the ribosome large subunit joins with
the small subunit. See page 104, Text. There are actually 61 types of t-RNA for all the
amino acids. As each codon appears on the ribosome, it's amino acid is attached to
the growing polypeptide chain. The antibiotic streptomycin binds to the small subunit
of the bacterial ribosome thereby stopping translation.

Exercise - Fill-in the appropriate three base codes (see page 88 of the text and the
Protein Synthesis exercise following):
 
 

Triplets of DNA	AAA	_____________	_____________	ATT
Codons of m-RNA	_____________	AUG	_____________	_____________
Anticodons of t-RNA	_____________	_____________	UCG	_____________
Amino acid translations	_____________	_____________	_____________	_____________

Sickle Cell Hemoglobin is produced when a GAA sixth position
codon (DNA triplet is CTT) translating for the amino acid glutamic
acid is changed to a GUA (DNA triplet is CAT) translating for valine.
Sickle hemoglobin with it's one abnormal amino acid, changes the shape
of the Red Blood Cell when it is deoxygenated. Normal RBCs are round,
sickle cells have the shape of the same name when deoxygenated.
Changes in DNA code and gene expression are called mutations.

DNA Base Pair Additions/Insertions and Deletions

If a DNA base pair is inserted or deleted into an existing functional DNA double
helix, a nonsense mutation will result if, for instance, a stop triplet code results
downstream from the insertion or deletion. Nonsense mutations are not translated
into functional polypeptides; i.e.,

AUG-CAU-UCC-UCA-ACU + 50 more codons (amino acids) ending with
UGA is changed to AUG-CAU-UCC-UAA (delete C), inserting a stop
(=UAA) code, and the chain stops at only 3 codons (amino acids). This is called
a nonsense mutation.

If a base pair is inserted or deleted and the result is a change in the reading frame
of the triplet/codon sequences that changes to the amino acid sequence
(primary and higher structure) expressed in the translated polypeptide,
a missense mutation results. The polypeptide resulting may have a reduced
function or different function altogether from the original unmutated polypeptide;
its secondary and tertiary structures may be changed. Inactivation of metabolic
enzymes by nonsense or missense mutations (originally) can cause inherited
diseases like PKU and albinism.

Cancer

Cancer is a disease characterized by three major changes in cell structure and activity.
Precancerous changes include a change in cell structure called (1) metaplasia
(one normal type to another normal type) or dysplasia (cells have "weird" structure),
and a lack of control over mitosis which results in abnormally high cell division rates and
(2) tumor formation or hyperplasia (tissue mass with too many cells). A tumor
becomes cancerous when the hyperplastic tissue invades other tissues. Cancerous
cells secrete a growth/migration factor, invade blood vessels and spread to other
tissues and on to other organs in a process called (3) metastasis. Mutations of DNA
are required for each stage (metaplasia, hyperplasia and metastasis).
 

Cancer occurs because of changes or mutations of DNA molecules/genes that regulate cell division.
Causes of mutations (mutagens, things that mutate DNA) include:

a. Radiation - both (atomic) particulate (i.e., alpha particles, electrons)
and short-wave electromagnetic (e.g., ultra-violet, x-rays, gamma rays).

b. Chemicals (Carcinogens, if epithelial tissues are effected) cause cancer.
Known carcinogens include nitrates and nitrites used as fertilizers and meat
preservatives; vinyl chloride used in the manufacture of plastic; asbestos fibers
used in brake pads and insulation; cyclamates formerly used as artificial
sweeteners, saccharin sweeteners, etc.

c. Viruses, One of the first viruses (at the time called filterable to distinguish
them from bacteria, which were also called viruses) discovered, the Rous Virus,
causes cancer in chickens. German measles virus causes birth defects.
Retroviruses, those that contain RNA and enzyme called reverse transcriptase
which allows the RNA to make a DNA complementary copy of itself, may
also insert viral DNA into nuclear DNA as a provirus, i.e., human genes!
Viral DNA "jumping'" into nuclear or lost DNA occurs frequently in bacteria
and occasionally in eukaryotic cells, the viral jumping genes are called transposons.
The new genes inserted are called oncogenes if they result in uncontrolled
reproduction and/or invasiveness. The oncogene may produce a protein
called a growth factor, normally produced by embryonic cells which are
rapidly multiplying.

d. Inherited Genes - genes that normally control mitosis can be mutated
and passed on to offspring. Examples include the p53 "colon cancer gene"
and the BRCA gene that causes breast cancer. Many of the cells that have
mutated genes are "steroid sensitive," that is steroid, such as estrogen for
the mutant BRCA cells increase their rate of cell division and therefore the
rate of the growth and spread of the cancer. In this case steroids are mitogens,
chemicals that increase the rate of cell division.

e. Mistakes in Replication - see above mitogen explanation. Nicotine,
you may remember from Biol 1611, is not a mutagen, but rather a mitogen.

Genetics - Chapter 29

Gregor Mendel

Mendel, a Roman Catholic Brother who elected to have a career in teaching
and research in Austria in the mid 1800s, used the approach which has
become the Hallmark of Modern Science, the quantification of research results.
He conducted crosses of various traits and kept large and meticulous records
of the results. Then, he did an elementary statistical analysis to develop
evidence for his conclusions about the Laws of inheritance. Secondly, Mendel
worked many years to arrive at pure breeding strains of plants for certain
traits like seed color, flower color and pod color.

i.e., green seeded plants (x) green seeded plants = all green seed plants,
generation after generation.

When pure breeding yellow seeded plants were crossed with pure
breeding green, all yellow plants resulted,

i.e., yellow parent (ovules) (x) green parent (pollen) = all yellow,
F1 (first generation).

But, Mendel assumed that the all yellow F1, had both yellow and
green genetic characters even though only one was showing.
The color appearing (phenotype) in the F1, yellow, Mendel
called dominant: the color hidden in the F1, green, Mendel named
recessive. See p. 1150-1151, 7th ed.; p. 1106, 8th ed.

Mendel then assumed that each parent of the F1 yellow contributed
dominant yellow (Y) and the recessive green (y). This explained the
occurrence of yellow and green-seeded plants in the second generation.

the F2, i.e., F1 yellow (x) F1 yellow = 6000 yellow seeds and 2000
green seeds (a 3/1 ratio of yellow/green)

Ratios are made by dividing the smallest number in a category into
the larger numbered categories.

Therefore, Mendel's results using genotypes (pairs of letter, representing
the genetic contributions of the parents of each genotype) were:

pure breeding yellow parent, YY (parent) (x) yy (parent),
pure breeding green seeded parent = Yy F1.

The parent genes are the same in the pair, if pure breeding.
When the genes in pairs are exactly the same like YY or yy, they
are called homozygous genotypes. Yy genotype pair has different
genes, therefore they are called a heterozygous pair of genotype.
Remember, genotypes use pairs of letters for representation,
phenotypes use a color or another term for appearance.
Genotypes are written in pairs because eukaryotic chromosomes
come in pairs and one gene (i.e., Y) occurs on one chromosome.
Remember also that one chromosome of each pair comes from
each parent, therefore, one gene Y or y comes from each plant
"parent":

Each parent contributed one of each pair of genes after Meiosis,
which you will remember as dividing the gene pairs into single genes,
i.e., Yy meioses into Y and y.

How was the F1 genotype Yy made from the pure breeding parents,
YY (x) yy?

This is Mendel's explanation of the Yellow and Green F2:

Yy (x) Yy = (Y in ovule + y in pollen grain) = Yy (F2 - second generation) = 2000 of 6000

yellow phenotype

(y in ovule + Y in pollen grain) = Yy F2 = 2000 of 6000

yellow phenotype

(Y + Y) = YY = 2000 of 6000 yellow phenotype

(y + y) = yy = 2000 green

The only genotype which will produce green is ____; it is ____-zygous
and ___________________ (dominant or recessive, choose one)?

GENETICS - Practice, See p. 1102, 8th ed.

The principles of genetics follow the laws of probability and for this reason
are highly predictable and lend themselves to statistical treatment. You will
be exposed to several math principles that should help you to understand genetics.

I. Independent Assortment: pairs of genes on homologous chromosome
pairs assort into gametes at random and recombine in zygotes at random.
In the following exercise, we will illustrate the phenomenon of independent
assortment by studying the independent action of two coins being flipped
simultaneously. The results of heads and tails simulate the chances of obtaining
a particular type of gene in each of the gametes which contribute to the zygote.
For example, if the genes available were A and a (any set of dominant and
recessive alleles, i.e. normal skin = A and albino = a or brown eyes = A, blue = a)
let heads stand for one of these alleles (each gene in a pair) and tails stand for
the other.

COIN FLIPPING EXERCISE

Record the results of your coin tosses in the space provided below.
The Law of Large Numbers states that the larger your sample, the closer
your numbers observed will be to your predicted numbers if your prediction
is correct.
 

Series of 8 each	Your count HH HT TT	Your ratios HH HT TT	Class Count HH HT TT	Class ratios HH HT TT
1st Series 2nd Series 3rd Series 4th Series Totals	__ __ __ __ __ __ __ __ __ __ __ __ __ __ __	__ __ __ __ __ __ __ __ __ __ __ __ __ __ __	__ __ __	__ __ __

-
II. STATISTICS: the mathematical study of probability.

Probability - the chance that one event will occur rather than (an)
other events (s), also called the p value.

A. Sum Rule: The sum of all probabilities equals one. The chance that a coin
will be heads is 1/2 and the chance that it would be tails is 1/2. The sum
1/2 + 1/2 = 1, accounts for all of the possibilities. For three possibilities the
rule still applies:

a + b + c = 1, where a equals the chance of having two heads with two
coins (1/4), b equals the chance of having one head and one tail (1/2:
the first could be heads and the second tails or the first tails and the second heads)
and c equals the chance of having two tails (1/4) 1/4 + 1/2 + 1/4 = 1.

B. Product Rule: is the product of each events probability - the probability
of two or more events occurring simultaneously or consecutively.

Let a be the chance of head and b be the chance of tails. The probability
when two coins are tossed that they both would be heads is the product
of the probability of head for one coin times the likelihood of heads for the
other coin (1/2 (x) 1/2 = 1/4), or, a (x) a = a².

Problem: What are the chances if three coins are tossed that they will
all be heads:

Errors in meoitic division of oogonia or spermatogomia can produce with
24 and 22 chromosomes instead of the normal 23. If such an error occured
two consecutive meiotic divisions, then 4 sex chromosomes could be
present. The following chart lists possible combinations in gametes and
zygotes when homologous sex chromosomes fail to separate
(nondisjunction) or a companion chromosome is lost (anaphase lag)
during one of the two meiotic divisions.

CONDITIONS ASSOCIATED WITH GROSS CHROMOSOME
DEFECTS

Errors in meiotic division of oogonia or spermatogonia can produce
gametes with 24 and 22 chromosomes instead of the normal 23. If
such an error occurred two consecutive meiotic divisions, then 4
sex chromosomes could be present. The following chart lists possible
combinations in gametes and zygotes when homologous sex
chromosomes fail to separate (nondisjunction) or a companion
chromosome is lost (anaphase lag) during one of the two
meiotic divisions. See Table #1 and #2.
 
 
 
 

Table #1
 

OVUM XX X O O XX X X X X X

SPERM CELL Y Y Y X X X O XX YY XY

ZYGOTE XXY XY OY XO XXX XX XO XXX XYY XXY

Syndrome Kleinfelter's normal male lethal Turner's Poly X normal female Turner's Poly X Super Male? Kleinfelter's

Hermaphrodites

During embryogenesis, errors in mitosis (an anaphase lag loss of a Y
chromosome followed by a nondisjunction (failure of an X double chromosome to separate in meiosis I)
of the zygote or early generations of daughter cells may produce
Mosaics - people with two genotypic cell lines; i.e., XX - XY (above), X0-XY.
Chimaeras also have two different cell lines but these possibly
arose from an error in fertilization such as both X and Y sperm cell
nuclei penetrating an ovum with two haploid nuclei, or perhaps a
chimaera is fusion of two zygotes.

True hermaphrodism is defined as the presence of both ovarian and
testicular tissue in either the same or opposite gonads. About 30% of
these are XX - XY mosaics or chimaeras. Fifty-percent are XX and 20%
are XY genotypes. The latter are probably due to mutations of the
testicular determining factor ( (SRY = Sex Determining Region Y gene) on chromosome y or
mutations of the maleness genes on chromosome x.

Gonadal Dysgenesis

Patients with Turner's syndrome have an XO genotype and exhibit
several features, especially, ovarian hypoplasia (or gonadal dysgenesis).
Germ cells die in the genital ridges. They have prominent neck folds
and absent secondary sexual characteristics. Sixty percent of female
gonadal dysgenesis are XO or mosaics with XO predominating over
other genotypes. The Lyon theory states that 2X chromosomes are
necessary for the first 16 days of female development. After that, only
on is active, the other (s) deactivated, appearing as Barr bodies or
"drumsticks."

The other 40% of gonadal dysgenesis patients are XX genotype and
have one of two conditions. Some may have one of several aberrations
of one X chromosome.

Pure gonadal dysgenesis can occur with normal XX chromosomes.
Apparently an autosomal recessive condition is responsible for same
defect in germ cell development and ovary formation.

The XY form of pure gonadal dysgenesis or vanishing testis syndrome
(male dysgenetic pseudohermaphrodism) also exhibits a normal karyotype
but testes fail to develop in the embryonic stage. Therefore, these are
phenotypically female infantile due to absence of androgenic stimulation
and no inhibition of the Muellerian duct system. An X-linked mutant or
abnormal Y chromosome suppresses the y-linked testicular determining
gene. XO - XY mosaicism or abnormal X may appear similarly.

Kleinfelter's syndrome (XXy genotype) or seminiferous tubule syngenesis
is rarely diagnosed before puberty. The typical adult is phenotypically male
with small firm testes. The seminiferous tubules contain no spermatogonia.
Gynecomastia, decreased muscle mass, and small gonads are also
characteristic. XX - XXY mosaics show more extensive virilization and
may be fertile. Forty-eight XXXY and 49 XXXXY individuals are
eunuchoid and typically profoundly retarded. The more extra X
chromosomes the more chance of mental retardation.

The Poly-X syndrome (XXX or XXXX) are occasionally sexually
infantile, and profoundly retarded. Poly-X used to be called super female!

XYY syndrome was once but not now correlated with antisocial
behavior. The phenotype is normal, but most are above 6 ft tall.

ADDITIONAL GENETICS PROBLEM KEYS

Blood Types

ABO System (multiple alleles)
 

TYPE	ANTIGENS	ANTIBODIES	GENOTYPE(S)
A B AB O	A B A,B neither	b a neither a,b	AA or Ai BB or Bi AB ii

Rh for Rhesus antigen (simple dominance)
 

Rh+	Rh+	n/a	Rh+Rh+ or Rh+rh-
rh-	no Rh+	anti Rh(anti-D) after exposure to Rh+ antigen	rh-rh-

Sickle cell hemoglobin (codominance)
 

Normal Sickle Cell Disease Sickle Trait	all normal Hemoglobin all sickle Hb 1/2 normal	n/a n/a n/a	HbnHbn             HbsHbs HbsHbn

OTHER SELECTED GENETIC TRAITS/DISEASES

Dominant

Huntington's Chorea (dementia)

Achondroplastic Dwarfs (homozygous dominant is lethal!)

Polydactylism (6 fingers/toes, not always expressed)

PTC tasting

Kinky hair

For eye color genetics see: http://en.wikipedia.org/wiki/Eye_color   Brown and green are dominant over blue.

Recessive

albinism

PKU Disease

xeroderma pigmentosum

blue eyes (or gray)
 
 

Incomplete Dominance

hair in caucasians: curly, wavy (heterozygous), straight

Multiple Genes

several pairs of genes contributing to height

skin color variations (three pairs of incompletely dominant genes)

color with six M genes, i.e., MM M¹M¹ M²M² = dark black

5 M genes, in any combination = less black

4 M genes, in any combination = brown

3 M genes in any combination = mid color (beige)

2 M genes in any combination = yellowish color

1 M gene = almost white

0 M genes, i.e., mm m¹m¹ m²m², = white

Can two mid-color parents to have a dark black or a white skin baby?
Show your work.

Y-linked

testicular determining gene (SR-Y)

hair turfs in ear canal

X-linked

hemophilia (recessive)

colorblindness (recessive)

testicular feminization (recessive)

ALIFF'S RECIPE FOR GENETICS PROBLEMS

1. Write the mating cross in plain language; i.e.,

brown eyed female (x) brown eyed male = blue eyed male child.

Leave plenty of space around the statement.

2. Determine the type of inheritance by answering the following questions:

a. How many pairs of genes are involved in the adult genotype?

(1 pair Bb = monohybrid (usually one trait like eye color with blue or brown as alternatives);
Aa Bb = dihybrid, combining inheritance of skin color and eye color, 4 phenotypes are possible)

b. Are the genes autosomal, x-linked or y-linked?

c. Is there dominance/recessiveness, or no dominance of genes?

3. Write a letter key for the genes.

4. Fill in the genotypes directly above the cross written in plain language.

5. Construct a Punnett checkerboard.

6. Do a phenotypic summary of the checkerboard.
 
 

Study Questions

1. Explain the blood types possible if an AB crosses with an O.

2. Explain the blood types possible if an heterozygous A crosses
with a heterozygous B?

3. How come Queen Victoria’s grandchildren had hemophilia when
her ancestors and her daughters didn't have the condition.

4. Predict the offspring of a cross between two sickle trait parents.

5. What is the probability of another albino child from parents who
have normal skin and had a previous child with albinism? Give
the probability of three in a row (separate pregnancies).

6. Compare and contrast Down’s, Turners and Kleinfelter’s syndromes.

7. Compare testicular feminization with xx-xy hermaphrodism.
 

Email:john.aliff @ gpc.edu