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A Biology A-level > T2 - Genes and health > Flashcards

T2 - Genes and health Flashcards

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1
Q

Factors increasing gas exchange

A
  1. Surface area increases
  2. Diffusion distance decreases
  3. Diffusion gradient is steeper
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2
Q

What is Fick’s Law

A

Rate of diffusion is directly proportional to surface area X concentration difference divided by the thickness of the surface.

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3
Q

How are the lungs adapted for gas exchange?

A
  1. large surface area - many alveoli
  2. steep concentration gradient - high concentration of oxygen and low concentration of carbon dioxide is maintained by ventilation and blood circulation
  3. short diffusion distance - alveoli are just one cell thick with many capillaries around them
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4
Q

Define diffusion

A

Net movement of small, non-polar molecules (CO2/O2) from an area of high concentration to an area of low concentration.

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5
Q

Define facilitated diffusion

A

requires a channel protein in the cell membrane to transport polar, charged and water-soluble molecules across the membrane. Down the concentration gradient.

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6
Q

Define osmosis

A

Net movement of water molecules from an area of low solute concentration to an area of high solute concentration through a partially permeable membrane.

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7
Q

Define active transport

A

Movement from an area of low concentration to an area of high concentration usually via a carrier protein using energy supplied from ATP

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8
Q

Define hypertonic

A

higher solute concentration outside the cell.

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9
Q

Define hypotonic

A

lower solute concentration outside the cell

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10
Q

Define isotonic

A

equal solute concentration inside and outside the cell

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11
Q

Process of active transport via carrier protein

A
  1. The molecule/ion binds to the receptor sites on the carrier protein
  2. ATP binds to the carrier protein where it is hydrolysed to ADP and Pi (inorganic phosphate)
  3. As a result of the hydrolysis of ATP, energy is released which causes the carrier protein to change shape and open on the opposite side of the membrane
  4. Another ion binds to the second binding site, e.g. K+
  5. When the Pi (inorganic phosphate) is released, the carrier protein reverts to its original shape
  6. The second ion is released to the opposite side of the membrane.
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12
Q

Define endocytosis

A

particles are enclosed in vesicles made from cell surface membranes and transported into the cell

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13
Q

Define exocytosis

A

vesicles containing large particles are fused with the cell surface membrane and transported out of the cells

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14
Q

What leads to the symptoms of cystic fibrosis

A
  1. Cystic Fibrosis (CF) patients possess two copies of a faulty CFTR (cystic fibrosis transmembrane regulator) allele (recessive).
  2. The ribosomes of CF patients only synthesise non-functioning chloride ion channel proteins.
  3. Chloride ions can not move from within cells out into the mucus (secreted by goblet cells)
  4. Water is drawn out of the mucus into the cells by osmosis due to the solute concentration being higher within the cell (water potential of the mucus is too high).
  5. The mucus becomes very thick (viscous).
  6. Ciliated epithelial cells are unable to push the mucus out of the respiratory tract.
  7. The mucus builds up in the airways and reduces airflow in and out of the lungs.
  8. The reduced airflow leads to reduced diffusion of gases between the air in the alveoli and the blood.
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15
Q

Effects of CF on the respiratory system

A

● Build-up of mucus in the lungs traps bacteria, thus increasing the risk of infection.
● Build-up of mucus in the airways decreases the surface area of alveoli involved exposed to fresh air, therefore reducing the surface area for gas exchange, slower rate of diffusion.

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16
Q

Effects of CF on the reproductive system

A

● Cervical mucus prevents the sperm from reaching the egg.
● In men, the sperm duct is blocked with mucus, meaning that sperm produced cannot leave the testes.

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17
Q

Effects of CF on the digestive system

A

● The pancreatic duct which connects the pancreas to the small intestine can become blocked with mucus, so the digestive enzymes do not reach the small intestine. As a result food is not properly digested, so fewer nutrients are absorbed.
● The mucus lining intestine is very thick, thus reducing the absorption of nutrients.
● Mucus can cause cysts to form in the pancreas and damage the insulin-producing cells, thus leading to diabetes.

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18
Q

How are amino acids formed

A
  • contain an amine group and carboxyl group
  • joined by peptide bonds formed in condensation reactions
  • dipeptide = 2 amino acids
  • polypeptide = 3 or more amino acids
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19
Q

What is the primary structure of proteins

A

The specific sequence of amino acids in a protein

20
Q

What is the secondary structure of proteins

A
  • The 2D arrangement of the chain of amino acids
  • alpha helix or beta pleated sheet
21
Q

What is the tertiary structure of proteins

A
  • The 3D folding of the secondary structure into a complex shape.
  • The shape is determined by the type of bonding present between different R groups:
    1. hydrogen bonding
    2. ionic bonds
    3. disulphide bridges
22
Q

What is the quaternary structure of proteins

A

The 3D arrangement of more than one polypeptide.

23
Q

What are fibrous proteins

A
  • Long parallel polypeptides
  • Very little tertiary/quaternary structure
  • Occasional cross-linkages which form microfibres for tensile strength
  • Insoluble
  • Used for structural purposes - such as collagen.
24
Q

What are globular proteins

A
  • Complex tertiary/quaternary structures
  • Form colloids in water
  • Many use e.g. hormones, antibodies, carrier proteins, for example haemoglobin.
25
Q

What are enzymes

A

Enzymes are biological catalysts and increase the rate of reaction by lowering the activation energy of the reactions they catalyse

26
Q

What is the lock and key model

A

Enzymes are specific to substrates they bind to, as only one type of substrate fits into the active site of the enzyme.

27
Q

What is the induced-fit theory

A
  • When the enzyme and substrate form a complex, the structure of the enzyme is distorted so that the active site of the enzyme fits around the substrate.
  • The conformational change puts strain on the bonds within the substrate(s) molecule, lowering the activation energy, (it does not take much more energy to break the bonds when they are under strain)
28
Q

How does enzyme conc. affect rate of reaction

A
  • The rate of reaction increases as enzyme concentration increases as there are more active sites for substrates to bind to.
  • However, increasing the enzyme concentration beyond a certain point has no effect on the rate of reaction as there are more active sites than substrates so substrate concentration becomes the limiting factor.
29
Q

How does substrate conc. affect rate of reaction

A
  • As concentration of substrate increases, rate of reaction increases as more enzyme-substrate complexes are formed.
  • However, beyond a certain point the rate of reaction no longer increases as enzyme concentration becomes the limiting factor.
30
Q

How does temperature affect rate of reaction

A
  • Rate of reaction increases up to the optimum temperature.
  • Rate of reaction decreases beyond the optimum temperature because enzymes denature.
31
Q

How does pH affect rate of reaction

A
  • Enzymes work within a narrow range of specific pH values, optimum pH
  • Above or below optimum pH denature the enzyme
32
Q

How does pH affect rate of reaction

A
  • Enzymes work within a narrow range of specific pH values, optimum pH
  • Above or below optimum pH denature the enzyme
33
Q

Extracellular enzymes

A
  • Enzymes that carry out their activity outside of cells
  • Digestive enzymes
34
Q

Intracellular enzymes

A
  • Enzymes that carry out their activity inside of cells
  • ATP synthase in mitochondria
35
Q

Process of transcription in protein synthesis

A
  1. DNA helicase separates the two strands of DNA by breaking h-bonds between bases
  2. RNA polymerase lines up free RNA nucleotides upon the template strand through complementary base pairing with the DNA bases.
  3. RNA polymerase catalyses the formation of phosphodiester bonds between the RNA nucleotides through condensation reactions.
  4. mRNA detaches from the DNA template strand.
  5. Pre mRNA gets spliced, removing non-coding sections called introns, leaving the coding exons to form the mature mRNA molecule.
36
Q

Process of translation in protein synthesis

A
  1. Mature mRNA leaves the nucleus and is bound to by a ribosome.
  2. Complementary base pairing occurs between a codon of mRNA and an anticodon upon a tRNA molecule.
  3. The tRNA carries a specific amino acid, adding this to the growing chain of amino acids (polypeptide). The amino acids are joined by peptide bonds through condensation reactions.
  4. The specific sequence of codons possessed by mRNA, determines a specific sequence of amino acids in the synthesised polypeptide, until a stop codon is reached on mRNA.
37
Q

Features of the genetic code

A
  • The genetic code is non-overlapping, meaning that each triplet is only read once and triplets don’t share any bases.
  • The genetic code is degenerate, meaning that more than one triplet codes for the same amino acid.
  • The genetic code is a triplet code - each three bases codes for one amino acid. It also contains start and stop codons which either start or stop protein synthesis.
38
Q

What is semi-conservative replication of DNA

A

The semi-conservative replication of DNA ensures genetic continuity between generations of cells meaning that genetic information is passed on from one generation to the next.

39
Q

What experiment provided evidence for semi-conservative replication of DNA

A
  • Meselson and Stahl grew DNA in a culture containing N15 - (an isotope of nitrogen) for several generations, so all the bases contained this isotope.
  • They then grew the DNA in a culture of N14 for one generation.
  • During this generation, the DNA contained one strand containing 15-N and one strand containing 14-N.
  • After another generation, half of the DNA molecules were the same as in generation one, and the other half contained entirely 14-N (where the 14-N strand from generation one had been used as a template).
  • This provides evidence for the semi-conservative model.
40
Q

The steps of DNA replication

A
  1. DNA helicase unzips the DNA double helix by breaking the hydrogen bonds between the complementary base pairs.
  2. DNA polymerase matches free DNA mononucleotides to complementary bases upon the leading strand in a 5’ to 3’ direction.
  3. On the lagging strand, an RNA primer provides a starting point for DNA polymerase to begin the process of replication. Called Okazaki fragments.
  4. DNA polymerase catalyses the formation of bases between the adjacent mononucleotides.
  5. Phosphodiester bonds form through condensation reactions.
  6. Hydrogen bonds form between the bases on the newly synthesised strand of DNA and the conserved template strand that is opposite, creating one new strand of DNA.
  7. DNA replication leads to conservation of the original DNA strand.
  8. DNA replication is regarded as semi-conservative as each new DNA molecule is formed from one existing strand (which acted as the template) and one newly synthesised strand.
41
Q

Compare transcription with DNA replication

A
  • Both transcription and DNA replication store genetic information
  • Both transcription and DNA replication are made from four types of nucleotides
  • Transcription requires DNA helicase and RNA polymerase, whilst DNA replication requires enzymes like DNA helicase and DNA polymerase
  • Transcription requires no primer to start, whilst DNA replication requires RNA primer to start replication
  • Transcription can create mRNA tRNA rRNA etc.., whilst DNA replication creates two daughter strands of DNA
  • Transcription doesn’t involves unwinding and splitting of the entire DNA molecule, whilst DNA replication involves unwinding and splitting of the entire DNA molecule.
42
Q

Define incomplete dominance

A

A form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele.
This results in a third phenotype in which the expressed physical trait is a combination/mixture of the dominant and recessive phenotypes. One allele doesn’t mask the effect of the other allele.

43
Q

Define codominance

A

A form of intermediate inheritance in which one allele for a specific trait is not completely expressed. This results in a fourth phenotype in which the expressed physical trait is both of the dominant and recessive phenotypes.

44
Q

Social and ethical issues with genetic screening

A
  • There’s a risk of harm to foetus or miscarriage.
  • The outcome of testing might lead to an abortion - right to life.
  • The cost of bringing up a baby with a genetic disorder.
  • Emotional and mental issues surrounding caring for a baby with a genetic disorder.
45
Q

What is chorionic villus sampling

A
  • This test is carried out at 8 to 12 weeks of pregnancy.
  • A sample of embryonic cells is taken from the placenta and the DNA is analysed.
  • This form of testing is quicker than amniocentesis.
  • With this method there is a 2% chance of miscarriage.
46
Q

What is amniocentesis

A
  • Carried out at 14-16 weeks.
  • A sample of amniotic fluid, which contains fetal cells, is obtained using a needle.
  • The DNA is then analysed. Results are available after 2-3 weeks, as fetal cells need to be grown in culture first.
  • With this method there is a 1% chance of miscarriage.
47
Q

What is pre-implantation genetic diagnosis

A
  • Embryos created through IVF are tested for genetic disorders before they are implanted into the woman’s uterus.
  • One out of eight embryo cells are taken to screen.

A Biology A-level (9 decks)

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