Topic 2

Question 1

The rate of gas exchange by diffusion is increased by:

Answer 1

• surface area increase
• diffusion disatnce decrease
• diffusion concentration gradient made more steep

Question 2

Fick’s law:

Answer 2

rate of diffusion (is proportional to) surface area x concentration difference / diffusion distance

Question 3

How are mammalian lungs adapted for rapid gas exchange?

Answer 3

• large surface area due to the presence of many alveoili
• steep concentration gradient maintained due to good supply of circulating blood provided by the capillary network and mechanical ventilation
• short diffusion distance as the alveoli and capillaries are only one cell thick

Question 4

5 functions of cell membrane:

Answer 4

1. is flexible- can break and fuse easily
2. compartmentalisation- e.g. lysosome controls cell death, phagocytosis
3. partially permeable
4. cell signalling- glycoproteins are antigens (cell recognition), glycolipids can send and recieve messages (e.g. a hormone receptor)
5. site of cell reactions- have enzymes allowing reactions to take place

Question 5

explain the term: fluid mosaic model:

Answer 5

fluid= components of the cell membrane are able to move freely laterally through the bilayer
mosaic= many different components making up the cell membrane e.g. phospholipids, proteins, cholesterol, carbohydrates etc.

Question 6

Types of passive movement of molecules into and out of the cell:

Answer 6

1. Diffusion- passive movement of small, non-polar molecules from an area of high concentration to low concentration (down a conc. grad.) The molecules move through the phospholipid bilayer.
2. Facilitated diffusion- requires channel proteins to tranport small, polar molecules across the membrane.
3. Osmosis- the movement of water molecules from an area of low solute concentrations to high solute concentrations through a partially permeable membrane.

Question 7

Types of active movement of molecules into and out of the cell:

Answer 7

1. active transport- can transport all types of molecules through carrier proteins. Movement may be up or down the concentration gradient. Up the gradient reaquires the hydrolysis of ATP to provide an accesible store of energy for biological processes. The phosphorelation (reforming of ATP) requires energy.
2. Endo/exocytosis- transport large molecules. In endocytosis, particles are enclosed in vesicles made from the cell surface membrane and transported into the cell. In exocytosis, vesicles containing large particles are fused with the cell surface membrane and transported out of cells.

Question 8

DNA structure:

Answer 8

• double stranded, alpha doubel helix with a sugar-phosphate backbone on each strand

Question 9

DNA-
bases:
pairing:
sugar:

Answer 9

bases:
purine (two nitrogen-containing rings): adenine and guanine
pyrimidine (1 nitrogen-containing ring): cytosine and thymine
pairing: A-T, C-G
sugar: deoxyribose (hydroxy group replaced by hydrogen on carbon 2)

Question 10

Bonding in DNA:

Answer 10

-Phosphodiester bonds between phosphate group and carbon 5 (and between adjacent nucleotides forming the sugar-phosphate backbone)
-Hydrogen bonds between bases (3 between G-C and 2 between A-T)

Question 11

mRNA structure:

Answer 11

single stranded, not folded, carries codons (triplets of bases) which attach to tRNA via H bonds

Question 12

mRNA-
bases:
pairing:
sugar:

Answer 12

bases:
purine: adenine and guanine
pyrimidine: cytosine and uracil (thymine is too unstable to leave the nucleus)
pairing: A-U, C-G
sugar: ribose

Question 13

tRNA structure:

Answer 13

single stranded, folded into a specific pattern held together by H bonds, carries anticodons complementary to mRNA codons, bonded via H bonds.

Question 14

Transcription process:

Answer 14

mRNA made in the nucleus:
1. DNA helicase unzips the DNA, breaking the H bonds between the complementary bases. The DNA uncoils, thus separating the two strands.
2. One of the DNA strands is used as a template to make the mRNA molecule. The template is called the antisense strand.
3. Free nucleotides line up by complementary base pairing and adjacent nucleotides are joined by phosphodiester bonds, thus forming a molecule of mRNA. This is catalysed by RNA polymerase.
4. mRNA moves out of the nucleus through a nuclear pore and attaches to a ribosome (in cytoplasm or RER) which is the site of protein synthesis

Question 15

Translation process:

Answer 15

Amino acids join together to form a polypeptide chain:
1. mRNA attached to a ribosome, tRNA binds to specific amino acids from the cytoplasm depending on its anti-codon (activation)
2. complementary anticodons of tRNA bind to mRNA codons and are bonded with H bonds
3. The ribosome joins the amino acids attached to two tRNA molecules with a peptide bond and the tRNA molecules detach from the amino acids.
4. Process is repeated leading to the formation of a polypeptide chain until stop codon is reached on mRNA.

Question 16

definition of a gene:

Answer 16

A gene is a sequence of bases on a section of DNA coding for a sequence of amino acids in a polypeptide chain.

Question 17

names for non-coding and coding regions:

Answer 17

non-coding = introns
coding = exons (expressed)

Question 18

3 features of the genetic code:

Answer 18

1. the genetic code is non-overlapping- each triplet of bases is only read once and triplets don’t share any bases
2. the genetic code is degenerate- more than one triplet codes for the same amino acid
3. the genetic cose is a triplet code- each three bases codes for 1 amino acid. It also contains start and stop codons (starting or stopping protein synthesis)

Question 19

Proteins: monomers

Answer 19

Amino acids:
-contains one amine group, a carboxyl group and a variable R group (carbon-containing chain). There are 20 different amino acids with different R groups.
- joined by peptide bonds formed in condensation reactions

Question 20

Primary and secondary structure of proteins:

Answer 20

Proteins are folded in the Rough Endoplasmic Reticulum.
1. Primary structure of a protein is the sequence of amino acids, making up a polypeptide chain.
2. The secondary structure is the 2D arrangement of the chain of amino acids- either alpha helix or beta pleated sheet.

Question 21

Tertiary and Quaternary structure of proteins:

Answer 21

3. Tertiary structure of a protein is the 3D folding of the secondary structure into a complex shape. The shape is determined by the types of bonds present- -Hydorgen bonds (between partially charged atoms in R groups)
   -ionic bonds (salt brigdes, form between oppositely charged groups on R groups)
   -disulfide bridges (covalent bonds between sulfur atoms on cysteine)
   - hydrophilic and hydrophobic interactions (polar and non-polar properties in R groups)
4. Quaternary structure of a protein is the 3D arrangement of more than one polypeptide chain.

Question 22

Fibrous protein features:

Answer 22

• long, parallel polypeptides
• very little tertiary/ quaternary structures
  -occasional cross-linkages which form microfibres for tensile strength
  -insoluble
  -used for structural purposes- such as collagen

Question 23

Globular protein features:

Answer 23

-complex tertiary/ quaternary structures
-form colloids in water
-used for transport e.g. hormones, antibodies, carrier proetins (e.g. haemoglobin)

Question 24

The fibrous proetin- collagen:

Answer 24

• High tensile strength due to presence of both hydrogen and covalent bionds.
  -made up of three polypeptides which form an alpha triple helix which forms fibrils and strong collagen fibres.
  -forms the structure of bones, cartilage and connective tissue, is the main component of tendons

Question 25

The globular protein- Haemoglobin:

Answer 25

- water soluble - 4 beta polypeptide chains and a haem group - carries oxygen in the blood of O2+ can bind to the haem (Fe2+) group- oxygen is released when required.

Question 26

Definition of an enzyme:

Answer 26

Biological catalyst that increase the rate of reaction by lowering the activation energy and providing an alternate reaction pathway. The work for anabolic and catabolic, inter and extracellular reactions.

Question 27

the active site of an enzyme:

Answer 27

- The area on the enzyme where the substrate binds. - They are specific and complementary to the substrate they bind to (lock and key theory) - induced-fit theory: the enzyme and substrate form a complex, the structure of the enzyme is distorted in order for the active site to be complementary to the substrate.

Question 28

How do you determine the initial rate of reaction?

Answer 28

By calculating the graident of a concentration-time graph at 0 seconds. Draw a tangeant if necessary

Question 29

Factors effecting enzyme- controlled reactions:

Answer 29

Concentrations: 1. enzyme concentration- the ROR increases and the conc. increases as there are more active sites for substrates to bind to. However, increasing the conc. beyond a certain point has no effect as there are more active sites for the available substrates so the the substrates become a limiting factor. 2. The same goes for increasing the substrate concentration.

Question 30

Factors affecting enzyme-controlled reactions:

Answer 30

Temperature- ROR increases up to the optimum temperature (temp the enzyme works best at). Enzymes denature beyond optimum temp- active site changes shape, hydrogen bonds break, enzyme-substrate complex cannot form, no reaction. pH- enzymes work within a narrow range of a specific pH values, values above or below denature enzymes.

Question 31

purpose of semi-consrvative DNA replication:

Answer 31

ensures genetic continuity between generations of cells meaning that genetic information is passed of from one generation to the next

Question 32

3 types of DNA replication:

Answer 32

1. semi-conservative 2. conservative replication- conserves both strands of parents DNA 3. dispersive replication- individula DNA strands are a mixture of old and new DNA.

Question 33

Meselson and Stahl evidence for semi-conservative replication:

Answer 33

- originally grew a DNA culture containing N15 (isotope of nitrogen) for serveral generations , so all bases continued this isotope - they then grew the culture in N14 for one generation. - After this generation, the DNA contatined one N15 and one N14 strand. - after another generation, half of the DNA molecules where the same as in generation one, and the other half contained entirely N14 (where the N14 strand frp, feneration one had neen used as a template.

Question 34

steps of semi-conservative DNA replication:

Answer 34

1. The double helix unwinds and the hydorgen bonds between the complementary bases break, catalysed by DNA helicase, thus separating the strands of DNA. 2. one of the strands is used as the template and complementary strand and base pairing occurs between the template strand and free nucleotides. 3. Adjacent nucleotides are joined by phosphodiester bomds formed in condensation reactions, catalysed by DNA polymerase

Question 35

Mutations:

Answer 35

Mutations are permanent changes in the DNA of an organism. Gene mutations are changed in the arrangement of bases by: - substitution - insertion - deletion - duplication - inversion This change to the base sequence results in a change to the mRNA, tRNA, therefore to the primary stucture of the protein. Mutations also may occur in the formation of mRNA and tRNA themselves.

Question 36

Cystic Fibrosis

Answer 36

- cystic fibrosis is a genetic disorder caused by a mutation of a single gene which codes for the CFTR protein. -CFTR is a channel protein that transports chloride ions out of the cells of the respiratory tract and into the mucus. -Na+ dissuses down the electrochemical gradient forming a salt in the mucus. - this causes the movement of water by osmosis into the mucus, making it watery. Therefore, a mutation in this gene makes the mucus very thick and sticky due to the less efficient movement of chloride ions. -Sticky, thick mucus causes many problems in gas exchange, reproduction anf digestion.

Question 37

Systems affected by cystic fibrosis:

Answer 37

Respiratory system: -build up of mucus in the lungs traps bacteria, increasing the risk of infection -build up of mucus in airway decreases SA of alveoli, reducing the rate of gas exchange Reproductive system: -cervical mucus prevents the sperm from reaching the egg -sperm duct is blocked, sperm produced cannot leave testes. digestive system: -the pancreatic duct (connects pancreas to small intestine) can become blocked with mucus, so digestive enzymes are unable to reach SI. Food is not properly digested and fewer nuitrients are absorbed. -the mucus lining of the duodenum in thick, reducing the absorption of nuitrients -mucus can cause cysts to form in the pancreas and damage the insulin producing cells, leading to diabetes.

Question 38

allele definition:

Answer 38

-different version of a gene -in one particular locus

Question 39

genotype:

Answer 39

all of the alleles of an organism

Question 40

genome:

Answer 40

The complete set of DNA in an organism

Question 41

phenotype

Answer 41

the set of observable characteristics of an individual resulting from the interaction of its genotype with the environment

Question 42

dominant allele vs recessive allele

Answer 42

dominant: an allele that produces a feature even if only one copy of the allele is present recessive: an allele that produces a feature only if two copies are present

Question 43

incomplete dominance:

Answer 43

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 of the dominant and recessive phenotypes.

Question 44

homozygote vs heterozygote

Answer 44

homo- an individual having two identical alleles of a particular gene hetero- an individual having two different alleles of a particular gene

Question 45

monohybrid inheritance

Answer 45

the inheritance of just one characteristic. (e.g. a punnet square where each parent is heterozygous and there is a 25% chance of the offspring being a recessive homozygote)

Question 46

Purpose of genetic screening:

Answer 46

To determine if the DNA of an individual contains alleles for genetic disorders. It can be also used to identify carriers of a genetic disorder.

Question 47

3 types of genetic screening:

Answer 47

1. pre-implantation genetic diagnosis- embryos created through IVF are tested for genetic disorders before they are implanted into a woman’s uterus. 2. Chorionic villus sampling- 8-12 weeks of pregnancy. A sample of embryonic tissue taken from the placenta is analysed. (quicker than amniocentesis) 3. Amniocentesis- 14-16 weeks. A sample of amniotic fluid (containing foetal cells) is obtained using a needle. The DNA is analysed. Results are available after 2-3 weeks as foetal cells need to be grown in a culture first.

Question 48

Social/ ethical views surrounding genetic testing:

Answer 48

Disadvantages: - risk of harm to foetus or misscarriage - the outcome of testing may lead to an abortion- debate of right to life -invasive procedures may be stressful for the mother, potentially harming the foetus -controversial religiously Advantages: -the cost of bringing up a child with a genetic disorder, therefore a warning can prepare the family -emotional and mental issues surrounding caring for a baby with a genetic disorder, therefore the option of abortion may reduce stress for the family later.