Download Biology Section 1 - Altius Test Prep

Biology Section 1
The Cell

Organelles:
o
Nucleus: Where the DNA is; DNA cannot leave the nucleus (Q1. Although a small amount of
non-nuclear DNA is found in the
); surrounded by a double bilayer membrane, one bilayer being continuous with the ER; contains nuclear pores; nucleolus = site of rRNA
transcription and ribosome assembly.
o
Rough ER (RER): Covered with ribosomes that are actively translocating proteins into the ER lumen
as those proteins are translated; all proteins bound for the ER itself, the Golgi, lysosomes,
endosomes, the plasma membrane, or secretion outside of the cell (plus some proteins bound for
other organelles), are made at the RER. All proteins bound for the cytosol, plus some proteins bound
for other organelles, are made on free-floating ribosomes in the cytosol. Post-translational
modification starts here (disulfide bonds, glycosylation, etc.) and continues in the Golgi.
o
Smooth ER (SER): Lipid synthesis/modification (but NOT lipid metabolism; many students get this
confused. Lipids are made at the ER, but metabolized in the mitochondria).
o
Golgi Apparatus: The cellular “post-office” for proteins; organize, continue post-translational
modification, excrete in vesicles bound for the plasma membrane, back to the ER, or to organelles.
o
Mitochondria: Q2. Know the structure of a mitochondrion and be able to draw and label one.

Mitochondria have their own DNA and variations to the nuclear genetic code; mitochondrial
genes are passed down through the maternal line only.

The Endosymbiotic Theory suggests mitochondria evolved from aerobic prokaryotes that were
engulfed by a larger “host” prokaryote. The two bacteria began to form a symbiotic relationship.
The presence of the double bilayer membrane, the fact that mitochondria have their own DNA,
that they replicate their own DNA, and that they divide and replicate much like a bacterium, are
all pieces of evidence in support of this theory.

Q3. How do the pH values of the matrix and the intermembrane space compare?

Q4. An MCAT favorite is to suggest a disease or condition that interrupts the gradient across the
inner mitochondrial membrane. Predict the consequences of the insertion of hydrogen channels
within either of the two mitochondrial membranes.
o
Centrioles/Centrosome: The centrosome is an amorphous area of proteins and nucleating factors
within which the centrioles are located. It organizes microtubules, flagella, and cilia; it also plays a
key role in cell division.
o
Lysosomes: pH of 5, digest cell parts, fuse with phagocytotic vesicles, participate in cell death
(apoptosis), etc.; lysosomes form by budding from the Golgi.
o
Peroxisomes: self-replicate, detoxify chemicals, participate in lipid metabolism.
197 | P a g e
Biology1 Altius
Sample MCAT Question
1) A lab worker must inject a segment of DNA into the nucleus of a living cell. To access the nuclear
lumen, the microscopic needle must pierce a minimum of how many layers of lipid membrane?
A)
B)
C)
D)
2
3
4
6
Solution: To enter the nuclear lumen, the needle must pass through the cell membrane (2), plus the outer nuclear membrane
(2), plus the inner nuclear membrane (2), for a total of six single layers of lipids. Answer D is thus correct.

The Cytoskeleton:
o
Microtubules, Intermediate Filaments & Microfilaments

o
Q5. Provide a conceptual definition for each of the following: tubulin, microtubule, cytoskeleton,
spindle apparatus, actin, microfilament, intermediate filament, and myosin. Is myosin a
microfilament?
Flagella vs. Cilia: Q6. Highlight the key differences in structure and function between the two.

In humans, cilia are found exclusively in the:



o

Respiratory System (lungs)
Nervous System (ependymal cells)
Reproductive System (uterine tubes)

Microtubules are found in all the places listed above (because cilia contain them), in the flagella
of sperm, and in all cells as part of the cytoskeleton and spindle apparatus.

Q7. What problems would a disease that prevented microtubule production cause?
Eukaryotic vs. Prokaryotic Flagella

Eukaryotic = whipping motion; microtubules made of tubulin

Prokaryotic = spinning/rotating motion; simple helices made of flagellin
The Plasma Membrane:
o
Cell Membrane Structure & Function

Q8. Provide a conceptual definition for each of the following: phospholipids, integral proteins,
transport proteins, surface proteins, membrane receptors, cholesterol, fluid mosaic model,
exocytosis, endocytosis, phagocytosis, and pinocytosis.
198 | P a g e
Biology1 Altius
o
Types of Membrane Transport
1)
Simple Diffusion: No ATP required
2)
Facilitated Diffusion: No ATP required


3)
Active Transport: ATP required

4)

Osmosis: Diffusion of water across a semi-permeable membrane
Q9. Distinguish between a hypertonic, hypotonic, and isotonic solution.
Active transport is always required to move something against its concentration
gradient or against an electrical potential.
Secondary Active Transport: no direct coupling of ATP required
Cellular Junctions:
o
Tight Junctions: Water-proof barriers
o
Gap Junctions: Tunnels between adjacent cells allowing exchange
o
Adherens Junctions: Strong mechanical attachments
o
Desmosomes: The strongest of the cellular junctions; they weld cells together, protecting against
stress, but are NOT watertight (only tight junctions provide a watertight barrier).
Q10. Provide an example of where each of the above junctions is utilized in the human body.

Tissues and Intercellular Communication

Tissue Types:
o
o
o
o
Epithelial
Nervous
Connective
Muscle
Q11. Provide multiple examples of each of the above tissue types throughout the body. What
type of tissue is blood? What type is the dermis? What type are adipocytes?

o

Tissue Organization: Organ Systems > Organs > Tissues > Cells
Cell Communication: Cells, tissues, and organ systems can communicate with each other in several
ways:
1)
Endocrine: Hormone signaling. Hormones are manufactured and secreted by cells in the endocrine
glands, travel in the bloodstream, then bind to receptors either on the cell surface (in the case of
water-soluble hormones) or inside the cell (in the case of lipid-soluble hormones). We will cover
hormones in much greater detail in the Biology 3 lesson.

Second Messenger Systems: Water-soluble hormones or signaling molecules bind to
membrane receptors on the external surface of the plasma membrane. Only rarely does this
binding initiate immediate and direct response in the target cell. More often, it initiates a
cascade of events that magnifies the signal and, after multiple steps, stimulates the target
cellular response. G-proteins are part of a very common second messenger system that
has shown up on the MCAT multiple times.

Q12. Give a generalized description of a G-protein cascade. Include terms such as Gprotein-coupled receptor (GPCR), alpha/beta/gamma subunits, GDP, GTP, adenylyl cyclase,
cAMP, and Protein Kinase A (Hint: Many enzymes are activated by phosphorylation).
199 | P a g e
Biology1 Altius

Intracellular Receptors: Lipid-soluble hormones (e.g., steroids) do not require a plasma
membrane surface receptor. They dissolve through the membrane and bind targets in the
cytosol. In most cases, their activated target then acts inside the nucleus on the promoter region
of a gene.
2)
Paracrine: Signal molecules secreted by one cell bind to receptors on other cells in the local area.
Neurotransmitters acting in the synaptic gap are an example of a paracrine response.
3)
Autocrine: Signal molecules secreted by a cell bind to receptors on that same cell.
4)
Intracrine: Signal molecules (usually steroids) bind to receptors inside the same cell that produced
them, without ever being secreted outside of the cell.
5)
Juxtacrine: Signaling requires direct contact between two cells.
6)
Nervous System: Communication between cells via electrical potentials carried on neurons. Notice,
however, that these often involve cascades too. Neurotransmitters bind to receptors on the postsynaptic membrane which initiates a signaling cascade.
Cell Cycle and Cell Division

The Cell Cycle:
o



Q13. Draw and describe the cell cycle pie chart—include G1, G2, Go, S and M phases.

Go is of particular significance. Q14. Why? Which cells are suspended in Go?

Q15. How many chromosomes do humans have?: a) before replication, b) after replication, c)
during interphase, d) before S-phase, e) after S-phase, f) in a diploid cell, g) in a haploid cell.
Describe how the mass of DNA differs for each of the above scenarios.

Apoptosis: Programmed cell death featuring autolysis of cell contents by lysosomes. Initiated in
cells exposed to extreme heat, radiation, viral infection, DNA damage, etc. or to remove healthy
but unwanted cells (e.g., interdigital tissues during embryological development of the fingers).
Chromosomes:
o
The major purpose of chromosomes is to efficiently package the very, very long DNA strands so they
can easily be stored between divisions and moved during division.
o
Q16. Provide a conceptual definition for each of the following: histones, nucleosomes, chromatin,
diploid, haploid, homologues, sister chromatids, centromere, and kinetochore.
o
Q17. Where is DNA found in the cell?
Mitosis:
o
Q18. Draw and describe the following: prophase, metaphase, anaphase, telophase, spindle
apparatus, centriole, and spindle fibers. Be able to identify the events of each phase, a picture or
diagram of each phase, and the number and type of chromosomes in each phase.
o
Mitosis Yields: Two (2) genetically identical, diploid daughter cells, which are also genetically
identical to the mother cell that produced them. The centromeres split.
Meiosis:
o
Q19. Draw and describe Prophase, Metaphase, Anaphase, and Telophase for both Meiosis I and
Meiosis II.
o
Q20. What is nondisjunction and when can it occur? What are the ramifications?
o
Q21. When does crossing over occur and why is it important? (Very frequent MCAT topic!)
o
Meiosis Yields: Four (4) genetically distinct, haploid daughter cells! Centromeres do NOT
split during meiosis I, but do split during meiosis II.
200 | P a g e
Biology1 Altius
Sample MCAT Question
2) A karyotype is somewhat like a photographic list of all of the chromosomes found in a cell. If
homologous pairs are present, they appear next to one another on the exposed film. All of the
chromosomes for the entire cell are presented on the same slide, making differences in the relative
length, size and orientation of the chromosomes readily apparent. If a karyotype were produced for a
human cell that had just undergone Telophase I of Meiosis, the entire karyotype slide should contain
(assume that the chromosomes are still condensed and have not reverted back to chromatin):
A)
B)
C)
D)
23 pairs of homologous chromosomes, all of similar size
46 chromosomes, increasing in size from 1 to 23
92 chromosomes, decreasing in size from 1 to 92
23 chromosomes, decreasing in size from 1 to 23
Solution: Meiosis I takes a cell with 23 pairs of homologous chromosomes, or 46 total chromosomes, and creates two cells,
each with 23 non-paired, non-homologous chromosomes. You should also know that chromosomes generally decrease in
size, with chromosome One being by far the largest. Thus answer D is the best answer.
Molecular Genetics

DNA: A polymer of deoxyribose nucleotides
o
o
Nucleotides:

Q22. What are the three components of a nucleotide and the bonds, elements, and connectivity
involved? Draw an example of a nucleotide.

Q23. What other biomolecules, besides DNA and RNA, are nucleotides?

Q24. Draw a DNA nucleotide. Label the phosphate, sugar and base moieties.
Four DNA Bases:

Adenine (A), pairs with T via two hydrogen bonds

Thymine (T), pairs with A via two hydrogen bonds

Cytosine (C), pairs with G via three hydrogen bonds

Guanine (G), pairs with C via three hydrogen bonds
o
Q25. Draw a short DNA helix. Include the bases, sugar-phosphate backbone, and exact connectivity
(i.e., Which element bonds to which, via which type of bond?).
o
Q26. Which of the four DNA bases are purines and which ones are pyrimidines?
201 | P a g e
Biology1 Altius


Replication:
o
Q27. Provide a conceptual definition for each of the following: origin of replication, bidirectional,
semi-conservative, semi-discontinuous.
o
Q28. Tell the story of DNA replication from beginning to end. Include the terms origin of replication,
primer [NOT promoter], lagging strand, leading strand, Okazaki fragments, helicase, RNase H,
sliding clamp, primase, single-strand binding proteins, DNA polymerase, and DNA ligase.
o
Telomeres:

Every time a DNA strand is replicated, the new strand is always slightly shorter than
the parent strand. Q29. Why? What causes the daughter strand to be shorter?

Telomeres are long sections of repetitive DNA nucleotides found at both ends of each
chromosome. They provide a buffer region of non-coding DNA so that these repetitive losses in
length do not affect a gene sequence. Approximately 50 replication cycles will consume the
entire telomere region and any subsequent replications will result in the loss of gene sequences.

Telomerase is an enzyme that adds length to the telomeres. Q30. Telomerase is active in
somatic cells early in development, but is turned off in somatic cells thereafter. Why would a
mature somatic cell with telomerase activity be potentially harmful?
DNA Damage & Repair:
o
o
Causes of DNA Damage:

Spontaneous Hydrolysis: DNA reacts in solution without external stimuli or chemicals. For
example, amine groups on DNA bases can react with water to form a carbonyl; via hydrolysis
the entire DNA base can be replaced by a hydroxyl group.

Damage by External Chemicals or Radiation: When exposed to radiation, neighboring
pyrimidines react with each other to form a covalent dimer; various chemicals cause alkylation
of functional groups on the DNA base; carcinogens are often large polycyclic compounds that
bind to the DNA and create bulky side groups.

Mismatched Base Pairs: Results from errors during replication or methylation of guanine (a
specific form of methylated guanine pairs with thymine instead of cytosine).
Mechanisms of DNA Repair:

Proofreading: DNA polymerase exhibits a substantial proofreading function that catches and
repairs most mismatched base pairs on the spot.

Mismatch Repair System: Enzymes that scan newly copied DNA and locate, excise, and replace
mismatched base pairs missed by the proofreading of DNA polymerase.

Base Excision: The base portion only is excised first via a DNA glycosylase; other enzymes then
remove the sugar-phosphate backbone; then DNA polymerase and ligase replace the nucleotide.

Nucleotide Excision: Excision of an oligonucleotide that includes several bases on either side of
the error. DNA polymerase and ligase replace the missing segment.
202 | P a g e
Biology1 Altius

Recombinant DNA Techniques:
o
Molecular Cloning:

o

Polymerase Chain Reaction (PCR):

Q32. Provide a generalized step-by-step description of how to use PCR to amplify a segment of
DNA. What piece of information must be known in order to perform PCR?

Southern Blot: Used to verify the presence/absence of a specific DNA sequence. It will also
indicate the relative size of restriction fragments.

Northern Blot: Nearly identical to a Southern Blot; used on RNA instead of DNA.

Western Blot: Same basic procedure and concepts as Northern and Southern blots; used on
proteins segments instead of nucleotide segments. The probes used are radiolabeled antibodies
rather than nucleotide sequences.

Eastern Blot: Similar to a Western blot, but used to verify post-translational modification. The
probes used bind to lipids, carbohydrates and phosphates (i.e., the most common posttranslational modifications).
RNA:
o
o

Q31. Provide a conceptual definition for each of the following: restriction endonucleases,
recognition sequence, “sticky ends,” hybridization, vector, phage, plasmid and gel
electrophoresis.
Differences between RNA and DNA

RNA has a 2’ hydroxyl group; DNA does NOT

RNA is normally single-stranded; DNA is normally double-stranded

RNA contains uracil bases; DNA contains thymine bases

RNA exits the nucleus into the cytoplasm; DNA always stays in the nucleus
rRNA, tRNA & mRNA: Q33. Describe the functions of each of the three kinds of RNA. Which one
can be considered an enzyme?
Transcription:
o
Q34. Tell the story of transcription (include RNA
polymerase, promoter [not primer], pre-mRNA,
exons, introns, poly-A tail, and 5’ cap).
o
Q35. The strand of DNA that is NOT transcribed will
be an exact match to the pre-mRNA formed with
one important exception:
.
o
Post-transcriptional Processing:

Exons = exit the nucleus (coding sequences
retained in mature mRNA)

Introns = stay inside the nucleus (noncoding sequences spliced out)

Alternate Splicing: Q36. Explain the concept
of alternate splicing and how it contributes to
the diversity of gene products.
IMPORTANT NOTE
Be extremely careful with the terminology
used in reference to transcription. There are
five terms you must clearly differentiate:
“Template Strand,” “Anti-Coding Strand,”
and “Anti-Sense Strand,” all refer to the
strand of DNA that is transcribed. “Coding
Strand,” and “Sense Strand,” both refer to
the strand that is NOT transcribed. Always
remember that a transcription is NOT an
exact copy. It is complementary and
substitutes U for T. It also uses a different
sugar, ribose, instead of deoxyribose.
203 | P a g e
Biology1 Altius

Genetic Regulation:
o
Lac operon: Although human regulation is a bit more complex, you should be familiar with the lac
operon and how it regulates the expression and translation of lactase (the enzyme that digests
lactose) in bacteria.

o

Genes are regulated via three basic mechanisms:
1)
Rate of transcription: RNA has a short half-life, so gene products will only continue to be
expressed if DNA is continually transcribed.
2)
Activators and Repressors: Regulatory molecules may upregulate DNA transcription (e.g.,
lactose in the Lac Operon) or downregulate DNA transcription (e.g., glucose in the Lac Operon).
These regulatory molecules are often hormones, products of the reaction or cascade that is
catalyzed by the gene product, byproducts that build up when the concentration of the gene
product is low, etc.
3)
Permanent or Semi-Permanent Suppression: Methylation or other covalent modification that
prevents or dramatically decreases transcription.
The Genetic Code:
o
Know the start codon and the three stop codons. All other codons are NOT required knowledge and
will always be provided to you if you need them.
o
Mitochondria do NOT use the same genetic code to translate their own DNA.



Q37. If the following pairs of circumstances both exist within a bacterium, determine whether or
not the Lactase gene will be transcribed: a) glucose, no lactose, b) lactose, no glucose, c) no
glucose, no lactose, d) lactose and glucose.
Q38. The human genetic code is said to be both degenerative and unambiguous. What does this
mean?
Translation:
o
Q39. Tell the story of translation (include mRNA, tRNA, rRNA, small subunit, large subunit, initiation,
elongation & termination).
o
Q40. Where does translation occur?
o
Post-Translational Modification: Occurs at the endoplasmic reticulum and the Golgi apparatus;
usually includes addition of polysaccharides, lipids, or phosphates.
Mutations:
o
Definition: A mutation is any change in the DNA sequence.

o
Germ Cells vs. Somatic Cells

o
Q42. Why is this an important distinction to make when discussing mutations?
Chromosomal Mutations:




o
Q41. Provide a conceptual definition for each of the following types of mutations: point,
missense, neutral, silent, frameshift, and nonsense.
Duplications (non-disjunction)
Deletions (non-disjunction)
Translocations
Inversions
Cancer: Uncontrolled cell division due to failure of the cell’s normal regulatory mechanisms.

Q43. Provide conceptual definitions for each of the following terms related to cancer: malignant,
benign, metastasis, proto-oncogenes, oncogenes, tumor suppressor genes, and carcinogens.
204 | P a g e
Biology1 Altius
Sample MCAT Question
3) Assuming no errors during translation, the following segment of mature mRNA should produce a
protein with how many amino acids?
5’ AUAACGACUAUCGCCAUGCUAAUUCAUGAUGACAUCAUCCCGCAUGGAAUUCGC
UCAAACGUAUACCCAGCGCCGUAGAUCCCAGCCACUUGGGCAAAAAAA 3’
A)
B)
C)
D)
20
21
29
34
Solution: The first step is to scan the code looking for the start codon, AUG. This will be the first amino acid. Each 3member codon thereafter will produce another amino acid until a stop codon is reached, in this case, UAG. Counting the
codons between AUG and UAG (including AUG because it does code for an amino acid, and NOT including UAG because
it does not) there are 20 codons. This makes Answer A the best choice. If you included UAG, you would get 21 and if you
simply counted all of the codons without respect to start or stop codons you would get 34.
Mendelian Genetics

Terminology:
o


Q44. Provide conceptual definitions for each of the following: P1 generation, F1 generation, F2
generation, test-cross, true-breeding, Mendelian ratio, phenotype, genotype, heterozygous,
homozygous, gene, allele and locus.
Mendel’s Laws:
o
Law of Segregation: Alleles segregate independently of one another when forming gametes.
o
Law of Independent Assortment: Genes located on different chromosomes assort independently.
Making Predictions:
o
Always draw out the Punnett Square
o
Dihybrid Crosses: When two traits are being considered, draw out two independent Punnett Squares,
one for each trait. To calculate the probability of having two traits in the same individual, multiply
the individual probabilities for each trait. If the number of individuals with a specific genotype or
phenotype is asked for, multiply the total probability of having both traits by the total number of
offspring.
o
Probabilities: You must be very confident in working with probabilities. They are frequently required
to answer genetics problems and are an easy place to make errors. After reading the stem carefully
you will only have 30-40 seconds to arrive at the correct answer, so you must be relatively fast as
well.

BOTH (AND): If both events must occur simultaneously, multiply the probabilities of each event
occurring individually.

EITHER (OR): If either event occurring fulfills the requirement, add the probabilities of each
event occurring individually.
205 | P a g e
Biology1 Altius



Important Convention: Homozygous dominant is assumed for an individual with the dominant
phenotype. If an individual is a carrier for a recessive allele (heterozygote), or is affected
(homozygous recessive), that fact will always be clearly stated. Some questions may ask you to
predict the genotype of offspring when you are only given information on one of the parents. By
convention, you assume the other parent is NOT affected and is NOT a carrier.

Wild Type = The normal or typical phenotype.
Sex-Linked Inheritance:
o
Genes located on the sex chromosomes, X and Y, exhibit a unique inheritance pattern. For the MCAT
you will only need to understand X-linked inheritance. The easiest way to approach these questions
is to use a chromosome Punnett Square. The possible contributions from the father are X or Y, and
the possible contributions from the mother are X or X. Place an identifying label on any X that holds
the recessive allele. An example is provided below for an affected male crossed with a carrier female.
o
Q45. Explain why hemophilia is more common in males.
o
Q46. What fraction of an affected father’s male children will have hemophilia? What fraction of his
daughters will be carriers?
o
Have your tutor drill you on sex-linked probability problems. These are among the most difficult
genetics problems you will face and are almost guaranteed to be on your exam.
Non-Mendelian Inheritance Patterns:
o
Human genetics is usually far more complex than the simple dominant-recessive inheritance patterns
discovered by Mendel. Some human traits follow this simple pattern, but most do not. Because reallife human applications are limited, you may notice the same examples being used over and over
again in textbooks. Occasionally, MCAT questions will even feature examples that are known to NOT
follow Mendelian patterns, but the question stem will include a note that you should assume
dominant-recessive inheritance.
o
Q47. Provide a conceptual definition for each of the following non-Mendelian inheritance patterns:
incomplete dominance, co-dominance, incomplete penetrance, limited expressivity, polygenic,
pleiotropy, mosaicism, genetic imprinting, and epigenetic.
o
Linkage:

Q48. Describe the concept of linkage in terms of the distance between genes on the
chromosome. Why is relative location on the chromosome important?

Another MCAT Simplification: As far as the MCAT is concerned, any variance from expected
ratios or random assortment suggests linkage. For example, in a dihybrid cross you should get a
9:3:3:1 phenotypic ratio and in a monohybrid cross you should get a 3:1 phenotypic ratio.
Anything other than these ratios is evidence of some form of linkage. In reality, any number of
non-Mendelian factors could be at play, but based on previous AAMC questions it is clear that
when an unexpected result is obtained you had better think: LINKAGE!
206 | P a g e
Biology1 Altius
Evolution & Populations

Terminology:
o

Q49. Provide a conceptual definition for each of the following: gene pool, evolution, polymorphisms,
niche, survival of the fittest, natural selection, speciation, adaptive radiation, evolutionary
bottleneck, genetic drift and carrying capacity.
Natural Selection: This will be a frequent MCAT topic. In order for natural selection to occur:
1) One individual must have a polymorphism that provides an evolutionary fitness
advantage.
2) That advantage must result in the individual with the favored polymorphism differentially
producing more offspring.

Divergent vs. Convergent Evolution:
o
Q50. A bat and a bird both have wings. Is this an example of convergent or divergent evolution?
Sample MCAT Question
4) If a male with a sex-linked recessive disease marries a normal female:
A)
B)
C)
D)
all of his daughters will be carriers
all of his sons will be carriers
all of his sons will be affected
one-half of his sons will be affected and one-half will be unaffected
Solution: The affected male carries the defective gene on his X chromosome. When he has a son, we know for sure that he
gave that child a Y chromosome and his sons will therefore always be unaffected. Males cannot carry a sex-linked gene
without being affected, making both B and D impossible. A is thus the correct answer. When the affected male has a
daughter we know for sure he gave her his defective X chromosome. The normal woman can only give a normal X, so all
daughters must be carriers.
207 | P a g e
Biology1 Altius

HARDY-WEINBERG Equilibrium:
o
H-W Assumptions: Hardy-Weinberg equilibrium conditions almost never actually exist in any real
population. Thus, you should realize that all calculations are idealized based on the following
assumptions. The MCAT may very well ask you to conceptualize how H-W predictions will vary from
real life.
1)
2)
3)
4)
5)
o
Large population
No mutation
No immigration or emigration
Random mating
No natural selection
Formulas:

p2 + 2pq + q2 = 1

p+q=1

Q51. What do the terms p2, 2pq, and q2 represent? What do p and q represent? If 90 out of every
1,000 individuals in a population have a recessive phenotype, which of the above terms are
known, or can be calculated?
Taxonomy

Keep it simple! It would be easy to go overboard here. Taxonomy concepts are tested on the MCAT, but
they are quite rare. Focus on the following:
o
o
Classification Levels: (a.k.a., “taxa”) Kingdom, Phylum, Class, Order, Family, Genus, Species
o
How Scientists Determine Taxonomical Classifications:
Human Taxonomy: Q52. Provide the name of the group to which humans belong for each of the
classification levels listed above. For each classification level, note one or two distinguishing features
of the members of that group.

As simple as it sounds, the overarching principle is that the more two organisms have in
common with each other the more closely they should be classified. Individuals in the same
group must be more similar to each other than they are to members of different groups.

Tools: Biologists use several tools to compare organisms and make taxonomical determinations.


Embryology: Often two organisms have similarities that are only
embryological development (e.g., human embryos have tails and gill slits).
present
during

Phylogeny: A shared evolutionary history can reveal similarities.

Anatomy: Shared anatomy is a reason to classify two organisms more closely.

DNA Sequencing: Comparison of DNA sequences can reveal otherwise hidden connections or
help add legitimacy to current classification models.

Fossils: Fossil records often reveal traits that were once shared but have since been lost.
The Species Distinction: Organisms classified as different species should not be able to mate
with one another and produce viable, fertile offspring.
208 | P a g e
Biology1 Altius
Microbiology
IMPORTANT NOTE: The MCAT focuses almost exclusively on human biology. The specific microbiology topics
below are the only exceptions—and they are only included on the exam because they are pathogenic to, or
symbiotic with, humans. Some MCAT books review information on plants, photosynthesis, amphibians,
invertebrates, etc., when such topics have never been required knowledge for the MCAT. Your wisest course is to
focus all of your energy on mastering content within the boundaries we outline for you. That being said, we
remind you once again that anything is fair game as a passage topic. So, although marine animal biology is
NOT required information for the MCAT, you very well could see a passage written about a rare fish that lives at
extreme ocean depths. What you would be tested on, however, would not be prior knowledge about this rare
fish—or even fish in general—but on 1) fundamental biology concepts that could be applied to that obscure
situation, or 2) the processing, analysis and logical synthesis of information provided within the passage.

Fungi: Mushrooms, yeasts and molds.
o
All fungi are heterotrophs. Most are saprophytic (live off of dead, decaying matter), but a few are
parasitic (live off of live host, often killing it), or mutualistic (symbiotic relationship with living host,
as in mycorrhizae—a symbiosis between fungi and plant roots).

Q53. Distinguish between heterotroph, autotroph, chemotroph, and phototroph.
o
Q54. Fungi have cell
o
Fungal Reproduction:
Fungi spend the majority of their life as haploid.

Fungi grow via long, intertwining branches called hyphae. Hyphae are haploid and a mass of
hyphae is called a mycelium.

Yeasts reproduce almost exclusively by budding.

Most fungi can reproduce both sexually and asexually
Sexual Reproduction = when life is hard (stress, little food, bad environment, etc.)
Asexual Reproduction = when life is good


.




made of
Q55. Provide a possible explanation for why fungi alternate between sexual and asexual
reproduction in the manner described.
Symbiosis:
o
Q56. Provide conceptual definitions for: mutualism, commensalism & parasitism.
o
Lichen: Symbiosis between fungi and algae.
o
Mycorrhizae: Symbiosis between fungi and plant roots.
Viruses:
o
Q57. What is a virus? Is it alive?
o
Q58. List and describe the major components of all viruses. How does a virus such as HIV differ
structurally from a bacteriophage?
o
Lytic vs. Lysogenic Cycles

o
Q59. Distinguish between lytic and lysogenic viral reproduction. A man is found to be HIV
positive but has no AIDS symptoms. Is this an example of a virus in a lytic or a lysogenic cycle?
Vaccines

Q60. What is a vaccine and how does it work? Why do some vaccines tend to lose their
effectiveness from year to year and require modification?
209 | P a g e
Biology1 Altius

Bacteria: Remember that all bacteria are prokaryotes.
o
Basic structure: Capsule, peptidoglycan cell wall, plasma membrane, no complex membrane-bound
organelles, single circular DNA chromosome, tiny circular DNA molecules called plasmids.
o
Bacterial shape:

o
Q61. Describe the differences between bacilli, cocci and spirilla bacteria.
Bacteria reproduce via binary fission; NO mitosis or meiosis; distribution of extrachromosomal
DNA (i.e., plasmids) is random and daughter cells may or may not receive a copy; bacteria have
three ways of increasing their genetic variability:
1) Conjugation: One bacteria must have an F plasmid (F+); the F plasmid is a plasmid containing
the gene for a sex pilus. The recipient can be (F-).
2) Transformation: Bacteria pick up DNA from the environment.
3) Transduction: Viruses accidentally incorporate host genetic material into their nucleic acids.

Q62. What are the major differences between mitosis and binary fission?

Q63. List the primary differences between prokaryotes and eukaryotes (include ribosome
composition).
o
Bacterial Growth: Bacterial colonies grow exponentially, doubling each generation. However, there
is a limit to colony size as food and resources decrease and waste accumulates.
o
Gram Positive vs. Gram Negative

Gram Positive:





Stain purple
Very thick cell wall
Form endospores
Single cell membrane
Gram Negative:




Stain pink
Relatively thin cell wall
Do NOT form endospores
Contain two (2) cell membranes: one inside the cell wall and one outside the cell wall.
210 | P a g e
Biology1 Altius
Sample MCAT Question
5) A strain of bacteria being actively cultured in the lab develops a viral infection from the bacteriophage
L. Pylobarium. One hundred percent of the bacteria test positive for the infection. The infected colony
is plated out into separate colonies and cultured. After forty-eight hours, twenty percent of the colonies
contain at least some bacteria NOT infected with the virus, while the remaining colonies are one
hundred percent infected. This observation is best explained by which of the following?
A)
B)
C)
D)
Environmental impacts during the transfer killed some L. Pylobarium.
The immune systems of some of the bacteria were able to successfully destroy L Pylobarium.
Some bacteria developed antibodies in response to L Pylobarium.
Some bacteria underwent mutations that made them resistant to L Pylobarium.
Solution: Even if some of the virions were destroyed those remaining would continue to infect the bacteria, making Answer
A an implausible explanation. Answer B is impossible because bacteria are single-celled prokaryotes and do not have
immune systems. Answer C is false because bacteria do not develop secondary immunity and, even if they did, antibodies
are not “developed in response” to an infection. A matching antibody simply exists or does not exist as a result of mass
production of antibodies with myriad different shapes. Answer D is correct. Although it may seem that the time period is too
short for a mutation, remember that each bacterium may have replicated and divided multiple times during this period.
211 | P a g e
Biology1 Altius
212 | P a g e