NS2: BBCF

Question 1

Pathway of air

negative pressure respiration

Answer 1

Pathway of air: nostrils or oral cavity > pharynx > passes epiglottis and enters larynx > trachea > bronchi > bronchioles > alveoli

Negative-pressure respiration: during inhalation, the diaphragm contracts and thoracic cavity expands, causing a drop in the pressure in the lungs that pulls external environmental air to rush in thru the respiratory tract
– During exhalation, the relaxation of the diaphragm is enough for contract the volume of the lungs, increasing the pressure inside and expelling air

Question 2

lacteals

Answer 2

The lacteals are structures in the intestines associated with absorbing fat into the lymphatic system.

Question 3

Residual volume

Answer 3

minimum volume the lungs can have, and which is attained at maximum intrapleural pressure

Question 4

define the following:

• aneuploidy
• monosomy
• trisomy
• nondisjunction

Answer 4

Aneuploidy results from having too many or too few copies of a given chromosome. This results from nondisjunction in anaphase during cell division. – Having only one copy of a chromosome is known as monosomy, and having three copies is known as trisomy.
– Autosomal monosomies are not viable; trisomies are.

Nondisjunction during meiosis is the cause of aneuploidies such as Down syndrome (trisomy 21) or Turner syndrome (monosomy X). However, nondisjunction during mitosis can also occur, and this is extremely common in cancer cells.

Question 5

Chromosome arms

Answer 5

In eukaryotes, DNA is organized into linear chromosomes, which each contain a specific part of the genome. (In contrast, prokaryotes have a smaller genome that is stored in circular DNA.)

Each chromosome contains a centromere that separates the chromosome into two arms, generally known as the p-arms and the q-arms. These arms are used as part of the information needed to specify the location of a gene on a chromosome.

Question 6

Chromsome pairs

Answer 6

Somatic cells contain two copies of each chromosome, one inherited maternally and the other paternally; thus, in humans, we have 22 pairs of distinct chromosomes called autosomes and then 1 pair of sex chromosomes (females are XX and males are XY).

In total, humans have 46 chromosomes, w two copies of the 22 autosomes and then either two X chromosomes or one X chromosome and one Y chromosome.

Question 7

Robertsonian translocation effects on chromosomes

might not be needed for memorization?

Answer 7

While other people usually have 46 chromosomes, Robertsonian translocation carriers have 45. We can also see that in Figure 1, in order to create ROB two chromosomes are combined and create two more, one of which is lost. Thus we have a net loss of 1 chromosome.

Question 8

Phases of the cell cycle

Answer 8

divided into a resting phase (interphase) and cell division (mitosis or meiosis)

Passive Interphase → G0: normal cell activity; not preparing for cell division

Active Interphase: cell prepares for division via growth and DNA replication; takes up 90% of cell cycle

• • G1: cell grows
• • G1/S checkpoint: restriction point where cell commits to division
• • S: synthesis; chromatin is loosely packaged (euchromatin) to allow transcription and replication of DNA
• • G2: cell grows again
• • G2 checkpoint: right before cell division; checks for DNA damage after replication → if damaged, will “pause” cell division until damage is repaired

Question 9

Types of Cell Division

Answer 9

Mitosis: asexual replication where a diploid parent will yield two diploid daughter cells

Meiosis: sexual replication where a diploid parent will yield four haploid daughter cells (gametes); results from two-stage process with no replication in btwn

Question 10

Histones

• function
• composition

Answer 10

Histones have a (+) charged tail that interacts with the (-) charged DNA. This salt-bridge interaction causes DNA and histones to be tightly wrapped around each other forming heterochromatin and with LESS transcription. This complex formed by DNA and a histone is known as a nucleosome.

Histones are composed of various subunits known as H1, H2A, H2B, H3, and H4.
Core is made of two dimers of H2A and H2B, and a tetramer of H3 and H4, while H1 serves as a linking unit btwn the two

Question 11

effect of acetylation (of histone tail) vs. effect of methylation (on DNA)

Answer 11

Acetylation of the histone tail would get rid of the (+) charged histone tail and replace it with a acyl group. This DISRUPTS the salt-bridge interaction and now the DNA-histone complex opens up forming euchromatin allowing for MORE transcription.

Vs. Methylation of DNA makes it harder for transcription factors to bind, thus resulting in less transcription

Question 12

two forms of chromatin

Answer 12

Euchromatin is a loose configuration that is difficult to see under light microscopy and allows DNA to be readily transcribed; this is the form that DNA generally exists in during interphase.

Heterochromatin is the tightly coiled, dense form of chromatin that is visible during cell division and is present to a lesser extent even during interphase.

Question 13

what direction is the amino acid structure of a protein written

Answer 13

The amino acid structure of a protein is conventionally written from the N-terminus to the C-terminus, thus numbering begins at the N terminus.

aka 5’ to 3’ direction

Question 14

post transcriptional modifications

vs. protein modifications

Answer 14

Post-transcriptional modifications (those performed on mRNA) occur in the nucleus; these include the addition of the poly(A) tail, the addition of the 5’ cap, and splicing.

Vs. Protein modifications are post-translational modifications, which can occur in a variety of locations.

Question 15

conserved sequences

Answer 15

DNA sequences that are common among different species, phyla, or even kingdoms are called conserved sequences. Conserved sequences tend to remain that way due to the fact that they code for a vital function that is common among disparate species.

Question 16

anomers

Answer 16

Anomers are a form of epimer. Epimers are sugars that differ only at one stereocenter, with all other stereocenters being exactly the same. Thus, epimers are a type of diastereomer.

Question 17

elcosanoids
vs catecholamines

not sure if needa memorize for MCAT

Answer 17

Eicosanoids: large family of lipids derived from arachidonic acid, a 20-carbon omega-6 polyunsaturated fatty acid with four cis double bonds – thus, eicosanoids have 20 carbons and the additional characteristic feature of a five-carbon ring → function as signalling molecules in the body

• • Prostaglandins: involved w inflammation
• • Thromboxanes: involved w clotting cascade
• • Leukotrienes

Catecholamines: class of molecules derived from tyrosine

Question 18

Importance of Vmax

Answer 18

Vmax represents the rate that is reached when the enzyme is saturated with substrates. When the enzyme is not yet saturated, the rate can reasonably fall along a range of values that are all less than Vmax, even zero. However, even if the enzyme is oversaturated, the rate of reaction cannot exceed Vmax.

Question 19

Splicing in bacteria … does it happen or na

explain ur answer

Answer 19

Bacteria, aka a type or prokaryote, have genes that do not undergo splicing, which is the removal of introns. This is because bacteria already lack introns – thus, a large, eukaryotic gene without introns suggests a bacterial origin.

Question 20

when will cell membrane transport be affected

Answer 20

Membrane transport is most likely to be affected if the disruption occurs in components that span the entire membrane.

Transmembrane proteins are components that pass all the way through the cell membrane and facilitate membrane transport.

Question 21

cardiac contraction

• nodes
• pathway
• other structures

Answer 21

cooperation btwn various nodes that send out action potentials and the muscles that respond

Sinoatrial node (SA): aka pacemaker; located above right atrium; action potential propagates thru tissue via gap junctions, causing contraction of atria

Atrioventricular node (AV): receives action potential from SA and passes it thru ventricles after atria have contracted; takes over if SA node is damaged

• • Bundle of His: enhances the speed of AV action potential
• • Purkinje fibers: spreads action potential to all muscle cells of the ventricles

Question 22

Why are electrical signals faster than chemical ones?

Answer 22

Electrical signals can propagate quickly across gap junctions (about 2-3 nm distance); chemical signals must diffuse across chemical synapses (20-40 nm), which take longer.

Question 23

Allosteric inhibitors

Answer 23

Allosteric inhibitors (aka noncompetitive inhibitor) bind to an allosteric site on the enzyme that prevents the binding of the substrate to the enzyme site.

Question 24

define the following:

• promoters
• enhancers
• silencers

Answer 24

Promoters: regions of DNA that lie upstream to a given gene and initiate transcription by binding specific transcription factors that contribute to the binding of RNA polymerase

Enhancers: DNA sequences that can be located further from the gene of interest, and work by binding transcription factors that twist DNA into a hairpin loop, bringing distant regions into close proximity for transcription to begin – acts to upregulate expression of a certain gene

Silencers: opposite of enhancers; regions of DNA to which transcription factors known as repressors bind

Question 25

Define the following

• siRNA

- miRNA

Answer 25

both silence genes by interrupting expression between transcription and translation, thus preventing translation of mRNA into its corresponding protein

siRNA (small interfering RNA) requires perfect complementarity and will cleave it’s target every time.

miRNA (microRNA) can bind without being exactly complimentary but will only prevent translation. If the miRNA can bind perfectly it will also cleave the target

Question 26

Diabetes

• what is it
• types

Answer 26

Diabetes (mellitus): results from the dysregulation of insulin; commonly associated with high blood glucose levels

Type 1: caused by an autoimmune attack on the pancreatic beta cells, which produce insulin
– Individuals are dependent upon the administration of exogenous insulin, because they no longer produce the hormone in sufficient quantity to properly regulate blood glucose levels

Type 2: results from a more gradual breakdown of the degree to which target cells respond to insulin signaling, resulting in insulin insensitivity
– Individuals are initially treated with dietary modifications and/or anti-hyperglycemic medications, but they may eventually require insulin treatment as well.

Question 27

Diabetes: what happens elsewhere in the body

Answer 27

High blood glucose levels may lead to excess glucose in the urine

High BG levels can contribute to the accumulation of acetyl-CoA

• • Excess acetyl-CoA molecules cannot be shunted into the citric acid cycle because the intermediaries of the citric acid cycle, especially oxaloacetate, have been siphoned off to gluconeogenesis
• • This excess can be used to produce acetone, D-β-hydroxybutyrate, and acetoacetate, which are known as ketone bodies.

The latter two compounds are both acidic, meaning that when present in the blood at an excessively high level, they can cause the blood pH to drop, resulting in a condition known as ketoacidosis. In patients with underlying diabetes, this condition is known as diabetic ketoacidosis.
– Ketoacidosis can be smelled on a patient’s breath, because acetone accumulates to a noticeable level. It smells fruity.

Question 28

Types of Biochemical Analytical Techniques:

• Immunoassays
• Hybridization
• Electrophoresis

Answer 28

Immunoassays operate on the principle that antibodies cultured from antibody-producing cells can bind with great specificity to a protein antigen of interest.

Hybridization: binding of complementary nucleic acid strands

Electrophoresis: movement of molecules toward a charged electrode

Question 29

Types of Biochemical Analytical Techniques:

- ELISA (what is it + process)

Answer 29

ELISA (enzyme-linked immunosorbent assay) uses a solid-phase enzyme immunoassay that detects the presence of an antigen

1) First, a sample containing an unknown amount of antigen is applied to a solid-phase supporting structure.
2) After the antigen contained in the sample becomes attached to the solid support, a specific detection antibody is applied to, and binds with, the antigen.
3) The antibody is then covalently linked to an enzyme directly or through a secondary antibody that is conjugated with an enzyme.
4) Following addition of the enzyme’s substrate, a reaction occurs that produces a visualizable signal.
- - The intensity of this signal is related to the quantity of protein antigen present in the original sample.

Between steps, the plate is often washed with a detergent to rinse unbound proteins or antibodies.

Question 30

Types of Biochemical Analytical Techniques:

• Southern blotting
• Western blotting
• Northern blotting

Answer 30

Southern blotting: technique used to identify specific DNA sequences

Western blotting: identify protein sequences

Northern blotting: identify RNA sequences

Question 31

Types of Biochemical Analytical Techniques:

• PCR
• Edman degradation

Answer 31

PCR (polymerase chain reaction): essentially “laboratory DNA replication,” and uses a thermostable DNA polymerase and successive cycles of denaturation, annealing of primers, and extension of a new complementary strand to produce many copies of a sequence of interest

Edman degradation: technique used to sequence proteins via successive cleaving of terminal amino acid residues

Question 32

what can pass the blood brain barrier

- what’s the issue w this selectivity (2)

Answer 32

Only small molecules or any other nonpolar substance that is essential to brain function would be able to pass the blood brain barrier. Amino acids is one of these as they are necessary for the production of proteins.

Small bacteria can proliferate in the brain but medication to eradicate them would be stopped at the BBB.

Carbon dioxide (shown below) is both small and nonpolar, making it highly lipid-soluble. Thus, it will freely pass through the BBB and build up can lead to carbon dioxide poisoning.