IMMS Flashcards

Question

Explain the structure of lipids What are they mainly used for? How much energy is released per gram of lipid?

Answer 1

Exist as triglycerides (3 fatty acids and 1 glycerol) - Hydrophobic. Ester bonds They are amphipathic - (tail is hydrophobic, head is hydrophilic) Mainly used for long term energy storage. Also used for protection and waterproofing 9kcal/g

Answer 2

Nucleotide- pentose sugar, phosphate and a base Purine (A+G) - 2 bonds Pyramidine (C, T, U) - 3 bonds Phosphodiester bonds Nucleoside- pentose sugar and a base

Answer 3

- peptide bond O=C-N-H The variable group (R group) determines the polarity of the amino acid.

Answer 4

Primary - The linear sequence of amino acids held together by covalent bonds Secondary - Formation of either an alpha helix (spiral arrangement) or beta pleated sheets (directionally oriented) Super secondary - combination of both secondary structures Tertiary - 3D conformation of a protein. ionic bonds, disulphide bridges Quaternary - 3D structure of protein composed of multiple subunits. Consists of more than 1 polypeptide chain linked together. (haemoglobin)

Answer 5

Enzymes are tertiary and quaternary structures(functional globular) - binds via active sites to the substrates Co-enzymes bind to enzymes to aid their function

Answer 6

2 alpha and 2 beta chains Quaternary structure Mutation (in DNA NOT RNA) from GAG (glutamic acid) --> GTG (Valine) This causes the red blood cells to take on a more sickle shape rather than round, thus decreasing the surface area for absorption. It is also less flexible and more prone to damage. It protects against malaria

Answer 7

A semi-conservative process because Every double helix consists of 1 new strand and 1 old strand Takes place in the nucleus 1) Topoisomerase unwinds the supercoil (also relieves strain) 2) DNA helicase separates the double stranded DNA by breaking the hydrogen bonds between base pairs and creates a replication fork 3) Single stranded binding proteins coat the DNA strands after separation to prevent the strands from re-annealing 4) DNA primase synthesises short RNA primers which serve as an initiation point for DNA polymerase 5) DNA polymerase attaches to the 3' end of the primer (5' end of the DNA) and synthesises a new strand of DNA in the 5' to 3' direction 6) On the leading strand it continuously synthesises the new DNA strand towards the replication fork. On the lagging strand, it builds away from the replication fork and synthesises the DNA strand in pieces called okazaki fragments 7) DNA ligase joins the okazaki fragments together to form a continuous strand by covalently joining the sugar phosphate backbones together.

Answer 8

Takes place in the nucleus 1) Transcription factors such as the TATA binding protein control the binding of RNA polymerase to the promoter region. 2) RNA polymerase unwinds and separates the DNA strand into the antisense strand and sense strand 3) As RNA polymerase moves along the template strand from the promotor region, free mRNA nucleotides line up next to their complementary bases and an antiparallel strand is formed in the 5' to 3' direction 4) This occurs until the termination point of the gene is reached. RNA polymerase is released and the double strand DNA reforms into its helix shape. 5) Post transcription events occur before the mRNA strand leaves the nucleus via the nuclear pore to attach to the 80s ribosome

Answer 9

Capping- adding of a methyl group to the 5' end of the transcribed RNA to prevent it from degradation Polyadenylation- adding of a Poly A tail to the 3' end of the transcript to improve stability and facilitate it export from the nucleus Splicing- non coding sequences called introns are removed and coding regions called exons are joined together to form a continuous sequence (can be in any order that exons are joined)

Answer 10

1) The small ribosomal subunit binds to the 5' end of the mRNA and moves along until it reaches the start codon AUG. The complementary tRNA then attaches to the codon via its anticodon. 2) The large ribosomal subunit aligns itself to the tRNA molecules at the P site, forming a complex with the small subunit. 3) A second tRNA molecule pairs with the next codon in the A site. The amino acid in the P site covalently attaches to the amino acid in the A site via a peptide bone. 4) The ribosome complex moves along the mRNA strand in a 5' to 3' direction. The amino acid in the P site moves to the E site and is released. The one in the A site moves to the P site. 5) This happens till a stop codon is reached (UGA, UAG, UAA). Here, a release factor is recruited which signals for translation to stop. The polypeptide chan in released and sent to the golgi apparatus for modification.

Answer 11

Nucleus and mitochondria (only a small amount of maternal DNA)

Answer 12

G1- Rapid cell growth, synthesis of new organelles (DNA damage is checked before DNA replication (p53, p21) S- Synthesis- DNA replication, protein synthesis, centrosome replication G2- Mitochondria double, chromosomes condense (DNA damage is checked for pre mitosis by glycosylase enzymes) G0- fully differentiated cells that don't replicate further

Answer 13

Prophase - The nuclear membrane breaks down, chromosomes condense (visible), centrosomes form microtubule spindle fibres and move to opposite poles of the cell Prometaphase- microtubule spindle fibres invade the nuclear space and attach to the centromere of the chromosome Metaphase- spindle fibres contract and chromosomes align on the equatorial plate Anaphase- continued contraction of spindle fibres separates the sister chromatids and pulls them to opposite poles of the cell Telophase- chromosomes unfold into chromatin (invisible) and the nuclear membrane reforms. Cytokinesis begins.

Answer 14

Division of the cytoplasm and cell organelles become evenly distributed around each nucleus. 2 genetically identical daughter cells are formed with a nucleus in each and 46 chromosomes.

Answer 15

Meiosis 1 - the first meiotic division is a reduction division from diploid to haploid (2n --> n) Prophase 1 --> crossing over occurs resulting in genetic diversity Metaphase 1 --> random assortment occurs further resulting in genetic diversity Meiosis 2 - the second meiotic division separates the sister chromatids and haploid cells are produced. (2 chromatids --> 1 chromatid)

Answer 16

Begins a puberty and continues after. There is an EVEN distribution of cytoplasm Spermatogonia --> primary spermatocyte --> (first meiotic division) 2 x secondary spermatocyte --> (second meiotic division) --> 4 x spermatids (immature sperm) --> they differentiate into spermatozoa. 100-200 million sperm per ejaculation Process takes 60-65 days

Answer 17

Oogenesis begins at birth and is suspended at metaphase 1 till ovulation (puberty). There is an UNEVEN cytoplasmic division. Meiosis 2 completes at fertilisation Oogonia --> primary oocyte --> (first meiotic division) secondary oocyte and polar body --> (second meiotic division) --> ootid and 3 polar bodies

Answer 18

Non disjunction - failure of the chromosomes to separate in meiosis 1 and sister chromatids to separate in meiosis 2. Leads to Trisomy 21 --> down syndrome Monosomy --> Turner's syndrome where there is only 1X chromosome in females Gonadal mosaicism --> 1 of the healthy parents has a mutated germ line. Incidence increases with paternal age. e.g. duchenne muscular dystrophy --> some egg cells may carry the duchenne gene but the mother's blood cells do not that is why when tested, she appears negative for the gene but can still pass it on.

Answer 19

Non pathogenic variation at the locus from wild type (normal alleles)

Answer 20

The reproductive union of 2 relatives

Answer 21

The percentage of people with the expected phenotype from their genotype

Answer 22

Some people with the same genotype may express it differently

Answer 23

In Huntington's disease - There is a wider expansion of trinucleotide repeats (CAG trinucleotide) --> causes mutation of HTT gene Genetic anticipation means the number of repeats get bigger when they are transmitted onto the next generation resulting in earlier symptoms of greater severity

Answer 24

Getting the same mutation (genotype of a kind) from both sides of the family

Answer 25

Homozygous - two idential alleles at 1 locus Heterozygous- two different alleles at 1 locus Hemizygous- presence of a single copy of the gene. Males only have 1 x chromosome and 1 Y chromosome (so they would only have a single copy of whatever gene was present on the chromosomes)

Answer 26

1 female x chromosome is randomly inactivated preventing 2 sets of genes being expressed in men (you want 1x and 1 Y rather than 2 X chromosomes)

Answer 27

Gene defects only affect 1 sex e.g. BRCA gene

Answer 28

There is no male to male transmission (unlike autosomal dominant) Female carriers --> pass it on to men (females don't really get it as both x genes would have to have the pathogenic variant from each parent which is very rare)

Answer 29

Only affects males. From 1 dad to all the sons

Answer 30

Carrier frequency --> 1/25 Incidence --> 1/2500 Most common autosomal recessive gene affecting (middle aged) white in the UK F508 gene defect

Answer 31

Duplication - a portion of genetic material or chromosome is duplicated resulting in multiple copies of that region Mis-sense mutation - a single nucleotide change results in a codon that codes for a different amino acid. A) silent mutation - different codon codes for the same amino acid B) can result in a non functional protein Nonsense mutation - a mutation that produces a stop codon, resulting in an incomplete usually non functional protein Deletion/Addition mutation - out of frame deletion - base is deleted, shifting the entire reading frame causing catastrophic effects. - in frame deletion - an entire codon is removed, so only 1 amino acid is lost, less catostrophic DNA damage --> UV light, radiation, chemicals Expansion of trinucleotide repeats

Answer 32

Shorter p arm Longer q arm

Answer 33

Trisomy 18

Answer 34

Carbohydrate - 4kcal/g Protein - 4kcal/g Alcohol 7kcal/g Fat - 9kcal/g 8 grams of alcohol is 1 unit (10ml).

Answer 35

They are the fat storing, stellate cells of the liver. They are the main place for vitamin A storage in lipid droplets

Answer 36

Fats- adipocytes, ito cells (liver) - triglycerides Carbohydrates - liver, skeletal muscle - glycogen Protein- muscle + liver

Answer 37

Glucose → glucose-6-phosphate via Hexokinase (but glucokinase in the liver)— ATP is converted to ADP (hexokinase minor limiting step) Glucose-6-phosphate → fructose-6-phosphate via Phosphoglucose Isomerase Fructose-6-phosphate → Fructose 1,6 bisphosphate via Phosphofructokinase-1 (1) – ATP is converted to ADP RATE LIMITING STEP Fructose 1,6 bisphosphate → split into 2, 3 carbon molecules. Dihydroxyacetone phosphate and glyceraldehyde-3-phosphate via aldolase Dihydroxyacetone phosphate must converted into glyceraldehyde-3-phosphate via triose-phosphate isomerase to be used in further reactions Glyceraldehyde-3-phosphate → 1,3- bisphosphoglycerate via glyceraldehyde-3-phosphate dehydrogenase (triose phosphate dehydrogenase) – 2NAD+ → 2 NADH + H+ 1,3 bisphosphoglycerate → 3 phosphoglycerate via phosphoglycerate kinase - 2 ADP is converted to 2 ATP (the phosphate is removed to convert ADP to ATP) 3 phosphoglycerate → 2 phosphoglycerate via phosphoglycerate mutase 2 phosphoglycerate → phosphoenolpyruvate via enolase – 2 H2O is removed Phosphoenolpyruvate → pyruvate via pyruvate kinase - 2 ADP is converted to ATP Good gracious father franklin did get by picking pickles Harry Potter’s parents are incredibly good people, pretty even powerful

Answer 38

2 Pyruvate, 2 NADH, 2 H+, 2 ATP (2H2O)

Answer 39

Glucose stimulates hexokinase Glucose-6-phosphate inhibits hexokinase (allosteric inhibition- binds to non catalytic site causing conformational changes) Insulin increases synthesis of hexokinase, increasing glycolysis Glucagon inhibits hexokinase, decreasing glycolysis

Answer 40

Phosphofructokinase-1 (PFK-1) - most highly regulated step in glycolysis ATP allosterically inhibits PFK-1 (too much ATP formation signals to reduce glycolysis) Acidosis also inhibits it. AMP is an activator of PFK-1 (When ATP is used up, ADP accumulates and is converted into AMP by adenylate kinase) Citrate (first product of the kreb’s cycle) allosterically inhibits PFK-1 (Kreb’s cycle is too active) Fructose -2,6- bisphosphate - generated from fructose 6 phosphate allosterically activates PFK-1 → converts more fructose -6 -phosphate to fructose 1,6 bisphosphate Insulin stimulates the conversion of fructose 6 phosphate to fructose 2,6 bisphosphate Glucagon inhibits this

Answer 41

1,3 bisphosphoglycerate - phosphoglycerate kinase - 3 phosphoglycerate Phosphoenol pyruvate - pyruvate kinase - pyruvate

Answer 42

Pyruvate is converted into lactate by lactate dehydrogenase The hydrogen that is added to the pyruvate is derived from NADH + H+ → NAD+ NAD+ produced funnels back to be used in glycolysis to oxidise more glyceraldehyde-3-phosphate (occuring in the cytosol) → if NAD+ is in short supply, glycolysis cant continue 2 lactate + 2ATP + 2H+ (NAD is recycled to be used in glycolysis)

Answer 43

Pyruvate produces more energy for the cell via decarboxylation and oxidation (removal of H+ picked up by NAD+) forming an acetyl group which combines with coenzyme A to form Acetyl-CoA which can enter the TCA (tricarboxylic acid) cycle for more energy production (in the mitochondrial matrix) Pyruvate dehydrogenase aids in this process to produce Acetyl-coA + CO2 + NADH + H+

Answer 44

Oxaloacetate + Acetyl coA --> citrate --> isocitrate --> alpha ketoglutarate (NAD+ --> NADH + H+) + CO2 --> succinyl CoA (NAD+ --> NADH + H+) + CO2--> Succinate (GTP-->GDP) --> Fumarate (FAD --> FADH2) --> Malate --> Oxaloacetate (NAD+ --> NADH + H+) CAN I KEEP SOME SUCCINATE FOR MY OXALOACETATE Enzymes Citrate synthase --> aconitase --> isocitrate dehydrogenase --> alpha ketoglutarate dehydrogenase --> succinyl coA synthetase --> succinate dehydrogenase --> fumarase --> malate dehydrogenase Citrate And Isocitrate Keep Some Substrate For My Oxaloacetate Take note Acetyl coA can also be produced from beta oxidation of fatty acids Alpha ketoglutarate can be formed from the oxidative deamination of glutamate

Answer 45

Allosterically inhibited by ATP and NADH Competitively inhibited by succinyl CoA Citrate inhibits Citrate synthase ADP activates it

Answer 46

ADP activates it ATP and NADH inhibits it

Answer 47

ATP, Succinyl CoA and NADH inhibits alpha ketoglutarate dehydrogenase Ca2+ stimulates it

Answer 48

Glycolysis - It is used as an emergency energy producing pathway when oxygen is limiting (it produces a small amount of ATP) and it generates precursors for biosynthesis Krebs cycle- used to produce NADH and FADH2 used in oxidative phosphorylation

Answer 49

1) NADH and FADH2 releases H+ and Electrons. 2) Energy from the electrons in the electron transport chain is used to pump H+ across the mitochondrial membrane. 3) H+ ions return via ATPase channels, triggering molecular rotation and ATP production. Hydrogen ions join with oxygen to form water. Oxygen is the terminal electron acceptor.

Answer 50

Fatty acids are converted into Acyl adenylate (ATP --> ADP) --> converted into Acyl CoA via acyl CoA synthetase --> shuttled into the mitochondrial matrix via the carnitine shuttle (rate limiting step) --> carnitine shuttle used if fatty acid chain is longer than 12 carbons. --> acetyl coA is then produced after a series of reactions (oxidation, hydrolysis, oxidation, thiolysis).

Answer 51

When there is excess acetyl Coa, acetylcoA is converted into ketones such as Acetate, acetoacetate, etc. (inactive storage form) (the brain can adapt to use ketones for energy) When acetyl coA is needed again, ketones are converted back to AcetylCoA

Answer 52

Occurs when there is a high concentration of ketones in the blood (low insulin level very high blood sugar- insulin causes the uptake of glucose into cells to be used for energy, without insulin there is a lack of energy so the body breaks down fat for fuel --> ketogenesis via acetyl CoA )

Answer 53

pH = 6.1 + log10[ HCO₃ / (0.03× PaCO₂) ] pH= pKA + log10 {A-/HA]

Answer 54

- O2 transport -CO2 transport to the lungs - Mops up excess H+ as a buffer to prevent conditions from getting too acidic - Nitric oxide transport around body for vasodilation

Answer 55

Under 7.35 pH = acidosis Over 7.45 pH = alkalosis

Answer 56

ROME - Respiratory opposites, metabolic equal (pH and HCO3 goes in the same direction) Metabolic acidosis - - pH low, HCO3 low Compensatory --> deep hyperventilating, Co2 is low Metabolic alkalosis -- pH high, HCO3 high Compensatory --> hypoventilation, Co2 is high Respiratory acidosis -- pH low, CO2 high Compensatory --> Increased renal reabsorption of HCO3, high HCO3 Respiratory alkalosis-- pH is high, CO2 low Compensatory --> Increased excretion of HCO3, low HCO3 80% of bicarbonate is reabsorbed in the PCT

Answer 57

[Na+ + K+] - [Cl- + HCO3-] 10-16

Answer 58

They can damage cells and DNA. Very dangerous. Function- Respiratory burst (immunological defence mechanism) It is the rapid release of reactive oxygen species, predominantly from neutrophils for pathogen killing.

Answer 59

Cytosine methylation - causes transcriptional silencing Acetylation of the histone tail makes DNA less tightly coiled and increases transcription Methylation of histone tail makes DNA more tightly coiled and decreases transcription Light, pH and temperature can also affect gene expression

Answer 60

6 NADH 2 FADH 2 ATP

Answer 61

36-38 NADH - 3 molecules FADH2- 2 molecules

Answer 62

Major - PFK-1 Minor- hexokinase, pyruvate kinase The processes catalysed by hexokinase, PFK-1 and pyruvate kinase Hexokinase - glucose stimulates, glucose-6-phosphate inhibits PFK-1 - AMP stimulates it, ATP, fructose -2,6-bisphosphate and citrate inhibits it Pyruvate kinase -ATP and acetylCoA inhibits it

Answer 63

Major - isocitrate dehydrogenase Minor - Citrate synthase, alpha ketoglutarate Isocitrate dehydrogenase - ATP and NADH inhibits it, ADP activates it Citrate synthase- ATP and NADH, Citrate and succinyl CoA inhibits it. ADP activates it Alpha ketoglutarate- ATP, succinyl CoA and NADH inhibits it, calcium activates it

Answer 64

The sum of chemical reactions that take place within each cell of a living organism that provide energy to maintain life.

Answer 65

Autosomal recessive

Answer 66

Huntington's disease

Answer 67

A DNA sequence variation occurring when a single nucleotide (i.e., A, T, C, G) in the genome sequence is altered, with the alteration present in at least 1% of the population