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Biology > Consolidation 69.69 > Flashcards

Consolidation 69.69 Flashcards

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1
Q

Nucleus

A
  • Presence of nuclear envelope: store hereditary material, protect chromatin from rxn in cytoplasm
  • Nuclear Pore: regulates exchange of substances between nucleus and cytoplasm
  • Nucleolus: site of rRNA synthesis and site of assembly of proteins and rRNA to form ribosomal subunits
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2
Q

Rough endoplasmic reticulum

A
  • To transport of proteins which are synthesised by the ribosomes on its surface to the Golgi apparatus via transport vesicles to be modified and packaged

presence of membrane bound ribosomes

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3
Q

Smooth endoplasmic reticulum

A
  • Synthesis of lipids
  • Detoxification of drugs and poisons
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4
Q

Golgi apparatus

A

Consists of cisternae
* Glycosylate to proteins and lipids to form glycoproteins and glycolipids respectively
* Further modifies, sort and package glycoproteins and glycoplpids into vesicles for secretion out of the cell
* Synthesis of lysosomes

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5
Q

Mitochondrion

circular DNA 70S ribosome

A

Inner membrane is highly folded to form a cristae: to increase SA to VR to embed more electron carriers to relase more energy for ATP production during oxidative phosphorylation

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6
Q

Chloroplast

circular DNA 70S ribosome

A

contain chlorophyll which convert solar energy to chemical energy through photosynthesis via light independent reactions and light dependent reactions

contains thylakoid that stack tgt to form intergranal lamella

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7
Q

Centrioles

A

A pair of hollow cylinders made up of 9 triplets of microtubules
* act as microtubule organising centre (MTOC) during spindle formation in cell division

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8
Q

exp how CSC support the fluid mosiac model?

SIGMA IMPT

A
  • PL molecules arranged in a bilayer
    Fluid:
  • weak HPI b/w PL molecules
  • allow PL to move laterally within the bilayer
  • kinks in unsaturated fatty acid tails of the PL hinder close packing
  • chlestrol b/w the PL hinder close packing
    Moasic:
  • Proteins are embedded in the PL bilayer in a random arrangement
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9
Q

S F of transport protein allow it to carry out its function

A
  1. Protein is transmembrane (S) to provide hydrophillic channel for polar molecule to pass across bilayer.
  2. Exterior of protein has aa residues containing non-polar R groups (S) interact with fatty acid tails of PL molecules via weak HPI
  3. Contains binding site complementary to ie. glucose for specific transport of the substance
  4. glucose polar, pass via facilitated diffusion down conc. gradient

Polar molecule repelled by the hydrophobic core of PSB

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10
Q

S F of starch/glycogen

A
  • OH group are face interior of the molecule: Insoluble in water, does not exert osmotic effect on the cell
  • Glucose residues joined by alpha−1,4-glycosidic bonds: results in helical coil, more compact for storage. bond can also be hydrolysed by enzymes
  • Amylopectin highly branched: provides many sites for enzymes to act on, allows rapid release of glucose
  • Anomeric carbon involved in glycosidic bond formation: Stable compound.

MUST LINK TO therefore GOOD STORAGE MOLECULE!!!!

Amylopectin is alpha-1,6-glycosidic bond

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11
Q

S F of Cellulose

A
  • Large Molecule: insoluble in water stored without changing the osmotic potential
  • alternate beta glucose molecules inverted, forming linear chians: allows bundling into microfibrils and macrofibrils to increase tensile strength through the crosslink of H2 bonds
  • OH grp facing exterior: allow for crosslink of H2 bond b/w adjacent cellulose chains, no OH interact with water, cellulose insoluble
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12
Q

S & F of Haemoglobin

A
  1. contains 4 subunits each with haem group contains an iron (II) ion that can bind reversibly with oxygen, allows oxygen binding (in oxygenrich lungs) and release (in metabolically active tissues);;
  2. four polypeptide chains / subunits allows for cooperative binding as affinity of subunit to oxygen increases after each binding to maximize both oxygen loading and release;;
  3. hydrophilic amino acid residues present on exterior of each subunit, results in haemoglobin being soluble in red blood cell;;
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13
Q

S & F of Collagen

A
  1. every third amino acid of each polypeptide chain is glycine, has H atom as r group, small r group able to fit into centre of triple helix, allows forming of tight compact coil in tropocollagen;;
  2. extensive hydrogen bonding b/w C=O and N-H group of adjacent amino acid residues to form triple helix –> increase tensile strength
  3. covalent cross links b/w lysine residues at N and C terminus of of adjacent tropocollagen molecules in staggered arrangement , results in the formation of microfibrils –> increase tensile strength
  4. microfibrils assembled to form long collagen fibres;;
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14
Q

State the function of DNA and describe how its properties allow it to perform this function.

A
  1. DNA stores genetic information;;
  2. DNA must be chemically stable to encode information without being easily degraded
  3. DNA must be able to replicate accurately so that the information can be pass down to the next generation
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15
Q

Purpose of cell membrane

A
  • acts as a boundary between interior and exterior of the cell that is selective permeable
  • allows for compartmentalisation
  • Provides localised environment to facilitate metabolic processes occurring simultaneously
  • allows for graident across membranes for the transport of substances
  • isolate harmful substances from the rest of the cell
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16
Q

significance of mitosis

A

Produce genetically identical daughter cells with same number and type of chromosomes for:
* growth by increasing no. of cells
* regeneration and cell repair by replacing damaged cells
* asexual reporduction to produce genetically identical offspring

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17
Q

why need to regulate cell cycle?

A
  • important for growth and development of organism
  • prevent uncontrolled cell division and cancer due to dysregulation of cell cycle checkpoints
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18
Q

why is there genetic variation in organism?

A
  • Crossing over between non-sister chromatids of homologous chromosomes in prophase 1 of meiosis: new combinations of alleles
  • Independent assortment of homologous chromosomes during metaphase I & random separation of homologous chromosomes during anaphase I: different combinations of paternal chromosomes
  • Random fusion of gamete during sexual fertilisation: different genotypes in offspring
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19
Q

how does mutation lead to sickle cell anaemia

take note of the spelling of anaemia!!!

A
  • base pair substitution of the DNA coding for beta globin gene
  • Thymine replaced by Adenine
  • change in corresponding codon transcribed from GAG to GUG
  • Amino acid translated change from hydrophilic glutamic acid to hydrophobic valine
  • generates a sticky patch on the surface of the beta-chain subunit
  • low oxygen concentrations, sticky patch on haemoglobin is exposed
  • adjacent haemoglobin polymerise with each other to form crystalline structures, results in sickle shape
  • carries less oxygen, obstruct blood vessel, deprive organ of O2, lead to organ damage
  • haemoglobin also haemolyse easily.
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20
Q

fuction of glycoproteins and glycolipids

A
  • acts as cell markers in cell to cell recognition and adhesion for tissue formation
  • function as receptors for signal molecules to bind to
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21
Q

effect of non compatitive inhibitor on activity of enzyme

A
  • bind to a site away from the active site of enzyme
  • induce a change in the 3d confromation of the enzyme
  • active site no longer complimentry to the substrate, subtrate cannot bind to AS and no formation of enzyme substrate complex
  • enzyme non functional
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22
Q

competitive inhibitor on activity of enzyme

A

inhibitor has a 3d conformation complementry to the active site of the enzyme, binds to it, competes with substrate to bind to active site, however can be overcame with high concentration of substrate, maximum rate of reaction reached.

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23
Q

how increase in temp to optimum affect the temp of enzyme

A

increase temp, increase KE of enzyme and substrate, increase rate of effective collisions b/w substrate and enzyme, increase in rate of formation of enzyme substrate complex, increase in product formation

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24
Q

explain the production of okazaki fragments

A
  1. DNA molecule is antiparallel;;
  2. DNA polymerase is specific and can only catalyse the addition of incoming deoxyribonucleoside triphosphate (dntp) to existing 3’ OH end of newly synthesised DNA strand from 5’ to 3’ direction
  3. As the replication bubble opens behind newly synthesised DNA
  4. DNA replication occurs discontinuously;
  5. new RNA primer synthesised everytime forming okazaki fragments;
25
Q

role of NAD in aerobic respiration

A
  1. coenzyme of dehydrogenase
  2. reduced to form NADH;
  3. carries high energy electrons and protons from Krebs cycle, Link reaction and Glycolysis to ETC
  4. reoxidised/ regenerated NAD+ ;
  5. 3 ATP molecules per molecule of reduced NAD;

coenzyme as undergoes reduction to become enzyme to reduce others

26
Q

Explain dominant epistasis

A
  1. Phenotypic expression controlled by 2 genes
  2. Epistatic gene determines the presence of characteristic (e.g. flower colour)
  3. Hypostatic gene determines the nature of phenotype (e.g. type of colour)
  4. Presence of at least 1 dominant allele of epistatic gene mask the phenotypic expression of hypostatic gene
  5. Homozygous recessive genotype at epistatic gene locus allows the phenotypic expression of alleles at hypostatic gene locus

Phenotypic expression = Characteristic

27
Q

Explain recessive epistasis

A
  1. Phenotypic expression controlled by 2 genes
  2. Epistatic gene determines the presence of characteristic (e.g. flower colour)
  3. Hypostatic gene determines the nature of phenotype (e.g. type of colour)
  4. Homozygous recessive genotype at epistatic gene locus mask the phenotypic expression of alleles at hypostatic gene locus
  5. Presence of at least 1 dominant allele of epistatic gene results in phenotypic expression of alleles at hypostatic gene locus
28
Q

Explain duplicate recessive epistasis

A
  1. Phenotypic expression controlled by 2 genes
  2. The genes are epistatic to each other
  3. Both genes are necessary for the presence of phenotype (e.g. production of pigment)
  4. The recessive allele of either gene encodes a defective enzyme
  5. If homozygous recessive for either of the 2 genes, absence of phenotype (e.g. pigments cannot be synthesised)
29
Q

how picture of different animal bones explain darwin theory of evolution.

A
  • different animal bones even though have diff function, but have similar structural arrangement
  • originated from common ancestor
  • shows decent with modification over time by natural selection due to diff selection pressures
30
Q

adv of using molecular methods

A
  • Molecular methods are objective and quantitative as genetic code is universal
  • all life come from nucleic acids and amino acids
  • nucleotide sequence and amino acid sequence able to be accessed from electronic database for comparison and classification of all life
31
Q

Biological species concept

A
  • group of organisms able to interbreed and produce fertile viable offsprings
  • members of the same species reporductively isolated from each other
32
Q

ecological species concept

A
  • group of organisms sharing the same ecological niche
  • difference b/w species due to difference in ecological resources they depend on
33
Q

Describe the role of NADP in photosynthesis

A
  • NADP + is an activated carrier found in stroma of chloroplast
  • acts as a final electron acceptor in non-cyclic photophosphorylation
  • carry high energy electrons from light dependent stage to light independent stage **in the form of NADPH **
  • NADP + accepts electrons and hydrogens from electron transport chain to form NADPH
  • Electrons from NADPH reduces carboxyl group on 1,3-bisphosphoglycerate to glyceraldehyde-3-phosphate
  • NADPH gets reoxidised back to NADP + to participate in light-dependent reaction
33
Q

morphological species concept

A
  • group of organisms sharing similar body shape, size and structural feature
  • definition can be applied to ALL ORGANISMS
33
Q

Locus

A

position of a gene on a chromosome

34
Q

Allele

A
  • alternative form of the same gene and responsible for determining phenotype.
  • occupy the same locus of a pair of homologous chromosomes
35
Q

how molecule structure of cellulose related to high tensile strength?

A
  1. alternate glucose monomers are inverted;
  2. glucose residues linked by β-1,4-glycosidic bond;
  3. resulting in straight chains;
    related to high tensile:
  4. hydroxyl (OH) groups project outwards from cellulose chain in all directions;
  5. allows hydrogen bond to be formed b/w C=O and N-H group of adjacent amino acid residues glycine and prolin forming triple helix –> tensile strength
  6. Covalent cross-links between lysine residues at C and N terminus of adjacent tropocollagen molecules results in the formation of microfibrils
  7. microfibrils associate to form long collagen fibres –> tensile strength
36
Q

How Sickle Cell Anemia arises?

A

At low O2 conc, sticky patch on haemoglobin is exposed, adjacent haemoglobin polymerise with each other to form crystalline structures, results in sickle shape carries less oxygen, obstruct blood vessel, deprive organ of O2, lead to organ damage. haemoglobin also haemolyse easily.

37
Q

Explain the need for the tight control of the mitotic cell cycle

A
  1. To prevent uncontrolled cell division (resulting in cancer);

Tight control via checkpoints including

(at G1 checkpoint)

  1. To ensure cell enters cell cycle only in presence of/sufficient growth factors;
  2. and no DNA damage before DNA replication in S phase;

(at G2 checkpoint)

  1. To ensure all DNA are replicated completely and accurately;
  2. And appropriate cell size before mitosis occurs;

(at M checkpoint)

  1. Ensures that all chromosomes are attached to microtubules at their kinetochores before anaphase;
  2. For proper separation of chromosomes during anaphase;
  3. To ensure genetic stability during cell division;
38
Q

describe antigen presentation

A

antigen-presenting cell engulfs bacterial cell via phagocytosis forming phagosome

phagosome fuses with lysosome;

lysosomal enzymes break down antigens on surface of bacterial cell into short peptides;

antigen peptides bind to Class II Major Histocompatability Proteins (MHC) proteins;

inside the vesicle from Golgi apparatus forming peptide-MHC complexes;

peptide-MHC complexes are transported to the** cell surface** of the antigen-presenting cells for antigen presentation;

39
Q

function of compartmentalisation

A
  • Provides localised environment to facilitate metabolic processes occurring simultaneously
  • allows of development of proton gradient across membrane for the transport of of substances
  • isolate harmful substances from the rest of the cell
40
Q

Biological Species Concept

A
  1. A group of organisms capable of interbreeding;
  2. To produce viable, fertile offspring;
  3. Organisms in one species is reproductively isolated from organisms of another species;
  4. Due to reproductive barriers such as physiological or behavioural isolation
41
Q

Ecological Species Concept

A

A group of organisms sharing the same ecological niche;

And are adapted to the same set of resources; in the same habitat;

No two species can share the same ecological niche;

42
Q

Morphological Species Concept

A
  1. A group of organisms sharing a unique set of structural features;
  2. Organisms of one species is morphologically distinct from organisms of another species
43
Q

G1 checkpoint

A

To ensure cell enters cell cycle only in presence of/sufficient growth factors;

and no DNA damage before DNA replication in S phase

44
Q

G2 checkpoint

A
  1. To ensure all DNA are replicated completely and accurately;
    ​
  2. And appropriate cell size before mitosis occurs;
45
Q

M checkpoint

A

Ensures that all chromosomes are attached to microtubules at their kinetochores before anaphase;

For proper separation of chromosomes during anaphase;

To ensure genetic stability during cell division;

46
Q

Exp how binding of insulin to the insulin tyrosine kinase receptor triggers a response.

A
  1. (bind of insulin causes) intracellular receptor subunits to associate with each other to form a dimer
  2. Activates the tyrosine kinase in the receptor subunits and results in
  3. crossphosphorylation of tyrosine residues;
  4. relay protein activated after binding to activated tyrosine kinase
  5. Relay proteins phsophorylate other kinases triggers the phosphorylation cascade;
  6. which causes activation of glycogen synthase;
  7. resulting in increase glycogenesis (glucose converted to glycogen)
47
Q

How natural selection occurs?

A
  1. Mutations → genetic variation → phenotypic variation in population
  2. Environment exerting differnet selection pressure
  3. phenotype best suited is selected for
  4. individuals survive and reproduce
  5. passing on favourable alleles to offspring
  6. changing allele frequencies in gene pool over time
48
Q

allopatric speciation

A

gene flow disrupted due to geographical isolation

49
Q

sympatric speciation

A

gene flow is disrupted by physiological and behavioural isolation even though no geographical isolation

50
Q

Locus

A

position of a gene/allele on a chromosome or within a DNA molecule

51
Q

Allele

A

alternative form of the same gene and occupy the same locus of a pair of homologous chromosomes responsible for determining characteristics of a gene

52
Q

describe the role of NADP

A
  • NADP + activated carrier found in stroma of chloroplast
  • acts as a final electron acceptor and reduced to form NADPH in non-cyclic photophosphorylation
  • carry high energy electrons from light dependent stage to light independent stage in the form of NADPH
  • reduces 1,3-bisphosphoglycerate to 3-phosphoglyceraldehyde (which stores more potential energy)
  • NADP + regenerated to participate in light-dependent reaction
53
Q

State the features of stem cells

A
  • undifferentiated and unspecialised, able to undergo proliferation and self-renewal
  • able to differentiate into specialised cell types when triggered by molecular signals due to differential gene switching
54
Q

Describe the main differences between meiosis and mitosis.

A

Mitosis will result in the production of 2 genetically identical daughter cells whereas meioisis will produce 4 genetically different daughter cells

After mitosis, the daughter cells will have the same number of chromosomes as parent cell but in meiosis the daughter cells have half the number of chromosome as parent cell;

There is only 1 nuclear division in mitosis but there are 2 nuclear divisions in meiosis;

At prophase of mitosis, there is no pairing of homologous chromosomes, no chiasma formation and no crossing over but pairing of homologous chromosomes, chiasma formation and crossing over occur in prophase 1 of meiosis;

55
Q

ADvantages of using molecular methods

A
  • it is objective as nucleotide bases are universal and are common in all organisms
  • quantitative as Molecular data are easily converted to numerical form for statistical analysis
  • accessed from electronic databases for comparative study & classification of all life.
56
Q

advantages of operon model

A
  • efficient use of cellular resources, prevent wastage of resources as genes are turned on only when necessary
  • coordinated gene expression for quick response to the environment to provide a selective advantage to survive and adapt to changes in the environment.
57
Q
A

Biology (7 decks)

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