Genes and health questions and answers

Q1.

Fick's Law of Diffusion can be used to calculate the rate of diffusion across gas exchange

surfaces.

Use Fick's Law of Diffusion to explain the adaptations of mammalian gas exchange surfaces.

The rate of diffusion is proportional to the surface area-alveoli have large surface area

rate of diffusion is proportional to difference in concentration-breathing maintains a difference in gas concentrations

rate of diffusion is proportional to difference in concentration-blood flow maintains a difference in gas concentrations

rate of diffusion is inversely proportional to diffusion distance-walls of alveoli and capillaries are one cell thick

diffusion distance is reduced due to flattened cells forming alveoli and capillary walls

rate of diffusion is proportional to to diffusion constant-cell membranes are relatively permeable to non-polar gas molecules

ms obtain oxygen from the water through the surface of their bodies.

The diagram below shows the structure of flatworms.

(a) Using the diagram and your knowledge of gas exchange surfaces, explain how the structure

of a flatworm is adapted to obtain oxygen from the water.

Large surface area to volume ratio or that it is thin.This helps diffusion e.g short diffusion distance, faster diffusion(Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. Diffusion happens in liquids and gases because their particles move randomly from place to place. Diffusion is an important process for living things; it is how substances move in and out of cells.)

(b) The table below shows the relationship between the temperature of water and the solubility

of oxygen in water.

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Edexcel (A) Biology A-level 2 PhysicsAndMathsTutor.com(i) Describe the relationship between the temperature of the water and the solubility of oxygen

in water.

Solubility of oxygen decreasess as temperature increases

(ii) Using the information in the table and your knowledge of gas exchange and enzymes, suggest why flatworms are often found in water at a temperature of about 15 °C .

There is quite alot of dissolved oxygen in the water at this temperature-sufficient oxygen

oxygen gradient concentration(between water and flatworms cells.

enzyme activity is is temperature dependent=15 degrees is optimum for their enzymes

water below 15 degrees would be too cold for enzymes/metabolism to work effectively

balance between oxygen availability and enzyme activity/kinetic energy

) Flatworms do not have a heart or a circulatory syst Explain why many animals need a heart and a circulatory system.

)Heart needed to pump blood around the body

animals have small surface are to volume ratio

circulatory system is needed to overcome limitations of diffusion, cardiovascular system =circulatory system-The cardiovascular system works with thermoregulation in order to keep the body at a healthy temperature and be able to warm or cool the body whenever it is needed., animals have a high metabolic rate .Complete double ciruculatory systems allow for higher metabolic rates to be maintained as there is no mixing of oxygenated and deoxygenated blood.

(Diffusion takes place throughout the bodies of humans and other organisms, and is crucial to many of the metabolic processes that allow us to live our lives and stay healthy. The molecules that move during diffusion can be solids or gases, nutrients or wastes. Many organs, such as the lungs and small intestine, can be observed to find examples of diffusion.)(transport of molecules within a number of organs including the lungs, kidneys, stomach and eyes.)

The movement of materials into and out of a cell needs to be controlled. Describe what is meant by the term fluid mosaic with reference to cell membranes.

Fluid refers to the movement of of phospholipids(a lipid containing a phosphate group in its molecule, e.g. phosphatidylcholine)

mosaic refer to the random association of proteins (of different shapes and sizes) within the membrane.

Many animals have specialised organs for gas exchange and transport. *(a) The diagram below shows the lungs of a mammal.

Describe and explain how the lungs of a mammal are adapted for rapid gas exchange.

alveoli one cell thick(any of the many tiny air sacs of the lungs which allow for rapid gaseous exchange.)/thin epithelium (the thin tissue forming the outer layer of a body's surface and lining the alimentary canal and other hollow structures.)

walls of capillaries are one cell thick(They are only one cell thick, and they are the sites of the transfer of oxygen and other nutrients from the bloodstream to other tissues in the body;)(The walls of capillaries are just one cell thick. Capillaries therefore allow the exchange of molecules between the blood and the body's cells - molecules can diffuse across their walls.)

alveoli covered with capillaries

short diffusion distance(distance - if the diffusion distance is small, diffusion happens faster because the particles do not have as far to travel)

large surface area provided by alveoli/capillaries

concentration gradient maintained by ventilation/breathing

large number of red blood cells=idea that oxygen combines with haemoglobin

concentration gradient maintained by blood flow

ficks law(rate of diffusion is proportional to difference in concentration-blood flow maintains a difference in gas concentrations)

) Daphnia have a circulatory system with a heart that pumps blood into cavities surrounding their organs. The photograph below shows the location of the heart in a Daphnia.

(i) Suggest how the heart of a Daphnia enables organs to carry out effective gas exchange.

blood carries oxygen/carbon dioxide

blood moving maintains concentration gradient

reference to mass flow(is the movement of fluids down a pressure or temperature gradient, particularly in the life sciences. As such, mass flow is a subject of study in both fluid dynamics and biology. Examples of mass flow include blood circulation)

organs have large surface area to volume ratio(? Gas exchange surfaces tend to have the adaptation a large SA:volume ratio. This is advantageous as it increases the area over which gases and other materials can be transported into and out of the organism, via diffusion and active transport.)(surfaces and body organs are specialised for exchanging materials. The ability to maximise exchange of substances in as short a time as possible in plants and animals)

) In mammals, blood passes through the heart twice for each circulation of the body. Suggest how this type of circulation enables mammals to carry out effective gas exchange. (

One side of the heart transports blood to the lungs, other to the body

seperation of oxygenated and deoxygenated blood

idea of maintaining concentration gradient

blood pressure lower in the lungs(The advantage is your lungs will work. If there was high pressure, the narrow walls of the endothelial cells in the alveoli would rupture, creating leaking in the lungs.

The resistance in the lungs is also low, therefore, the flow of blood to the lungs is no different than the flow to the rest of the body. Therefore, low bp in the lung does not make blood flow slower.

(formula is F = P/R) Increasing the pressure in the lungs would result in faster oxygenation to a point, and this does occur under stress (exercise) but the primarily reason why low bp in the lungs is present is because it has to be.), higher to the body

reference to mass flow/supply of o2 to body cells maximised

good supply of oxygen as mammals are very active/high rate of metabolism/warm blooded

The fluid mosaic model has been developed from the knowledge of the structure and properties of cell membranes. It can explain how molecules can enter and leave a cell. (a) Describe the structure of a cell membrane. (You may use a labelled diagram to support your answer).

phospholipid bilayer(phospholipid bilayer. noun. a two-layered arrangement of phosphate and lipid molecules that form a cell membrane, the hydrophobic lipid ends facing inward and the hydrophilic phosphate ends facing outward)

correct orientation and structure of phospholipids in the bilayer

phosphoilipids are oriented in the way they are=heads are attracted to water or tails repelled by water

proteins in membrane shown or described

There are globular proteins in between this bilayer that help transport things in and out of the cell.

1. Ion channels. Ion channels are a very narrow tube-shaped protein that help establish a tiny pore in the cell membrane. They are only large enough to allow an ion to go through. Each ion channel is specifically for specific ions (Na+, K+, Cl-, Ca+2, etc). They can open and close and are very important in understanding the rest of physiology. Some of you may have heard of a class of drugs called "calcium [ion] channel blockers" which are important in regulating blood pressure, for example.

2. Transporter or Carrier Proteins. They are embedded in the cell membrane to help transport glucose and amino acids across the membrane. They are too large to go through ion channels. Remember a glucose molecule (C6H12O6) is 24 atoms large, so it needs help getting across and amino acids are even bigger than that. When energy is required to transport/carry a protein across, that's known as active transport. When no energy is needed, that's called passive transport. Amino acids are actually transported by active transport. Interestingly, the transport of glucose across a cell membrane uses a transport protein but doesn't require ATP, so it's called passive transport, or alternatively, facilitated diffusion.

3. Enzymes are chemicals that catalyze biochemical proteins. These are found in the cell membrane as well.

4. Linker proteins are always facing the cytoplasmic fluid. The linker proteins are always attached to the cytoskeleton of the cell. The cytoskeleton refers to the matrix of proteins in the cytoplasm. These proteins are associated with movement and affect the shape of the cell.

5. Receptor site proteins (this is the most important part). The receptor sites are always found on the outer cell membrane surface (unless it's for a protein hormone, which we'll discuss soon). This is where hormones and neurotransmitters and other chemicals attach to the surface of the cell. We know hormones affect the activity of the cells. These receptor sites are how they affect the cells. These signal molecules (hormones, neurotransmitters, chemicals) are called ligands. (Etymology of ligand: From Latin ligandus, gerundive of ligare 'to bind'.)

two different locations of proteins

extrinsic, intrinsic,transmembrane

glycoproteins/glycopolids(Describes/shown

idea of cholesterol within the membrane(shown or described)(Within the cell membrane, cholesterol also functions in intracellular transport, cell signaling and nerve conduction. Cholesterol is essential for the structure and function of invaginated caveolae and clathrin-coated pits, including caveola-dependent and clathrin-dependent endocytosis.)

Suggest two properties of molecules that enable them to enter a cell by diffusion. (2) 1 .................................................................................................................................................... 2 ....................................................................................................................................................

small

nonpolar

lipid soluble

they are recognised by specific protein receptors

(c) Facilitated diffusion and active transport are two ways in which molecules are transported across cell membranes. Describe one similarity and one difference between facilitated diffusion and active transport. (i) Similarity

Active transport (the movement of ions or molecules across a cell membrane into a region of higher concentration, assisted by enzymes and requiring energy.)moves molecules against a concentration gradient/facilitated diffusion allows molecules to move down a concentration gradient/eq

active transport requires energy/diffusion does not require energy

similarity=use carrier or channel proteins

transport hydrophillic/molecules/named molecule

The consistency of the mucus is determined by the movement of water, by osmosis, from the cells lining the bronchi. Explain how the partial permeability of the surface membrane of the cells lining the bronchi allows osmosis to take place.

Partially permeable membrane is a barrier to some solutes but not water

enables a concentration gradient of solutes/water

Cell membranes are involved in the transport of molecules. The diagram shows the structure of a cell membrane. Describe how the structure labelled B is involved in passive transport

B is a channel protein(Channel protein - A protein responsible for mediating the passive transport of molecules from one side of the lipid bilayer to the other. Transport is carried out by its membrane-spanning hydrophilic structure which, when open, allows molecules to pass through.)

which allows the movement of large/charged/polar molecules

by diffusion from high concentration to lowconcentration/down concentration gradient

Muscle cells contain globular and fibrous proteins. Compare and contrast the molecular structures of globular and fibrous proteins.

Both chains of amino acids joined by peptide bonds

both contain named bonds(holding molecule in its th three dimensional shape)

globular proteins have hydrophillic groups on the outside whereas fibrous proteins have hydrophobic groups on the outside

globular have tertiary or quaternary structures whereas fibrous have little or no tertiary structure

globular are folded into compact shapes where a fibrous have long chains.

State why enzymes are described as biological catalysts.

Proteins which reduce activation energy f biological reactions=also increase rate of biological reactions

) Explain the effect of changing enzyme concentration on the initial rate of reaction.

enzymes reduces activation energy=decreases energy needed for the reaction=provides an alternative reaction pathway.

As enzyme concentration increases, the rate of reaction increases since more active sites will be available to bind to substrate molecules. This means that there will be more frequent collisions between the enzyme and substrate, so there will be more formation of enzyme-substrate complexes.

) Describe an experiment that could be carried out to investigate the effect of enzyme

concentration on the initial rate of reaction.

idea of range of concentrations of enzyme(at least 5), idea of substrate concentration not limiting, reference to mixing, description of how to measure dependent variable with time decription of how o measure the initial rate of reaction, referance to appropriate named control variable reference to replicates/repeats at each enzyme concentration, control (Described/used as a comparison)

THE EFFECT OF CHANGING ENZYME CONCENTRATION ON THE RATE OF REACTION

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CORE PRACTICAL 4 - FROM TOPIC 2 (GENES AND HEALTH)

Contents hide

1 Core Practical 4 - From Topic 2 (Genes and Health)

1.1 Aim

1.2 Independent Variable

1.3 Dependent Variable

1.4 Control Variables

1.5 Equipment

1.6 Control

1.7 Method

1.8 Results & Calculations

1.9 Conclusion

1.10 Evaluation Points

experiment equipment on table

AIM

To investigate the effect of enzyme concentration on the initial rate of reaction.

INDEPENDENT VARIABLE

Protease concentration (measured in percentage)

DEPENDENT VARIABLE

Time taken for enzyme to breakdown substrate (in seconds)

CONTROL VARIABLES

Temperature - measure temperature of water bath with thermometer and make sure it is constant for repeats

Volume of enzyme solution - 2 cm³ of the protease enzyme solution will be used each time

Volume of substrate - 5 cm³ of casein solution will be used each time

Concentration of substrate - same concentration of casein should be used for all repeats

EQUIPMENT

Range of protease (trypsin) concentrations (e.g. 0.2%, 0.4%, 0.6%, 0.8% and 1%)

Casein solution

Thermometer

Marker pen

Water bath

Stopwatch

Test tubes

Test tube rack

Distilled water

CONTROL

Enzyme concentration of 0.0% (distilled water) can be used and added to the casein solution. Results can be compared to this control result.

METHOD

Set up a water bath so that the temperature can be kept constant.

Mark an 'X' on one side of a test tube using a marker pen. Fill this test tube with 5 cm³ of casein solution and place it into the water bath alongside a second tube containing 2 cm³ of the lowest concentration trypsin (e.g. 0.2%).

Allow both substances to acclimatise for 3 minutes so that at they are both at the same temperature.

Add the test tube of trypsin to the casein and start the stopwatch. Time how long it takes for the casein solution to turn transparent. This is when you can clearly see the shape of the 'X' mark.

Repeat this a further 2 times and then repeat for the other concentrations of trypsin (e.g. 0.4%, 0.6%, 0.8% and 1%).

RESULTS & CALCULATIONS

Any recordings should be noted down in an appropriate table and a graph can also be drawn. Mean values should be calculated using repeat data. The initial rate of reaction can be calculated using the following equation:

Rate = 1 ÷ Time (seconds)

CONCLUSION

The initial rate of reaction is directly proportional to the enzyme concentration because the more enzymes that are present, the greater the number of active sites that are available to form enzyme-substrate complexes. The increase in rate will continue in this linear fashion assuming that there is an excess of substrate. This trend is shown on the image below.

Only a specifically shaped substrate will induce the correct change in shape of an enzyme's active site. The slight change in shape of the active site enables the substrates to react. This is known as the induced fit theory and is currently more widely accepted than the lock-and-key theory.

Q4.

Proteins, such as collagen, are made from amino acids joined together.

(i) Which of the following is the name of the bond used to join amino acids together?

peptide

Lysozyme is an enzyme found in tears. Lysozyme can destroy some bacteria by

breaking down the polysaccharide chains that form part of their cell walls.

The primary structure of lysozyme is a specific sequence of 129 amino acids.

Two of the amino acids that make up the active site are in positions 35 and 52 in

the primary structure.

Suggest how these two amino acids could be brought closer together to form

part of the active site of this enzyme

idea of formation of secondary or tertiary structure

idea of bonding of r groups

named bond e.g ionic, disulfide, hydrogen

Lysozyme is an enzyme found in tears. Lysozyme can destroy some bacteria by breaking down the polysaccharide chains that form part of their cell walls. Temperature affects the activity of lysozyme. Suggest why increasing the temperature above 45 °C causes a decrease in the activity of lysozyme.

idea that increasing the temperature changes the bonding in the enzyme

active site is denatured/changes shape

substrate no longer fits into the active site, the enzyme no longer catalyses the reaction/lowers the activation energy

Haemophilia is a disease that affects blood clotting. People with haemophilia are sometimes

given a protein called factor VIII. Factor VIII is an enzyme that is involved in the process of blood

clotting.

Explain how a change in the primary structure of factor VIII could cause difficulties with blood

clotting

different primary structure results in a different sequence of amino acids

change in the R groups changes bonding, secondary structure/tertiary structure

changing the shape of the active site prevents the substrate from being able to bind

stopping the production of fibrin

Some species of bacteria have developed resistance to antibiotics.

This has led scientists to investigate many molecules for antimicrobial properties.

Peptides extracted from broad bean plants and cowpea plants have been studied.

Describe how a peptide bond is formed.

a peptide bond is formed from a condensation reaction

between the amine group(nh2) and the carboxyl group(cooh) of adjacent amino acids.

Cystic fibrosis is a genetic disorder caused by mutations in the CFTR gene. One aim of somatic gene therapy is to overcome the effects of defective genes. (a) (i) Describe the difference between somatic gene therapy and germ line gene therapy. (2

Somatic involves (body/somatic)cells and germ line involves (gametes/ovaries/testes/eq)

somatic cannot be inherited/germ line can be inherited

somatic legal/ germ line illegal

somatic temporary treatment/germ line could be cure

*(ii) Suggest how somatic gene therapy could enable cells lining the lungs to function normally in people with cystic fibrosis.

idea of inserting a (functional)gene that codes for CTFR protein

method of getting into lungs e.g nebuliser

CTFR protein made via (Transcription/translation)

allows chloride ions to leave cells

water leaves by osmosis

nucleus is less sticky

(b) Rhythmical tapping of the chest wall during physiotherapy can relieve the symptoms of cystic fibrosis in the lungs. Suggest an explanation for this.

loosens mucus

mucus expelled from lungs

clearer airways/better breathing

Cystic fibrosis is an inherited condition. Explain why people with cystic fibrosis can have breathing difficulties

Produces thicker more /viscous/stickier mucus

blocking of trachea/bronchi or bronchioles

cillia are unable to move mucus out of lungs

idea of reduced reduced flow of air/oxygen to alveoli

reduced concentration gradient for oxygen carbon dioxide (in alveoli)

idea of loss of surface area/elasticity

idea of reduced gaseous exchange

trapped bacteria may result in more respiratory infection

The photograph below shows a cleft iris, a rare condition in humans. Cleft iris may be due to the inheritance of recessive alleles. Explain the meaning of the term recessive allele

idea that both of these alleles need to be present in order for the recessive phenotype to be expressed

different form of a gene

same locus/position

different base sequence

Mutations to DNA can affect the structure of proteins produced in the cell. Removing one base from a DNA sequence will affect the primary structure of a protein. Changing one base for another may not affect the primary structure of a protein. Explain why these two types of mutation have different effects on protein structure.

deletion could affect every codon (on the mRNA/substitution will affect only one codon)

deletion more likely to affect the position of (stop codon or start codon)

deletion results in a different sequence of amino acids/substitution may not affect the sequence of amino acids

substitution may code for the same amino acid

(Same amino acid )due to the degenerate anture of the genetic code

Leptin is a protein hormone with a role in the control of appetite in humans. Several mutations of the leptin gene have been identified. All these mutations are frameshift mutations that result in shortened primary structures. A frameshift mutation involves the insertion or removal of one or two nucleotides from a gene. Describe how a frameshift mutation could result in the production of leptin with a variety of shorter primary structures

adding or removing ne or more nucleotides, changes the triplet code

introducing a new (start/stop) codon

coding for a shorter sequence of amino acids.