Genes and health Flashcards
2.3) Explain, with the use of diagrams, how changing the size of an organ or
organism affects the surface area to volume ratio
As an organism increases in size, surface area to volume ratio decreases. Larger organisms have evolved to have a larger surface area to volume ratio in two different
ways. They may be long and thin thereby providing a larger surface area to volume ratio than being spherical or the may have special organs which increase their surface area to volume ratio such as lungs or intestines.
2.4) State the formula for Fick’s Law
Rate of diffusion surface area x difference in concentration
thickness of gas exchange surface
2.8)Describe how Fick’s Law can be used to explain the rapid rate of diffusion in a lung
The rapid rate of diffusion in the lungs occurs because:
A large surface area:
A branched system of tubes
ending in many alveoli.
Many capillaries surrounding
alveoli.
A large difference in concentration (concentration gradient):
Ventilation (breathing in and
out) replaces air that has lost
some oxygen.
Continuous blood flow, replaces oxygenated blood with deoxygenated blood.
A small distance for diffusion to occur over:
Alveolar walls are one cell thick and these cells are thin.
Capillary walls are one cell thick and these cells are thin.
Capillaries are only just wide enough for Red Blood cells to pass through.
2.10)Draw a diagram of an amino acid including the following labels: amino group, carboxylic acid group, R group
2.11)Identify the part of an amino acid molecule which is variable
All amino acids have a amino group.
There are 20 different types of amino acids and each one has a different R-group.
All amino acids have a carboxylic acid group.
2.12)Draw a labelled diagram
demonstrating the condensation and hydrolysis of peptide bonds: include amino acid monomers, dipeptides and polypeptides
The diagram shows a condensation reaction to produce a dipeptide (2 amino acids joined together).
Reversing this reaction – by adding water in to break apart a dipeptide is a hydrolysis reaction.
Adding many amino acids in a chain using multiple condensation reactions produces a polypeptide (a molecule made up of many amino acids).
2.13)Describe what is meant by the “primary structure of a protein”
The “primary structure of a protein” is the number and sequence of amino acids in a polypeptide chain
The amino acids are joined by peptide bonds
2.14) Describe what is meant by the “secondary structure of a protein”, including sketches to show the two different types, and the bonding involved
Secondary structure:
The most common is an extended
spiral spring, the alpha-helix. The helix shape is held by hydrogen bonding only. Most proteins have at least part of their structure in the form of an alpha-helix.
Alternatively, parallel chains may be
held by hydrogen bonds in an
arrangement known as a β-pleated
sheet. Within one protein molecule
there may be sections with α-helices
and β-pleated sheets.
2.15)Describe what is meant by the “tertiary structure of a protein”, including all the possible bonds involved
The “tertiary structure of a protein” is the folding into an overall 3- dimensional shape. The shape is determined by the location of bonds between R-groups of amino acids (and so determined by the primary structure of the polypeptide chain).
The bonds that hold the tertiary structure together are:
* Disulphide bonds (between two R-groups that contain sulphur)
* Ionic bonds (between positively and negatively charged R-groups)
* Hydrogen bonds (between d+ and d- R-groups)
* Hydrophilic / Hydrophobic interactions (water loving tend to be
external part of enzyme and water hating on interior of protein)
2.16) Give an example of a tertiary protein, and explain why denaturation may cause the protein / polypeptide to not work
An example of a tertiary protein is an enzyme.
Denaturing by heat – the kinetic energy causes the hydrogen bonds
in the protein to break, so changing the shape of the active site. The
substrate can no longer bind so no ESCs can be formed. The enzyme
is denatured.
Denaturing by pH – ionic bonds in the protein break, so changing
the shape of the active site. The substrate can no longer bind so no
ESCs can be formed. The enzyme is denatured.
2.17) Describe what is meant by the “quaternary structure of a protein”, including all the possible bonds involved
Quaternary structure
Only found in complex proteins. Two or more polypeptide chains curl together to form the complete protein molecule. The different polypeptide chains are held
together by
Hydrogen bonds – a weak bond, but if it occurs frequently it can add to molecular stability. Easily broken down, by too high temperatures for example.
Ionic bonds – susceptible to changes in pH (e.g. enzyme structure is affected by
changes in pH).
Disulphide bonds – a strong bond between amino acids which have sulphur in their R groups.
Hydrophobic interactions - some R groups are non-polar, and so are arranged so they face the inside of the protein molecules, as they are water-repelling. Water is thus excluded from the centre of the protein molecules.
2.18) Give an example of a quaternary protein, and explain why denaturation may cause the protein to not function
Haemoglobin is a quaternary protein.
During denaturation, the hydrogen bonds break, causing the haemoglobin to lose its 3-d shape.
This means that the oxygen cannot bind and so the haemoglobin cannot transport oxygen.
2.19) Using a diagram describe the structure of the fibrous protein collagen and relate the structure of collagen to its function as a structural protein
Collagen is the most abundant protein in you. It is found in all the connective tissues of the body.
A collagen fibre is made of 3 long parallel polypeptide molecules wound around each other to make a rope-like strand.
Each fibre cross links (hydrogen bonding) with other fibres to produce a molecule with very great strength.
Apart from being very strong, fibrous proteins are insoluble in water and metabolically inactive.
Collagen is mostly made of secondary structure proteins.
2.20) Using a diagram describe the structure of the globular protein
haemoglobin and relate the structure of haemoglobin to its function as an oxygen transport pigment
Quaternary structure
Only found in complex proteins. Two or more polypeptide chains curl together to form the complete protein molecule.
The shape is held by bonds:
* Hydrogen bonds
* Ionic bonds
* Disulphide
* Hydrophobic interactions
Non-amino acid components may be included in the protein molecule of tertiary and quaternary proteins (e.g. haemoglobin has 4 polypeptide chains and 4 haem groups,
containing iron).
These are called prosthetic groups.
Oxygen binds to the haem prosthetic groups.
Water-soluble, metabolically active, mostly made of tertiary and quaternary protein
structure.
2.21) Draw and label a diagram of a phospholipid molecule, including regions which are hydrophobic and hydrophilic
A phospholipid molecule:
A hydrophilic head:
the head is attracted to water and so points out of the membrane (either in contact with the watery tissue fluid or the watery cytoplasm).
Two hydrophobic tails:
The tails are repelled by water and so point into the centre of the membrane (away from the watery tissue fluid or the watery cytoplasm). They form a fatty barrier around the cell.
2.24) Describe and explain the effects of changing temperature on the permeability of cell membranes
Excess heat causes the molecules within the channel / carrier proteins to vibrate (due to increased kinetic energy).
Hydrogen bonds in the proteins break, and the proteins lose their shape.
The proteins now allow molecules to freely pass through them (e.g. the pigment in beetroot cells).
2.25) Describe and explain the effects of changing alcohol concentration the permeability of cell membranes
Alcohols, such as ethanol, dissolve the phospholipid bilayer, so
allowing molecules to freely pass through the membrane.
2.26) Describe a safe, reliable method for measuring the effect of temperature on the permeability of membranes
Use a cork borer to cut 1cm length cylinders of beetroot
1) Rinse the cylinders in distilled water to remove all dye on the surface of the cylinder
2) Place one beetroot cylinder in each of boiling tubes
3) Add 20cm3 of water to each of the 5 boiling tubes
4) Place in thermostatically controlled water baths for 30 mins (0oC,20oC,40oC,60oC,80oC)
5) Swirl a boiling tube to evenly distribute the pigment throughout the liquid
6) Pour some of the liquid into a cuvettte
7) Zero the colorimeter using a cuvette of distilled water
8) Place the cuvette into the colorimeter and measure the % absorbance of the liquid
9) Repeat for all boiling tubes
10)Repeat each temperature 10 times.
2.26) Describe a safe, reliable method for measuring the effect of alcohol concentration on the permeability of membranes
Use a cork borer to cut 1cm length cylinders of beetroot
1) Rinse the cylinders in distilled water to remove all dye on the surface of the cylinder
2) Place one beetroot cylinder in each of 6 boiling tubes
3) Add 20cm3 of the appropriate alcohol to each of the 6 boiling tubes
(distilled water 0%, 20%,40%, 60%, 80%, 100%)
4) Leave for 30mins at room temp 20oC
5) Swirl a boiling tube to evenly distribute the pigment throughout the liquid
6) Pour some of the liquid into a cuvettte
7) Zero the colorimeter using a cuvette of distilled water
8) Place the cuvette into the colorimeter and measure the % absorbance of the liquid
9) Repeat for all boiling tubes
10)Repeat each ethanol concentration 10 times.
2.27)Draw a diagram explaining
how substances can be moved across a membrane by diffusion
Diffusion is the passive, net movement of molecules from a higher concentration to a lower concentration down its concentration gradient.
Small, non-polar molecules (e.g.
oxygen, carbon dioxide, water) can
diffuse through a membrane by passing between the phospholipid molecules.
This is also known as “simple diffusion”.
2.28)Define the term “osmosis”
The passive, net movement of free water molecules across a partially permeable membrane from a high water concentration to a low water concentration down the diffusion gradient.
2.29) Explain how osmosis is different to diffusion
Diffusion is the net movement of liquid or gas molecules or ions
from where they are in a high concentration to where they are in
a lower concentration.
Osmosis is a special form of diffusion involving the movement of water molecules across a partially permeable membrane. It is the net movement of water molecules from a region where the water molecules are at a high concentration to a region where the water molecules are at a lower concentration, through a partially permeable membrane.
2.30)Draw a diagram explaining
how substances can be moved across a membrane by facilitated diffusion, including protein specificity
Facilitated diffusion is the passive, net movement of molecules across a
membrane from a higher concentration to a lower concentration through a channel protein or a carrier protein.
Ions (small, charged) pass through
channel proteins.
Larger polar molecules (e.g. glucose,
amino acids) pass through carrier
proteins by binding to them. The carrier proteins changes shapes to let them through.
Channel and carrier proteins are a
specific shape so they only allow one
type of molecule through.
2.31)Draw a diagram explaining
how substances can be moved across a membrane by active transport including protein specificity
Active transport is the active movement of molecules across a membrane from a lower concentration to a higher concentration through a protein carrier.
Active means the process requires energy in the form of ATP.
Larger or polar molecules (e.g. glucose, amino acids, ions) can pass through membranes in this way.
The molecule attaches to a binding site on the protein in order to be transported. The binding site is a
specific shape so they only allow one type of molecule to attach and so only one type of molecule is
transported through.
The protein carriers are specific to one type of molecule due to their shape.
2.32, 2.33)Draw a diagram explaining how substances can be moved across a membrane by endocytosis and exocytosis
Endocytosis is the active movement of molecules into a cell using vesicles. Molecules are packaged into vesicles formed from an infolding of the cell surface membrane.
Active means the process requires energy in the form of ATP.
Large particles (e.g. bacterial cell in pohagocytosis) can pass through membranes in this way.
Exocytosis is the active movement of molecules out of a cell using vesicles. Molecules are packaged into vesicles
formed by the Golgi apparatus. The vesicle travels to the cell surface membrane and fuses with it, thereby releasing the molecules out of the cell.
Active means the process requires energy in the form of ATP.
Large molecules (e.g. hormones, enzymes) can pass through membranes in this way.
2.34) Identify which of the transport mechanisms require an input of energy
Transport mechanisms requiring energy in the form of ATP are
* Active transport
* Endocytosis and exocytosis
In active transport the energy from ATP is used to change the shape of the carrier protein molecule causing the substance to be released on the other side of the membrane.
2.36) Describe the symptoms of Cystic Fibrosis including the effects on the function of the gas exchange system, digestive system and reproductive system.
Gas exchange system
* Mucus accumulates in bronchioles and can block bronchioles – coughing to remove
* Bacteria trapped in mucus increase the possibility of infection
* Mucus reduces air flow into alveoli
* Number of alveoli in contact with fresh air is reduced
* Reduction of the total surface area for gas exchange – breathlessness, tired, lack energy
Digestive system
* Mucus blocks pancreatic duct
* Digestive enzymes cannot reach the small intestine
* Food is not properly digested leading to tiredness and difficulty gaining weight
* Enzymes trapped with the pancreas causing damage
* Insulin producing cells damaged, leading to diabetes
Reproductive system
* In men the sperm duct is missing or blocked with mucus so sperm cannot leave the testes
* In women a mucus plug blocks the cervix so sperm cannot reach the egg
2.37) Explain, using diagrams, the role of the CFTR protein in the cell
membrane of a healthy person and a person affected by Cystic Fibrosis
CFTR is a transmembrane channel protein found in the membranes of
epithelial cells.
CFTR allows the facilitated diffusion of Clions out of the cell into the mucus.
In a healthy person
* Clions diffuse out of the cell, through the CFTR channel, into the mucus.
* This increases the solute concentration/lowers the water potential of the mucus
* Water moves into the mucus by osmosis
In a person affected by CF
* CFTR channel is non-functional/missing
* Clions cannot diffuse out of the cell, through the CFTR channel, into the
mucus.
* Water does not move by osmosis into the mucus – in fact water is drawn out of mucus and into cells
* Mucus is thicker and stickier
2.39) Describe both the structure of genes and their roles in cells
Gene definition:
A sequence of bases on a DNA molecule coding for a specific sequence of amino acids in a polypeptide chain, and found in a particular locus on a chromosome.
Role of genes:
They carry the genetic code to make proteins.
2.40)Draw a labelled diagram of a nucleotide of DNA and a nucleotide of RNA
A nucleotide is made up of a phosphate group bonded to a pentose sugar (Deoxyribose in DNA and Ribose in RNA) which is bonded to a base (A,C,G,T in DNA and A,C,G,U in RNA)
2.41Label and annotate a diagram of DNA to explain how the nucleotides are linked to form the double helix. Show how the bases pair, the strands are antiparallel and include the following: purine, pyrimidine and hydrogen bond.
Hydrogen bonds between complementary bases on opposite strands.
The larger bases (A and G) are the purines.
The smaller bases (C and T) are the pyrimidines.
There are 2 hydrogen bonds between A and T.
There are 3 hydrogen bonds between G and C.
2.42)Construct a table to show the differences between DNA and RNA
RNA:
Single polynucleotide chain.
Pentose sugar is ribose.
Organic bases are A, U, G, C.
Made in the nucleus, but found throughout the cell.
Amount varies from cell to cell.
2 basic forms: mRNA and tRNA.
mRNA – no complementary base pairing within the molecule and so no hydrogen bonds.
tRNA - the single strand folds back on itself and there is complementary based pairing between sections of the strand by hydrogen bonding.
Smaller: mRNA is the length of one gene, tRNA is a fixed, relatively short length.
DNA:
Double polynucleotide chain (always a double helix).
Pentose sugar is deoxyribose.
Organic bases are A, T, G, C.
Found in the nucleus.
Amount is constant for all cells of a
species (except in gametes).
Hydrogen bonding between complementary base pairs.
Larger: whole chromosomes (many
genes).
2.44) Explain why the genetic code is referred to as ‘universal’, ‘nonoverlapping’ and ‘degenerate’
The genetic code is the way in which information about the sequence of amino acids which make up a protein is coded for by the bases on a molecule of mRNA.
It is a triplet code. Each amino acid is coded for by a sequence of 3 bases.
It is a non-overlapping code. Each base can only be in one triplet.
It is a degenerate code. One amino acid may be coded by more than
one triplet.
It is a universal code. A triplet codes for the same amino acid in all
species.
2.45) Draw a flow chart to show
the stages of transcription in the
nucleus
- The DNA is in the nucleus
- DNA unwinds
- Hydrogen bonds between the two DNA strands are broken by the enzyme DNA helicase
- The template (or antisense strand) is used as a template to construct the
mRNA - Complementary RNA nucleotides link with the exposed DNA nucleotides by hydrogen bonds
- RNA polymerase binds to the DNA (at the promoter region)
- The enzyme RNA polymerase joins the RNA nucleotides together by
phosphodiester bonds to form mRNA - Hydrogen bonds attaching the mRNA to the template strand of DNA are broken
- mRNA is released from the DNA
10.mRNA leaves the nucleus via a nuclear pore (and enters the cytoplasm)
11.Complementary bases of the DNA rejoin once the mRNA has detached
2.47) Use a diagram to explain the process of translation - include the roles of mRNA, tRNA and ribosomes.
- In the cytoplasm the mRNA attaches to a ribosome
- The first codon (start codon) of the mRNA enters the ribosome
- A tRNA molecule with the complementary anticodon, brings the appropriate amino acid
- The anticodon of the tRNA binds to the codon of the mRNA using hydrogen bonds
- The ribosome moves along one codon and the next appropriate tRNA (with its amino acid) binds to the next
codon - The two amino acids join by a peptide bond (in a condensation reaction)
- The first amino acid detaches from its tRNA
- The ribosome moves along one more codon and the process is repeated
- tRNAs that have detached from their amino acid detach from the mRNA and are reused by attaching to another specific free amino acid (the one that is appropriate for the tRNA’s
anticodon)
10 The process is repeated again and again and the chain of amino acids gets longer, folding up as it grows - Eventually a stop codon is reached and the process stops. The ribosome detaches from the mRNA and the polypeptide chain detaches from the ribosome and the last tRNA.
The process of protein synthesis is now complete
2.48) Describe the role of start and stop codons during translation
The start codon is the first codon of a messenger RNA (mRNA) molecule. The start codon always codes for the
amino acid methionine in eukaryotes. The most common start codon is AUG.
In the genetic code, a stop codon is a nucleotide triplet within mRNA that signals an end to translation.
The 3 stop codons are UAA, UAG, UGA.
Note: you do not ned to remember which bases make up the
start and stop codons.
2.50) Use diagrams to explain
how replication of DNA takes place
including the role of DNA polymerase
DNA replication
1) The double helix uncoils into two
separate strands as hydrogen bonds
between the polynucleotide strands are broken
2) Each strand acts as a template for the formation of a new complementary strand
3) Free DNA nucleotides bind to each
template by complementary base
pairing.
4) Adenine pairs with Thymine, and
Guanine pairs with Cytosine
5) The new nucleotides are joined
together by phosphodiester bonds by
the enzyme DNA polymerase to form
a polynucleotide strand. The 2 new
DNA double helixes coil up into this
double helix shape
6) The two new DNA molecules are
identical to each other and to the
original DNA molecule
7) Each newly formed DNA molecule
contains one of the original
polynucleotide strands. This is called
semi-conservative replication
2.51) Explain, using diagrams, how the results of Meselson and Stahl’s
experiment demonstrated the method by which DNA replication takes place and how it shows that other methods did not take place
They grew the bacterium Escherichia coli in a medium that contained only the heavy isotope of nitrogen, called 15N (rather than the “normal” common isotope 14N). So, at the start of the experiment, all of the DNA nucleotides in the E. coli were made up of 15N, which made the
DNA molecules denser and heavier than normal DNA. They then transferred the bacteria into a medium containing only 14N, so from that point onwards, any new nucleotides incorporated into the DNA would be made with 14N (“light”
nitrogen). They allowed the bacteria to replicate their DNA once (one cell division), then extracted and centrifuged their DNA.
DNA molecules made using nucleotides with 15N will be heavier than those made from 14N, so they will pass further down the density-gradient solution, and end up nearer the bottom. Alternatively, DNA molecules made using nucleotides with 14N will be lighter, and so will be
nearer the top of the tube. Finally, DNA molecules made from a mixture of nucleotides, some with 15N and some with 14N, will end up between the light and heavy DNA in the tube.
2.52) Define the term “semi-conservative” replication?
Each newly synthesised DNA molecule has one strand from the old DNA molecule and one newly synthesised strand
2.53) Explain what is meant by “mutation of a gene”
A mutation is any permanent change to one or more nucleotides of DNA which leads to a wrong sequence of
amino acids in the protein.
2.55) Describe the possible consequences of these mutations, and explain why some mutations are more significant than others
A mutation is basically any change to one or more nucleotides in DNA or organic bases (or any arrangement of the sequence), which leads to a wrong sequence of amino acids in the protein.
Changes in DNA structure are a source of genetic variation. New forms of alleles arise from changes (mutations) in existing alleles.
A mutation may result in a change in the way the protein functions. Many proteins are enzymes, thus they may have the wrong molecular shape, and hence reduce their efficiency or prevent them from reacting (metabolic block).
A mutation in a gamete can be inherited.
Addition and deletion mutations are more significant than substitution mutations as they cause a frameshift in the reading frame after the mutation. This means that all the amino acids following the mutation may be altered.
2.57) Define the terms: gene,
allele, genotype, phenotype, recessive, complete dominance, incomplete dominance, co-dominance, heterozygous, homozygous, and give an example for each
Gene: A sequence of bases on a DNA molecule that codes for a sequence of amino acids in a polypeptide chain.
Allele: An alternative form of a gene (e.g. the allele for blue eyes) found at the same locus on a chromosome.
Genotype: The alleles the individual has for a particular characteristic (e.g. BB, Bb, bb).
Phenotype: The appearance of the characteristic in an individual (e.g. Blue eyes, Brown eyes)
Recessive: The allele whose trait is not expressed in a heterozygote (e.g. b). To have the condition an individual needs two recessive alleles (e.g. bb).
Incomplete dominance: A cross between organisms with two different phenotypes, produces offspring with a third phenotype that is a blending of the parental traits so that the third phenotype is something in the middle (e.g. red x white = pink). The dominant allele is not expressed completely.
Codominace: A cross between organisms with two different phenotypes produces offspring
with a third phenotype in which both of the parental traits appear together. In codominance both alleles are expressed completely.
Heterozygous: A genotype in which both alleles are different (e.g. Bb).
Homozygous: A genotype in which both alleles are the same (e.g. BB, bb)
2.58) Use a genetic diagram,
including a Punnet square, to predict
the offspring of a monohybrid cross
This is an example of a question that requires the use of a Punnett square.
There are other variations but all usually require identification of:
1) the appropriate parental genotypes
2) the parental gametes
3) all possible offspring genotypes
4) the associated offspring phenotypes
5) the ratio of offspring phenotypes
6) a link back to what the question was asking
2.63) Draw and label a diagram to show the structure of an enzyme
Enzymes are globular proteins with extensive tertiary structure. The enzyme molecule is much larger than the substrate molecules. Therefore only a small part of it actually contacts with the substrate – this is the active site.
Active site – made up of only a few amino acids and the substrate needs to be in contact with all of them. They may not adjoin each other in the polypeptide chain, but can be brought together by folding of the chain.
2.64)Describe the function of an
enzyme (mentioning activation energy)
Enzymes are biological catalysts, as they speed up the rate of reaction by reducing the activation energy, and are made in cells.
Enzymes act as catalysts to speed up reactions inside (intracellular) or outside(extracellular) cells. They allow cells to control which reactions happen, when and how much. Enzymes have an active site that contains amino acids with R-groups that allow the substrate to bind. This active site is a complementary shape to the substrate.
At any given temperature more reactants have enough energy to undergo the reaction and so more reactions happen per second.
2.65) Describe the two different models of enzyme theory
Lock and key hypothesis (1890 Fischer)
Enzymes are very specific because they have a particular shape into which the substrate fits exactly. They have complementary shapes.
When the enzyme-substrate is formed, the activation energy is lowered (as the enzyme holds the substrate molecule(s) in such a way that they react more easily)
and the reaction will take place. The products do not fit in the active site, so they escape into the surrounding medium. Therefore the active site is available for more substrate molecules.
Induced fit hypothesis
A modification of the lock and key hypothesis (1959 Koshland)
It was found that the active site is often flexible. So the shape of the active site only corresponds generally to the shape of the substrate molecule. The substrate
molecule then produces changes in the active site, which results in the two fitting together exactly (analogies – hand in glove, person in wetsuit). In other words, the substrate induces the enzyme to fit. This gives a more efficient catalysis.
2.68) Describe and explain, with the help of a graph, how changing the concentration of an enzyme affects the rate of a reaction
As the enzyme concentration increases the rate of reaction also increases up to a point (assuming substrate concentration is not limiting).
The more enzyme is added the more active sites are available to bind to form ESCs, so the faster the rate of reaction.
If the substrate concentration is a limiting factor the rate of reaction can no longer increase.
2.69) Describe safe, reliable methods for measuring the effect of enzyme and substrate concentration on the initial rate of reactions
This experiment uses catalase (in potatoes) to breakdown hydrogen peroxide into oxygen gas and water. The rate of the reaction can be identified by collecting the volume of oxygen gas given off in a specific time e.g 5mins catalase
Hydrogen peroxide water + oxygen
1) Use a cork borer to cut 5mm thick discs from a potato
2) Add different numbers of discs to each of 5 boiling tube (5, 10, 15, 20, 25 discs)
3) Add 1cm3 of a pH 7 buffer to each boiling tube
4) Place the boiling tubes in a water bath at 20°C
5) Add 10cm3of 1M Hydrogen peroxide to one of the boiling tubes
6) Add a bung with a delivery tube to the boiling tube in order to collect any gas given off over water in an upturned measuring cylinder
7) Record the volume of oxygen given off after 5mins
8) Repeat for each of the boiling tube with potato discs in
9) Repeat for each number of discs 10 times
10)Plot a graph of the volume of oxygen given off in 5mins against the number of potato discs
2.71)Describe two methods for obtaining sample of cells for prenatal testing (on a pregnant
mother)
Genetic tests can be conducted before birth (prenatal) by obtaining cells containing foetal DNA.
Amniocentesis involves inserting a needle into the amniotic fluid to collect cells that have fallen off the placenta and foetus. This can be carried out at around 15-17 weeks of pregnancy and involves a 0.5%-1% risk of miscarriage.
The other method is Chorionic villus sampling (CVS). Here a small sample of placental tissue is removed either through the wall of the abdomen or through the vagina. This can be carried out earlier (between 10 and 14 weeks) but there is a 1%-2% risk of miscarriage.
2.72) Describe the process of preimplantation genetic diagnosis (PGD) and state under what circumstances this may be carried
out
This can be done as part of the IVF process, so the
mother is not pregnant yet. An embryo (a few days old)
has a cell removed for DNA testing, looking for conditions such as cystic fibrosis.
Only an embryo free of the condition will be selected for implantation into the mother.
There is no pregnancy to terminate, so no ethical concerns over abortion.
It is an expensive process, and unwanted embryos will still need to be destroyed.
2.73) Describe two social and two ethical issues raised by prenatal
genetic screening
Ethical Issues:
Terminating a pregnancy / destroying an embryo is murder in the eyes of some people.
People may not understand the risks of prenatal testing / there is a risk of a miscarriage.
May be a false negative (baby actually does have CF) or false positive (baby didn’t have CF but was aborted).
Testing allows for an informed decision to be made about the baby.
Emotional stress can put a heavy burden on couples and individuals.
Social Issues:
A person with severe genetic abnormalities may need special schooling, may be unable to hold down a job, may have higher life
insurance.
Cost of testing (if not free) may be too much for some people.
Cost to the NHS / social services of caring for the baby as it grows up and has certain needs.
