General Biology II
Third Quarter
Marie Christine T. Gacelo
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Chromosome
a. Etymology: chroma meaning “colored”, soma meaning “body”
b. Organizes very long strands of DNA
c. Ensures DNA transfer is accurately copied during cell division
i. Gene – short segment of DNA that contains the code, or
recipe, for a protein
d. Humans have 23 pairs of chromosomes, a total of 46.
i. 22 pairs – autosomes
ii. 1 pair – gender distinction
e. Parts
i. Centromere – pulls the chromatid during division; “centro”
ii. Telomere – protects the ends (i.e. Since DNA is sticky, it needs some sort of protection to
prevent them from sticking to one another)
iii. Arms
1. P arm – “petit”, shorter arm
2. Q arm
f. Disorders
i. Cry-du-chat Syndrome (“Cry of the Cat”)
1. Occurs at 5P: At Chromosome 5, the p arm is gradually deleted because of the faulty
telomere.
2. Signs: cries like a cat, high-pitched voice
ii. Down Syndrome
1. The mother, instead of giving one chromosome, gives 2 chromosomes.
2. When trisomy occurs in Chromosome 21, the baby is still formed,
3. Trisomy: XX + Y
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Definition:
a. Process of copying the cell’s DNA prior to cell division
b. Involves making exact copies of all 46 chromosomes
c. Product – new strands of DNA
d. Semi-conservative – one new & one old strand
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Process
1. Occurs in the “Origin of Replication”
(a) Prokaryote: 1 origin  bidirectional
(b) Eukaryote: Numerous origins  multidirectional
2. DNA is a double helix.
3. Strands are separated into two independent strands called parent/template/original strand, one
strand runs from 5’ to 3’ and the other from 3’ to 5’ – anti-parallelism
4. Helicase enzyme untwists and unzips the DNA/two daughter strands  H bonds are broken
5. The replication fork is the newly unzipped portion.
6. RNA Primase synthesizes RNA primer to start off pairing to guide DNA polymerase
7. DNA polymerase III, the major replicating enzyme, synthesizes and elongates the daughter DNA
strand, adds nucleotides and back reads/proofread exonuclease activity
8. Because of anti-parallelism, there are two types of DNA strand:
Leading Strand
Lagging Strand
Synthesized continuously
Fragmented  Okazaki Fragments
Replicated towards the replication fork Replicated away from the replication fork
Goes from 3’ to 5’
Goes from 5’ to 3’
Requires 1 RNA primer
Requires more than 1 various primers
pg. 1
Why is there a lagging strand? Replication starts at 3’. Therefore, replicating the 5’ to 3’ is continuous because the
replicate strands starts at the 3’. For the lagging strand, nucleotides are added discontinuously and slowly until the
whole 3’ to 5’ template strand is fully unzipped.
9. DNA Polymerase I – retrieves RNA primer and replace it with DNA material
10. DNA ligase – fills in gaps of lagging strands by joining all Okazaki fragment
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Mechanism of DNA Repair
a. Done by the DNA Polymerase
b. Enzyme recognize errors
c. Nucleotide excision – repairs a nuclease by cutting out and replaces damaged stretches of DNA
d. Mutation – alterations in DNA caused by:
i. Errors during base-pairing in replication
ii. Physical or chemical forces (i.e. exposure to harmful chemicals or physical agents such as
cigarette smoke & x-rays)
iii. Somatic cell mutations – may affect function of tissues or cause cancer, but not passed on
to descendants
iv. Gamete mutation – passed on to future generations
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Definition:
a. Process of creating a coding message of a single gene that can be carried out of the nucleus
b. Process of copying DNA of a gene into mRNA
c. Occurs within the cell nucleus
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Products
a. Messenger RNA (mRNA) – template for the synthesis of proteins by ribosomes
b. Non-coding RNA
i. Transfer RNA (tRNA)
1. Relatively small RNA molecules that escort amino acids to the ribosome
2. Transfer specific amino acids to growing polypeptide chains at the ribosomal site of
protein synthesis during translation
ii. Ribosomal RNA (rRNA) – a component of ribosomes; binding site
iii. Micro RNA – regulates gene activity
iv. Catalytic RNA– enzymatically active RNA
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Process
1. Initiation
a. The portion of the DNA molecule corresponding to the gene unwinds temporarily, and a
complementary strand of RNA is produced from one of the DNA strands.
b. Promoter – region of DNA that initiates transcription of a particular gene
c. TATA Box – DNA sequence that indicates where a genetic sequence can be read and
decoded; it is where transcription begins
2. Elongation
a. RNA Polymerase / RNA Pol has the following tasks:
i. Initiate transcription
ii. Helicase
iii. Proofreading
iv. Coding
v. Termination recognition
b. Triplet code – sequence of three genetic codes
c. Codon – sequence of three mRNA bases for amino acids
i. 64 different codons, but 20 different amino acids
ii. Start Codon: AUG or Methionine
iii. Stop Codons: UAA, UAG, UGA
d. Primary Transcript – strand of RNA released from DNA which contains:
i. Exons – sections that carry genetic information
pg. 2
ii. Introns – intervening sequences; sections that may allow different combinations of
genetic information and are edited out by enzymes to produce the final mRNA
3. Termination
a. Prokaryote:
i. Rho-dependent Termination – protein “Rho” disrupts the complex involving the
template strand, RNA Pol and RNA
ii. Rho-independent Termination – loop forms causing the detachment
b. Eukaryote: RNA Pol
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Pre-mRNA  Mature RNA
a. Splicing – spliceosome removes introns (noncoding), while joining together exons (coding)
b. Polyadenylation
i. Capping – 7-methylguanosine cap [1 G] at the 5’ end
ii. Tailing – Poly-A (adenosine) tail [250~ As] at the 3’ end
iii. Protects mRNA from enzymes that might break it down, especially outside the nucleus
iv. Signals the end of the mRNA
v. Guides the mRNA out of the nucleus
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Definition:
a. Process of converting the coded message or mRNA into proteins useful to the cell
b. Occurs in the cytoplasm at ribosomes
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Products: Proteins
a. Antibody – body’s protection against foreign elements
b. Enzyme – catalyzes reaction
c. Messenger – transmits signals
d. Structural – structure and support for cells
e. Transport / Storage – binds and carries atoms and small molecules
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Process
1. Initiation: Initiator tRNA (carrying methionine), start codon on mRNA, and ribosomal subunits form
an initiation complex
2. Elongation: tRNA brings specific amino acids to developing protein chain which elongates one
amino acid at a time
3. Termination: Stop codon terminates developing chain, protein is released from ribosome
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Cell Reproduction
a. A defining feature of life which involves the fusion of a mother’s egg cell and the father’s sperm cell
b. Unicellular organisms: cell division is the mechanism of reproduction
c. Multicellular organisms: cell division enables growth from a fertilized egg to a multicellular
individual — Mechanism of cell division is the same in all eukaryotes
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Definition: Creates new cells
a. Complete cell cycle takes 18 – 24 hours
Phase
Duration
G1
10 hours
S
9 hours
G2
4 hours
M
1 hour
Metaphase – 20 minutes
pg. 3
b. G0 / Non-Growing Stage – cells enter a non-dividing / dormant state temporarily or permanently
i. Neurons and osteocytes enter G0 after adolescence therefore degenerative diseases such
as Alzheimer’s Disease or osteoporosis occur.
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Two Major Phases
a. Interphase – “between cell division”
i. G1 (first gap) – primary period of cell
growth, very active growth, duplication of
organelles, 23 chromosomes
ii. S (Synthesis) – DNA synthesis and
replication, 46 chromosomes
iii. G2 – cell prepares for division, growth
continues slowly, finalization
b. Mitotic Phase – cell division that starts and ends
with 46 chromosomes
i. Nuclear division – duplicated DNA is
distributed between two daughter nuclei,
nucleus divides
ii. Cytokinesis – division of cytoplasm to form
two new daughter cells
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Two Types of Cell Reproduction
Mitosis
Somatic / Germ Cells / Autosomes
2 daughter cells are identical to the parent cell
Diploid cells (2n / 2 sets of chromosomes 
one from the mother and one from the father)
Starts with 46 chromosomes, ends with 46
Meiosis
Gametes / Gametocytes / Sex cells
4 daughter cells that are genetically different
Haploid cells (n / 1 set of chromosomes)
Starts with 46 chromosomes, ends with 23
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Process: Nuclear division (mitosis) followed by cytoplasmic division (cytokinesis)
pg. 4
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Prophase
a. Chromatid – a term used in Prophase; strands are condensed and compact
b. Chromatin – a term used in Interphase; strands are randomly distributed, loose, and at rest
c. Events
i. Pairing of sister chromatids.
1. Chromatids that are not sister or identical chromatids cannot pair up.
ii. Centrioles move to the opposite sides.
1. Centriole is the cylindrical organelle that develops spindle fibers.
2. Centrosome contains two centrioles.
iii. Spindle fiber forms and are attached at the centromere.
iv. Nuclear membrane dissolves or disintegrates so that the genetic makeup inside would be
liberated.
d. Metabolic activity decreases
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Metaphase
a. It is the longest phase because:
i. Centromere starts to develop 2 different structures for each chromatid,
ii. The breaking apart of the centromere takes a long time.
b. Duplicate chromosomes form single line at the equator between centriole poles
i. They align themselves at the center so that the centrioles exert an equal amount of applied
force on each chromatid, making the division more balanced.
c. Meta meaning “after”
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Anaphase
a. It is the shortest phase and expends the most ATP because the strength of the spindle fibers makes
it easy to pull the chromosomes toward the opposite poles.
b. Duplicate chromosomes separate
c. Cleavage furrow is first seen. For plants, we use the term cell plate.
i. Actin and myosin complexes pinch the cell membrane
d. Anaphase VS Telophase
i. If you can still see every arm, then it is still in Anaphase because there is still pulling.
ii. In telophase, everything is already at rest. No more expenditure of energy.
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Telophase
a. “Reverse of prophase”
b. Mitotic spindle disintegrates
c. Nuclear membrane reforms
d. Chromosomes uncoil and revert to chromatin
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Cytokinesis
a. Definition: successful breaking apart of cells forming two identical daughter cells (diploid)
b. Contractile ring of filaments forms at midsection of cell and tightens, forming a cleavage furrow
caused by protein actin and myosin complex
c. Two daughter cells formed as the contractile ring pinches them apart
pg. 5
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Prophase I
a. Synapsis – the pairing up of homologous chromosomes.
i. “Homologous” means they will be the same in function and characteristic in the future.
ii. A pair of homologous chromosomes is called a tetrad.
iii. The pairing of sister chromatids happened during the S phase in Interphase.
b. Crossing over – one chromatid / arm is literally crossing over the other arm of the chromosome
i. Requirement: Must be in equal size
ii. Function: Genetic diversity – important for natural selection
iii. Chiasma – the point of exchange of genetic makeup
iv. Result: Recombinant Genes
c. Centrioles move to two opposite poles
d. Formation of spindle fibers
e. Nuclear envelope dissolves
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Metaphase I
a. Homologous pairs of chromosomes line up in the center
b. Aligning of the tetrad in the middle – each pole is exerting equal force
c. Takes a shorter time than Metaphase in Mitosis because there are already 2 centromeres.
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Anaphase I
a. Pairs of chromosomes separated, but duplicated chromosomes stay intact
b. “Disjunction” – separation
c. Law of Segregation – distribution of paternal and maternal chromosomes is purely random
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Telophase I
a. Two haploid daughter cells but chromosomes are still in a duplicated state
b. Cleavage furrow is evident
c. Nuclear membrane needs to develop in order to seal or compartmentalize them in one space or else
chromosomes would be floating around and possibly be unequal in number.
d. Spindle fibers disintegrate.
e. Cytoplasm starts to divide.
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Interphase II – nothing happens; it is mostly at rest; simply waits for Meiosis II
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Meiosis II
a. Prophase II and Metaphase II is exactly the same as Mitosis
b. In Anaphase II, the chromatids split. There is no chromatid that is similar due to crossing over.
c. In Telophase II, there is a discrepancy in genetic makeup in each haploid cell.
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Males
a. Four sperm produced from each cell entering meiosis
b. All viable or functional
c. Sperm cells are produced numerously because most of them are weak and die when reaching the
cervical wall. — Sperm cells regenerate but as the male ages, his capability to produce reduces.
d. Requirement for Fertilization
i. Female Reproductive environment should be welcoming — Sperm cell should be able to
penetrate the cervical wall. Seminal fluids are basic and are used to neutralize the very
acidic environment of the cervical wall. But, for 24 to 30 hours, during the ovulation day, the
ovum lowers it pH to be more “welcoming” to sperm cells. But, during the safe days or 5 days
before and after the last period, the acidity spikes.
ii. Sperm should be able to swim — motility
iii. All structures of the sperm cell must be present. If there is at least one defect, the sperm cell
would not be successful.
pg. 6
1. Acrosome – produces a digestive enzyme that destroys the Zona pellucida, the thick
follicle lining of the egg cell.
2. Mitochondria – ATP production for the tail to swim
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Females
a. Unequal cytokinesis during Meiosis I and II
b. One ovum and three polar bodies (“unfertilized”) produced from each cell entering meiosis
c. Only one egg, the ovum, is viable or functional because the egg cells expend a lot of energy on the
development of the cytoplasm of the ovum, creating a sustainable environment for its division in
Meiosis II. The other three polar bodies donate their cytoplasm to the ovum.
i. We use ovum instead of egg cell to avoid confusion. For mammals, the term is changed
every phase. For birds, reptiles, etc., it is called an egg until it hatches.
1. Egg cell (can be fertilized or not) – Ovum — Zygote — Embryo — Fetus — Baby
d. A definite number of egg cells is given to us and we just consume it until we age.
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Oogenesis VS Spermatogenesis
a. Dormant Gametes / Undifferentiated / STEM Cells: Oogonium and Spermatogonium
b. After 4 days, during the primary level, the cells become differentiated: oocyte / spermatocyte.
c. After differentiation, it is dormant for a decade because Meiosis II only happens during adolescence
because our hormones because active — estrogen / testosterone.
i. If a sperm is able to impregnate the ovum, it will undergo Meiosis II. But if not, the female
excretes it via menstruation.
d. For females, the original number of egg cells given are reduced by three-fourths. For males, the
original number of sperm cells regenerate.
pg. 7
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Regulation of Cell Reproduction
a. Unequal cytokinesis during Meiosis I and II
b. Not all cells divide at the same rate
c. Internal surveillance and control mechanism
d. Several key checkpoints where “go ahead” signals must be received in order for the cycle to
progress to the next phase – G1, G2, M checkpoints
e. Outside influences can modify cell cycle (i.e. Hormones, growth factors, presence of other cells)
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Differentiation
a. Assigning functions to new cell
b. Process by which a cell becomes different from its parent
or sister cell
c. Differentiation is based on different gene expression
d. All body cells have the same DNA, yet there are great
differences between the shape and function of different cell
types.
e. Stem cells – undifferentiated cells
i. The only source of stem cells are embryos.
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(1) Gamete Formation – Eggs form and mature in female reproductive organs. Sperm form and mature
in male reproductive organs.
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(2) Fertilization – A sperm and an egg fuse at their plasma membrane. Then, the nucleus of one fuses
with the nucleus of the other to form the zygote.
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(3) Cleavage – cell divisions carve up different regions of egg cytoplasm for daughter cells
a. Cleavage – cell division converts the zygote into a ball of cells
b. Morula – sixteen-celled zygote; uterus; means “mulberry” in Greek
c. Blastomere – each cell in the morula that forms during cleavage and will have different fates
pg. 8
(4) Gastrulation – Cell divisions, migrations, and rearrangements produce two or three primary tissues,
the start of specialized tissues and organs
a. Cell differentiation – newly formed cells become specialized for a certain function
b. Morphogenesis – “the beginning of form”; specific organs and tissues form
c. Gastrulation – Process of early development that produces the three germ layers
d. Germ layers – Three primary tissues that form as an early embryo develops.
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Ectoderm
Nervous system &
sense organs
Pituitary gland
Epidermis and
associated structures
(hair)
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Mesoderm
Cartilage, bones, muscles
Cardiovascular lymphatic
system
Urinary system reproductive
Outer layers of digestive
system
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Endoderm
Lining of digestive tube
Lining of respiratory
airways
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Implantation
a. Blastocyst
i. Embryonic stage that develops from morula and will eventually implant the uterine wall
ii. Secretes hCG (human chorionic gonadotropin)
iii. Happens when the morula enters the uterus
iv. Trophoblast — surface epithelium
v. Inner cell mass — small clump of cell where embryo develops
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(5) Organ Formation – subpopulations of cells are sculpted into specialized organs and tissues in
spatial patterns at prescribed times
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(6) Growth, tissue specialization – organs increase in size and gradually assume their specialized
functions
•
•
•
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Day 1 – 3: Cleavage
Day 4: Gastrulation
(Not unethical to
harvest stem cells
during Days 1-3)
Day 5: Implantation –
During Days 1-4, the
cell is swimming from
the ovary, to the
fallopian tube, to the
uterine wall.
External substances harmful to fetuses:
a. Cigarette smoke: retards growth
b. Alcohol: fetal alcohol syndrome
c. Medications (prescription/over-the- counter): pass through placenta
d. Illegal drugs: child born addicted
e. Environmental chemicals: in air, water, soil
f. Radiation: radon, X-rays
g. Intrauterine infections: HIV, syphilis, rubella
pg. 9
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Reproductive cloning
a. Producing a “copy” of an entire organism
b. Requires a completely undifferentiated cell as the starting point
c. Increase body count
d. Done on an industrial scale
e. Two Methods
i. Embryo Splitting
1. Definition: Producing Identical Offspring
2. Procedure
a. Egg is fertilized in vitro, and allowed to divide to the eight-cell stage
b. Cells of eight-cell stage are carefully separated and each is implanted into a
different surrogate mother in which it develops
3. Results: Clones are genetically identical to each other but not to either parent
4. Process has not been attempted with humans
ii. Somatic Cell Nuclear Transfer
1. Definition: Producing a clone of an adult
2. Somatic cell – any cell other than a gamete
a. Each somatic cell has a full diploid set of
chromosomes
3. Procedure
a. Somatic cell from the adult organism to be
cloned is inserted into an enucleated fertilized
egg
b. An electrical current is used to fuse the cells
c. Fused cell is implanted into uterus of
surrogate mother and allowed to develop
4. Result: Offspring is a clone of the adult organism that
provided the nucleus
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Therapeutic Cloning
a. The cloning of human cells (i.e. for treating patients)
b. Ideally, remove a single cell from a patient and nurture it to develop
and differentiate into the cell type needed to treat the disease
c. Potential for creating new cells, tissues, or organs—as yet unrealized
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On endangered species, we do not do reproductive cloning because we
respect the Laws of Natural Selection. They are endangered because
they are unable to adapt. Even if you do increase the body count, their genes will still be the same and
since genetic diversity and variation helps in making a species live, their species will be eradicated.
pg. 10
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Drosophila melanogaster
a. Common Name: Fruit Flies / Vinegar Flies
b. Widely used in genetics, physiology, microbial pathogenesis, etc.
c. Widely used because they are easy to cultivate, breeds quickly, and lays many eggs
d. Life Cycle  The higher the temperature, the faster the generation.
i. Day 1 – Eggs hatch
ii. Day 2 – First instar (one day in length)
1. Instar – larval stage wherein it sheds its outer covering (A common procedure for
arthropods or shelled insects)
iii. Day 3 – Second instar (one day in length)
iv. Day 5 – Third and final instar (two days)
v. Day 7 – Larvae begins roaming
vi. Day 11 – Eclosion (start of adulthood)
1. Females become sexually mature
2. Pupariation / Pupal Formation – 120 hours after egg laying
e. Males: Smaller, Short and round abdomen, dark end on the body, has a sex comb
f. Females: Larger, abdomen is curved to a point, striped lines on body
g. Has 4 pairs of chromosomes – 3 autosomes and 1 sex chromosome
h. Wild Type VS Mutant
i. Wild type – natural appearance of the offspring
ii. Mutant – mutations or changes in the genes; can be assumed as recessive to the wild type
Wild Type
Eyes
Red, oval-shaped, many-faceted
Bristles
Fairly long and smooth – proper
distribution on head and thorax
Wings
Smooth edges, uniform venation,
extend beyond the abdomen
Body Color
Basically gray with pattern of
light and dark areas
Mutant
White, black, apricot, scarlet red, pink or
brown, changes in shape and number of facets
Shortened, thickened, or deformed bristles –
changes in patterns of distribution
Changes in size and shape; absence of
specific veins; changes in position in which
wings are held when at rest
Black (in varying degrees), yellow, in doubtful
cases, color can often be determined most
clearly on wing veins and legs
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There are 3 billion base pairs in the human body. Genome-wide arrays (GWA) can determine the
body’s whole entire sequence by getting a small sample of DNA.
a. This can prove whether a person is positive for a certain disease, especially those that are
hereditary. (e.g. Oncogenes and proto-oncogenes – mutations in genes)
b. This can also determine the specific lifestyle or diet we need to have. But, it entails panic in people.
c. Social Issue: invalidation of Privacy, causes discrimination in the workplace or health insurance
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Terminologies
a. Genes – DNA sequences that contain instructions for building proteins
i. Even before the Central Dogma, there is already the list of proteins which will dictate certain
traits we will have because of the genes.
ii. Base pair sequences  specific amino acid chain  protein  trait
iii. Genes code for specific proteins, not for specific behaviors. They do not dictate how we
think or act but strongly influence them.
iv. Proteins have specific functions leading to phenotypes.
v. Protein functions:
1. Hormones
2. Enyzmes
3. Structural
4. Neutrotransmitters
5. Intracellular messengers
b. Genetics – study of genes and their transmission from one generation to the next (inheritance)
pg. 11
c. Genome – sum total of all an organism’s DNA; collective term for genes
d. Karyotype
i. A composite visual display of all of the chromosomes of an individual
ii. Shows all 23 pairs of chromosomes paired and lined up side-by-side
iii. If they are not identical, it means the person has abnormalities.
1. Chromosomes must have the same size, shape, locus, centromere location
iv. Standard test performed in hospitals wherein a nucleus is taken from one somatic cell.
v. Benefits
1. If chromosomes have alterations or mutations that may be hereditary, one can make
informed decisions how to move on about life.
2. To know if you are fertile or not
3. Can be done on fetuses to detect chromosome defects
e. Homologous Chromosomes
i. Code for the same genes
ii. Their loci fit perfectly together
f. Pair of autosomes – each autosome carries the same genes at the locus
g. Pair of genes – have the same structure and function
h. Gene locus (plural: loci) – location of a specific pair of genes
i. Alleles
i. Alternative versions of the same gene pair
ii. Different from one another and always in pairs because one is from the father and one is
from the mother.
iii. Codes for the same gene, but never genetically similar.
iv. ** Genes are a general term for sequences
v. ** Alleles are variants of genes.
j. Dominant Allele, D
i. Masks or suppresses the expression of its complementary allele
ii. Always expressed even if heterozygous
iii. Expresses physically
iv. Does not mean it is more in quantity, it is dominant in the sense that its presence is stronger
than the recessive.
k. Recessive allele, dd
i. Will not be expressed if paired with a dominant allele (heterozygous)
ii. Will only be expressed if individual is homozygous for the recessive allele
iii. Originated from mutations in our genetic history. Recessive alleles are mutations from our
long evolutionary history that are not eliminated in the gene pool by Natural Selection
because they are not harmful in any way.
l. Homozygous – two identical alleles at a particular locus
i. Homozygous Dominant – double Ds
ii. Homozygous Recessive – double ds
m. Heterozygous – two different alleles at particular locus
n. Linked Alleles
i. Physically located on the same chromosome
ii. The closer the genes are together, the lesser probability that they will be separated during
crossing over, the more likely they are to be inherited together
iii. May be shuffled during crossing over at meiosis
iv. Linkage Mapping – map positions of genes by studying how often particular linked alleles are
inherited together
v. Three major sources of genetic variability as a result of sexual reproduction:
1. Independent assortment of alleles located on different chromosomes
2. Partial reshuffling of linked alleles as a result of crossing-over between autosomes
3. Random fertilization of an egg by a sperm
o. Genotype – an individual’s complete set of alleles (the plan or blueprint)
i. Besides the genotype, environmental factors greatly affect the phenotype.
ii. For genetic diseases, we inherit the risk of developing the disease but those risks are
modifiable by environmental factors and own actions. Genotype is not the sole determinant.
pg. 12
p. Phenotype – observable physical and functional traits (the execution)
i. Determined by inherited alleles and environmental factors (i.e. effect of diet on body size)
q. Polygenic Inheritance
i. Inheritance of phenotypic traits that depend on many genes acting simultaneously
ii. Usually distributed within a population as a continuous range of values
iii. Alleles do not show dominance, instead they show an additive effect in expressing a specific
phenotype
iv. i.e. Eye color (blue – light blue, dark blue, etc.), skin color, height, body size & shape
v. i.e. There are three genes for eye color – color, shade, absorption of melanin.
vi. i.e. Height is controlled by about 400 genes.
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Gregor Mendel
a. Worked with pea plants in the 1850s in Austria
b. Did multiple genetic experiments to develop basic rules of inheritance
c. Law of Segregation
i. Stressed that there are dominant and recessive genes
ii. Genes exist in more than one form of allele
iii. Organisms inherit one from each parent
iv. Complete dominance – recessive, although is present, is not expressed (i.e. this is why
some traits skip generations)
d. Law of Independent Assortment
i. Randomly pass of genes
ii. Genes for different traits are separated from each other independently during meiosis
iii. This law applies only if the two genes in question are on different chromosomes
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Punnett Square Analysis
a. Heredity can be quantified using a Punnett Square.
b. Possible alleles one parent is placed on one axis, and the other parent on the other axis
c. Possible combinations of parental alleles are written in the squares within the grid
d. Monohybrid – one trait
e. Dihybrid – two different traits
i. i.e. You have a female who is short with blonde hair and a male who is a homozygous tall
person with heterozygous brown hair.
Legends:
T = tall
t = short
B = brown
b = blonde
SOLUTION #1: Dihybrid Cross
TB
Tb
TB
Tb
tb
TtBb
Ttbb
TtBb
Ttbb
tb
TtBb
Ttbb
TtBb
Ttbb
tb
TtBb
Ttbb
TtBb
Ttbb
tb
TtBb
Ttbb
TtBb
Ttbb
Questions:
(TB) Tall with brown hair: 8/16
(Tb) Tall with blonde hair: 8/16
(tB) Short with brown hair: 0/16
(tb) Short with blonde hair: 0/16
pg. 13
SOLUTION #2: 2 Monohybrid Cross
Height
T
T
Hair color
t
Tt
Tt
Tall – 4/4
Short – 0/4
t
Tt
Tt
B
b
b
Bb
bb
b
Bb
bb
Brown hair – 2/4
Blonde hair – 2/4
Questions:
(TB) Tall with brown hair: (4/4)(2/4) = 8/16
(Tb) Tall with blonde hair: (4/4)(2/4) = 8/16
(tB) Short with brown hair: (0/4)(2/4) = 0/16
(tb) Short with blonde hair: (0/4)(2/4) = 0/16
✓
❖
Complete dominance
a. In the presence of a dominant allele, the recessive gene is completely masked.
b. Mendelian Laws fall under complete dominance
c. Encompasses both monohybrid and dihybrid cross
❖
Incomplete dominance
a. Non-Mendelian Law
b. Proposed by Carl Corren
c. Does not totally negate complete dominance but proposes that both dominant and recessive can be
expressed
d. Interaction or blend between of both alleles  intermediate phenotype between two homozygous
phenotypes
e. No masking effect, only additive effect
f. i.e. A breed between a white and black dog will result to a gray dog.
g. i.e. A cross between homozygous straight and curly hairs will possibly result to a wavy-haired child.
h. i.e. A cross between a white and red flower will result to a pink flower.
i. i.e. Familial hypercholesterolemia
i. Homozygous – have blood cholesterol concentrations more than 6x the normal level
ii. Heterozygous – have blood cholesterol concentrations more than 2x-3x the normal level
❖
Codominance
a. Equal contributions of both alleles are visible in the phenotype
b. Both alleles are expressed, but individually
c. i.e. A breed between a white and black dog can result to a dog with white fur and black spots.
d. i.e. Genes for ABO blood types
i. A gene and B gene, in type AB, are codominant.
ii. An individual heterozygous for the A and B genes will be blood type AB, expressing both A
and B antigens on red blood cells.
Allele
IAIA
IA𝑖 o
IBIB
IB𝑖 o
𝑖 o𝑖 o
IAIB
Blood Type
A
A (Recessive)
B
B (Recessive)
O
AB
pg. 14
e. i.e. Sickle Cell Gene
i. Two different alleles of hemoglobin genes
1. HbA encodes for normal hemoglobin
2. HbS encodes for sickle cell hemoglobin  a mutation in one amino acid
ii. Anemic: HbSHbS (homozygous)
1. HbS will crytallize if O2 is slightly decreased, resulting in bending of red blood cells
into crescent shapes / sickled cells
2. Characteristics of Sickle cells
a. Rupture easily because they are more fragile
b. Less flexible so they tend to clog small blood vessels and block blood flow
c. Carry less oxygen
3. Characteristics of resultant oxygen deprivation of tissues and organs
a. Fatigue
b. Chest pain
c. Bone and joint paint
d. Swollen hands and feet
e. Jaundice
f. Visual problems due to retinal damage
4. Long-term complication of untreated sickle-cell anemia:
a. Stroke
b. Blindness
c. Pneumonia
d. Multi-organ damage
5. Primarily affects Africans and Caucasians of Mediterranean descent
a. Nearly 40% of the black population has the sickle cell trait
6. Sickle cell allele protects against malaria, a deadly parasitic disease common in
tropical climates (i.e. Africa)
a. Because sickle cell easily ruptures, when the malaria enters this red blood
cell, it is killed before it can reproduce.
iii. Trait / Carrier: HbAHbS (heterozygous)
1. Affected individual makes both types of hemoglobin
2. Rarely symptomatic – do not exhibit symptoms unless deprived of oxygen
a. Trigger factors include riding on an airplane or high altitudes.
3. At the biochemical level, sickle-cell trait displays codominance because normal and
sickle-cell hemoglobin are produced in equal amounts.
4. At the level of the whole person, it displays incomplete dominance because the
person exhibits a disease pattern that is intermediate between two homozygous
conditions.
iv. Normal: HbAHbA
✓
❖
Definition: Inheritance patterns that depend on genes located on the sex chromosomes
a. 23rd pair of chromosome
b. Not homologous
c. X and Y chromosome carry different genes
d. Males: have one X and one Y chromosomes
i. 50% X-carrying gametes, 50% Y-carrying genes
ii. Male determines gender of offspring
e. Females: have two X chromosomes
f. X-linked – located on the X chromosome (not related to sex determination: Red-green color
blindness, Duchenne muscular dystrophy)
g. Y-linked – located on the Y chromosome
i. Few genetic disorders occur in the Y chromosome because it is relatively small and many of
its genes relate to “maleness” (i.e. differentiation of male sex organs, production of sperm,
development of secondary sex characteristics)
pg. 15
❖
Female VS Male Susceptibility
a. More males than females express the disease
b. Passed to sons by mother  females usually carry the disease
c. Father cannot pass the gene to sons, but daughters can be carriers
d. Males are more susceptible to diseases associated with recessive alleles on the sex chromosome
because they only have 1 X chromosome.
❖
i.e. Hemophilia / Bleeder’s Disease – lack a blood-clotting factor encoded by an X-linked gene
a. Found only on the X chromosome
i. H = healthy; produces clotting factor
ii. h = sick / hemophiliac; cannot produce clotting factor
b. Many more males have the disease than females. Females can be protected by inheriting at least
one normal allele (XH) out of two, whereas males inherit only one allele because they inherit only
one X chromosome.
c. Females: 2/3 healthy, 1/3 hemophiliac
i. XHXH – healthy
ii. XHXh – carrier (healthy)  can pass on the allele but do not suffer from the disease
1. Half of the sons of carriers will have the disease
2. Half of the daughters of carriers will be carriers
iii. XhXh – hemophiliac
d. Males: 50% - 50% chance of each type
i. XHY – healthy
ii. XhY – hemophiliac
e. A father cannot pass the disease to a son, but his daughters will all be carriers. If the mother is also
a carrier or has the disease, the daughters may also have the disease.
❖
Sex-Influenced Phenotype
a. Definition:
i. Not inherited with sex chromosomes but influenced by them
ii. Located on the autosomal pairs of chromosomes
b. i.e. Baldness
i. b+ = normal
ii. b = baldness
iii. Present for both male and female
1. Recessive for females
2. Dominant for males - testosterone strongly stimulates expression of the baldness
allele
iv. Females who are homozygous recessive for the baldness gene + other factors will have
significant hair loss.
v. Male: homozygous recessive  very bald
vi. Male: heterozygous  significant pattern baldness
✓
❖
Nondisjunction
a. Definition: Failure of homologous chromosomes or sister chromatids to separate
b. A gamete may end up with two copies of a chromosome, instead of just one
c. Alteration of the number of sex chromosomes
d. Types:
i. Mitosis – Nondisjunction of sister chromatids
ii. Meiosis I – Nondisjunction of homologous chromosomes
1. This is more disastrous than disjunction in Meiosis II because there are 2 defective
products in Meiosis I, but only 1 in Meiosis II.
iii. Meiosis II – Nondisjunction of sister chromatids
e. If nondisjunction occurs in mitosis, the two daughter cells would probably die and be replaced. For
meiosis, disjunction has the potential to alter the development of the organism
pg. 16
f.
Down syndrome
i. Three types:
1. Three copies of chromosome 21 – Trisomy 21 (most common)
2. Altered autosomal chromosomal number- Edwards Syndrome / Trisomy 18
3. Patau Syndrome / Trisomy 13
ii. Have developmental abilities, respiratory complications, and/or heart defects
iii. High chance if the mother becomes pregnant at a later age (>35 years old)
1. It is because the production of the egg cell has reached its “expiration date.”
g. Jacob Syndrome / XYY Syndrome
i. Found at chromosome 23
ii. Evident for males only, not females
iii. Signs & Symptoms
1. Very tall
2. No facial hair
3. Breast development
4. Manifest osteoporosis
5. No testes
6. Impaired mental function
h. Klinefelter Syndrome / XXY Syndrome
i. Found at chromosome 23
ii. More symptoms and is worse than Jacob’s because it has 2 Xs thus it is exposed to more
genetic disorders.
iii. Signs & Symptoms
1. Very tall
2. Has breast development
3. Uneven fat distribution
4. Show mild mental impairment
5. Infertile / Sterile
6. Cryptorchidism – undescended testis (found in the abdomen instead of the scrotum
after 3 months)
a. Testes produce testosterone. Because they are not able to do their function,
estrogen is not balanced out thus they manifest female characteristics.
7. Hypospadias – opening of the urethra is not in the proper place
8. Micropenis – 1-2 inches in length, ring of fat around the penis
a. When a baby boy is born, he can also be diagnosed with micropenis. He will
then be injected with hormones.
i. Turner Syndrome / XO Syndrome
i. Only one X chromosomes is given hence the term X zero
ii. Signs & Symptoms
1. Abnormally short
2. Small breasts
pg. 17
j.
3. Sterile
4. Shortened life expetancy
5. Webbing of the neck
6. No development of ovaries
Trisomy-X Syndrome / XXX Syndrome
i. Female
ii. Normal except for a tendency towards mild mental retardation
❖
Deletions
a. Piece of chromosome breaks off (lethal for sex chromosomes)
b. Can be detected with Karyotyping because we get a visual picture of the chromatids to detect the
size
c. i.e. Cri-du-chat syndrome (French: “cat-cry”) – caused by a deletion from chromosome 5
i. Signs & Symptoms: mentally and physically retarded, kitten-like cry
❖
Translocations
a. Piece of chromosome breaks off and attached to a different chromosome
b. Crossing over is induced while translocation happens upon inheritance.
✓
❖
Phenylketonuria (PKU) – Recessive Alleles
a. Disease caused by a mutation of the gene on chromosome 1 that is responsible for producing the
enzyme phenylalanine hydroxylase (PHH), which is needed for the normal metabolism of
phenylalanine (amino acid)
b. Without PHH, build-up of phenylalanine which would lead to its conversion to phenylpyruvic acid
which is toxic at high concentrations. Too much can cause mental retardation, slow growth rate,
early death
c. Treatment: Proper diet (lessen meat and dairy products), provide supplements
❖
Tay-sachs Disease – Recessive Alleles
a. Disease caused by a mutation of the gene on chromosome 15 that is responsible for producing an
enzyme needed for the metabolism of sphingolipid which is found in the lysosome of the brain cells
b. Accumulation of sphingolipid  cerebral degeneration
c. Fatal disorder that occurs in baby boys
d. Progressive destruction of the nervous system
e. Can only be diagnosed when the child is 6 months old and by that time it is already too late
i. 4-8 months: motor and brain function begin to decline
ii. Seizures, blindness, paralysis
iii. Death by age 3 or 4
❖
Huntington’s Disease – Dominant Alleles
a. Progressive brain disorder that occur in children and adults
b. Caused by a dominant-lethal allele
c. Mutation in the HTT gene
i. HTT gene produces the Huntingtin protein that is vital for the health of our brain cells.
ii. A mutation will cause CAG trinucleotide repeat. A normal person can have this CAG
sequence for no more than 35 repeats. More than 39 repeats can cause overproduction of
huntingtin which will then cause degeneration of neurons.
pg. 18
❖
In the 1990s, there were a significant amount of studies regarding GMOs to answer problems of
malnutrition, pollution, and starvation. They were trying to find solutions using natural products. DNA
technology aims to new discoveries to help mankind and make use of organic sources that are readily
available. The use of microorganisms is ensued because they are easy to culture.
❖
Biotechnology
a. Definition:
i. Technical application of biological knowledge for human purposes
ii. Any technique that uses living organisms or substances from those organisms:
1. To make or modify a product
2. To improve plants or animals
3. To develop microorganisms for specific uses or to solve specific problems
b. Genetic Engineering
i. Manipulation of the genetic makeup of cells or whole organisms
ii. Direct altering of an organism’s genome to fulfill a specific purpose
iii. All organisms have a genome that rotate around the four nucleotide bases, A-T and C-G,
that is why genes are considered universal.
iv. Genetically engineering creates transgenic organisms
1. Transgenic bacteria have been developed to produce the following:
a. Insulin
b. Human growth hormone
c. Erythropoietin
d. Tissue
v. GMOs can be good or bad, depending on how much and how good the research was done.
c. Recombinant DNA Technology
i. Applied science that explores applications of cutting, splicing, and creating DNA
ii. Literal combination of two or more species
❖
DNA Sequencing
a. Determining the precise sequence:
b. DNA must be amplified to get adequate amount
c. Primers – enable initiation of DNA synthesis
d. A, G, C, T nucleotides
e. Process:
i. Extraction of DNA
ii. Induce replication by injecting it with enzymes:
1. Helicase to unzip the DNA  12 minutes
a. DNA is bound with the Hydrogen Bond which is the weakest bond.
2. Primer for the starting point  10 minutes
iii. Inject solution with:
1. Nucleotide bases that are randomly placed
2. Modified nucleotide bases / sequences:
a. Serve as a stopping point (instead of using another enzyme)
b. Contain fluorescent dyes
iv. Short complementary DNA sequences, each ending with a labeled nucleotide
v. Inject with dye
vi. Short DNA sequences produced by the first step are separated by gel electrophoresis
1. Inject solution to the wells
2. Subject to electricity
3. All colors will separate because they each have a different size and frequency.
a. Smaller particles will move faster in the downward direction
vii. Analyzed by a laser
f. Result: Printout of the sequence of the complementary strand of DNA
pg. 19
❖
Recombinant DNA Technology
a. Definition:
i. General name for taking a piece of one DNA, and combining it with another strand of DNA.
ii. By combining two or more different strands of DNA (usually from two different organisms),
scientists can create a new strand of DNA.
iii. Recombinant DNA - DNA assembled out of fragments from multiple sources.
b. Tools for manipulating DNA
i. Restriction enzymes – cut DNA at specific sites
1. Palindromic sequences – dictate where DNA will be cut and where the plasmid will
bind and naturally stick
ii. DNA ligases – join fragments of DNA (the glue)
iii. Plasmid – small circular pieces of DNA where the gene or DNA of interest is inserted
1. Use of microorganisms because they have an extra plasmid which only have few
genes. They function for the replication of the bacteria but they are indispensable
because the bacteria is fully-functional without it.
c. Process:
i. Decide what gene you want
ii. Cutting of genes and plasmid by restriction enzymes
iii. Fitting of DNA into plasmid
1. DNA is naturally sticky thus they will naturally stick together
2. DNA ligase is used to permanently stick the two together
iv. It is inserted again to a new host bacterial cell to clone the bacterial cells and make them
many.
❖
Polymerase Chain Reaction (PCR)
a. Definition:
i. Can be used to amplify DNA sequences to obtain millions of copies
ii. A technique to make many copies of a specific DNA region in vitro (in a test tube rather than
an organism)
iii. Repeated heating and cooling cycles allow for rapid implication of a sequence of DNA
b. Process (One cycle which takes about 1-2 minutes)
i. Denaturation (95°C): Heat the reaction to separate/denature the DNA strands without the
use of an enzyme. This provides single-stranded template for the next step.
ii. Annealing (55 - 65°C): Cool the reaction so the primers can bind to their complementary
base-pair sequences on the single-stranded template DNA and become starting points
iii. Extension (72°C): Raise the reaction temperatures so Taq polymerase extends the primers,
synthesizing new strands of DNA.
1. Taq – a marine bacterium, Thermus aquaticus, that is resistant to heat
c. This cycle repeats 25 - 35 times in a typical PCR reaction, which generally takes 2 - 4 hours,
depending on the length of the DNA region being copied. If the reaction is efficient, the target region
can go from just one or a few copies to billions.
d. i.e. Forensics – In crime scenes, we will only see a small drop of blood, saliva, or semen. So, we can
make more copies with PCR.
pg. 20
❖
DNA Fingerprinting
a. Often DNA sample must be amplified by PCR first
b. Different individuals will have different lengths of DNA between the restriction sites
c. DNA is separated by gel electrophoresis and fragment pattern is compared
d. Commonly used process for identifying process in criminal investigation
e. Process:
i. PCR
ii. DNA Fingerprinting
iii. Agarose Gel Electrophoresis – separation of DNA fragments according to size and charge
1. DNA samples are loaded into wells.
a. You inject the copies of the DNA from PCR, not the original strands of DNA.
2. Electricity is turned on and DNA fragments migrate through gel
3. The fragments are separated by size with the longest at the stop and the shorter at
the bottom because shorter strands travel faster.
a. Because DNA is negative in charge because of the phosphate backbone,
when electricity is introduced, like charges repel and they are attracted
towards the positive which is in the lower part of the mold.
b. The smaller the molecules are, the farther it will be from the wells because
the faster it will travel.
c. Gel mold: negative  positive
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pg. 21