1. Module 1 – Cellular control

2. 1) Coding for proteins from DNA GENE – length of DNA that codes for a polypeptide POLYPEPTIDE – many amino acids joined together by peptide bonds GENOME- the entire sequence of DNA of an organism PROTEIN – large polypeptide. Genetic code characteristics – 1) triplet code, 3 nucleotide bases = amino acid. 2) degenerate code – most AA have more than one code. 3) some = STOP codon, end of polypeptide chain. 4) Widespread but not universal, most 3 bases = the same AA in all organism but have a few exceptions. Transcription is the first stage of protein synthesis where mRNA is made. Made by copying the template strand (DNA) using free nucleotides with 2 extra Pi (releases energy). 1) gene unwinds and unzips, breaking the H+ bonds between complementary bases 2) RNA nucleotides bind to exposed comp. Bases. U=A, C=G, T(template)=A. Catalysed with RNA POLYMERASE 3)extra Pi are release for energy to bond adjacent nucleotides. 4)mRNA is complementary to template strand – COPY OF BASES ON CODING STRAND OF DNA. 5)mRNA released through nuclear envelope pores to ribosome. Translation is second stage of protein synthesis. Sequence is dictated by the codons. Sequence is essential as it forms primary structure of protein, this determines tertiary structure which is what allows it to function, where the substrates fit or channel protein wont allow ions to pass through on membrane. tRNA is also used to bring about the amino acid to form a polypeptide chain, it has a anti-codon, which binds to the complementary codon on the mRNA strand. 1) mRNA binds to ribosome and 2 codons (6 bases) exposed, first is ALWAYS AUG, and forms hydrogen bond between them. 2) second tRNA brings different amino acid and binds to 2 nd complementary codon. 3)peptide bond form between two amino acids, using enzyme in small ribosome subunit to catalyse 4)RIBOSOME moves along strand reading codons, then brings about another tRNA and forms peptide bond, first tRNA leaves. 5)polypeptide chain continues until stop codon is reached. Some proteins have to be activated by cAMP which activates proteins by changing their shape in order for a better fit to their complementary molecules.

3. (1)Mutations MUTATION- random change to the amount or arrangement of genetic material CHROMOSOME MUTATION – changes to the structure of a chromosome or number of chromosomes DNA MUTATIONS- changes to the genes because of changes in nucleotide base sequence. Mitosis mutations are SOMATIC and are not passed on to offspring, whereas meiosis mutations can be passed on to offspring. 2 types of DNA mutations – 1) Point mutations = one base pair replaces another, AKA substitutes. 2) Insertion/Deletion = one or more nucleotide bases inserted/deleted from length of DNA, can cause a FRAMSHIFT. Point mutation can equal – MISSENCE = changes the amino acid but polypeptide chain continues but slightly different - NONSENSE = STOP codon is reaches and polypeptide chain is stopped - SILENT MUTATION = although changes base, amino acid coded and polypeptide chain continues in the same way - FRAMESHIFT = base is removed and completely alters the sequence as all the amino acids are now different from original polypeptide chain. DNA mutations can cause genetic diseases such as 1) Cystic fibroses = triplet deleted, so amino acid is removed from chain of 1480. 2) Sickle-Cell Anaemia = point on triplet 6 of gene for B-polypeptide chains of haemoglobin. Valine instead of Glutamic Acid. 3) Protooncogenes (growth promoting) changed to oncogenes by point so they can’t be switched off, unregulated cell divisions leads to tumours. 4)Hungtington disease = expanded triple repeat (a stutter). Normal gene has repeating CAG but if above threshold protein is altered to disease. If gene altered it becomes an allele for the gene and may not produce any change if in non-coding region of DNA or is a silent mutation. But if is altered but has no advantages/disadvantages then is neutral. Mutations can have both harmful and beneficial effects. These include the pigment melanin (skin colour) which protects again UV. So in black people they could still get VIT D without getting skin cancer, and a mutation to white skin happed as migrate north and couldnt get enough VIT D, but those who did had the advantage and their children wouldn’t be born with defects such as rickets. THIS IS NATURAL SELECTION, AND WITHOUT GENETIC MUTATIONS THERE WOULD BE NO EVOLUTION.

4. 1) The LAC operon E.Coli normally respire in glucose, but can adapt to respire to metabolise which ever nutrients are available, hence being able to respire in lactose. When first placed in lactose (e.g milk) medium they do not metabolise at first as only small amounts of B-galactosidase ( hydrolysis of lactose to glucose) and lactose permease ( transports lactose into cell) but after few minutes rate of synthesis increased by 1000x. Lactose is the inducer for these proteins, but only triggered if no glucose is present. LAC operon is section of DNA within bacterium’s DNA and has number of parts... Structural Genes – z codes for B-galactosidase and Y codes for lactose permease. Each can be transcribed into length of mRNA. Operator Region – O, length of DNA next to structural and can switch them on and off. Promoter Region – p, length of DNA which enzyme RNA polymerase binds with to begin transcription of structural genes. LACTOSE ABSENT FROM GROWTH MEDIUM: 1)Regulatory gene expressed and repressor protein is synthesised. Has two binding sites, one to OPERATOR region and one to LACTOSE. 2)Repressor protein bins to O, but also covers P where RNA polymerase normally attaches. 3)RNA polymerase cannot bind so structural genes cannot be transcribed into mRNA. 4)Without this mRNA genes cannot be translated and B-galactosidase and lactose permease cannot be synthesised. LACTOSE PRESENT IN GROWTH MEDIUM: 1)Lactose molecule binds to repressor protein and alters its shape so cannot bind to operator region and breaks away from it. 2)Leaves the P region unblocked so RNA polymerase can now bind and cause transcription of Z and Y 3)Operator-repressor-inducer system to take up lactose from medium into cells. Glucose that have been converted and can now be used from respiration, thus gaining energy from the lactose.

5. 1) Genes and body plans Most of our knowledge of cell differentiation comes from study of Drosophila. Drosophila development. Eggs are laid and mitotic divisions are triggered at rate of one every 6-10 minutes (one of fastest known). At first no new cell membranes form and multinucleate synctium is formed (lots of nuclei) After 11 th division the nuclei form outer layer around central core. Division rate slows and nuclear genes switch from replication to transcription. Plasma membrane invaginates (fold inwards) and 6000 nuclei from single outer layer. Embryo then divides into series of segments which correspond to body plan. Three segments merge to produce head, three thoracic segments and 8 abdominal segments. At metamorphosis (larval forms to adult) the legs, wigs and antennae develops. Genetic control of Drosophila development: MEDIATED BY HOMEOBOX GENES -Some genes affect embryo’s polarity (maternal-effect genes) -Others specify polarity of segments -Mutations can occur, and seen in antennepedia ( antennae of Drosophila look more like legs) Homeobox genes in genomes of many organisms, including humans. Contain sequence of 180 base pairs and produces polypeptides of about 60 amino acids. Arranged in clusters known as HOX CLUSTERS. Roundworms = 1 HOX CLUSTER, Drosophila = 2 and Vertebrates = 4. They are activated in same order as expressed along body plan, from anterior to posterior. Retinoic acid can interfere with normal expression of these genes and can cause birth defects. It is a derivate of VIT A and activates the homeobox genes so its a MORPHOGEN.

6. 1)Apoptosis Apoptosis is programmed cells death. Cells should undergo 50 mitotic division (HAYFLICK CONSTANT) and should lead to a tidy cell death. Controlled by range of signals, including nitric oxide which can induce apoptosis by making inner mitochondia membrane more permeable to H+ ions and dissipating proton gradient. The normal sequence of cell death : 1) enzymes break down cell cytoplasm 2) cytoplasm becomes dense and organelles become tightly packed 3) Cell surface membrane changes and blebs form 4) Chromatin condenses and nuclear envelope breaks, DNA breaks into fragments. 5) cell then breaks into vesicles and taken up by phagocytosis. Debris disposed of and doesn’t damage any other cells or tissues. 6) whole process occurs VERY quickly. Apoptosis is extensive division of particular cell type followed by programmed cell death. Excess cells shrink and are phagocytosed so components are reused and nothing harmful released into surrounding tissue. Apoptosis tightly regulated during development and can weed out inefficient or harmful T Lymphocytes during development of immune system. During limb development it causes digits (toes/fingers) to separate from each other. the rate of cells dying should always balance the rate of cells being produced by mitosis, but if not balanced then NOT ENOUGH APOPTOSIS = formation of tumors and TOO MUCH APOPTOSOS = cell loss and degeneration. Cell signalling plays a crucial role in the maintaining the right balance.

7. 1)Mitosis Because two gametes fuse to make one cell and the chromosomes are combined they number of chromosomes in each gamete must be a haploid (1/2 original number). Meiosis occurs in two stages: I & II: Prophase I : chromatin condenses and chromosomes come together in homologous pairs (matching) to form a bivalent. Both have same genes at same loci. Non-sister chromatids wraps around at certain point (chiasmata) and cross over to swap sections. Nucleus disappears and envelope disintegrates and spindle forms. Metaphase I: Bivalent line up across equator and attach to spindle, bivalent arranged randomly with each member of homologous pair at each pole so segregate independently when pulled apart. Anaphase I: palled apart to each pole by spindle, centromeres do not divide by the chiasmata remain with them so are not separated. Telophase I: 2 nuclear envelopes form and briefly chromosomes uncoil. Prophase II: nuclear envelope has reformed by breaks down again, chromosomes condense and spindles form. Metaphase II: arranged at equator, the chromatids of each chromosome randomly arranged. Anaphase II: centromeres divide an pulls to poles by spindle fibres. Telophase II: Nuclear envelope reform around haploid daughter nuclei. Animal: 2/ = 4. Sexual reproduction increases genetic variation, increased by the crossing over (shuffles alleles), random mutation, random re- assortment. - Crossing Over : chiasmata are where they are broken so gene has a different allele to previously. -Re-assortment: chromosomes across the spindle equator at MET I, but also chromatids at MET II. -Fertilisation: 300 million sperm that are all genetically different so one can only be used for each egg. -Mutation: increases genetic variation as mutated gene present in every single cell of offspring.

8. 1) Important terms Genotype – entire genetic make up of organism Homozygous – organism with two identical alleles of particular gene Heterozygous – organism with two different alleles of particular gene Phenotype – characteristics expressed in the organism, features that can be observed. Dominant Allele – always expressed in the phenotype, even if different allele for same gene is present.(Bbaa) Recessive Allele – only expressed in the phenotype is in presence of identical allele ( aa ) or is absence of dominant (bbaa) Co-dominance – if two alleles on the same gene are both expressed in phenotype of heterozygote. (C w C r = roan colour, not red/white) Linkage – two or more genes that are located on same chromosome and are normally inherited together unless chiasmata form between them. Reduced the number of phenotypes from a cross. (e.g. pea plants have gene fall height and gene for texture of mature seeds both on chromosome 4. Sex Linkage – is gene that codes for it found on one of sex chromosomes (Y/X), most on X chromosome. Females in human are homogametic sex – XX, but birds, butterflies and moths males are homogametic.

9. 1) Using genetic diagrams 1) to solve problems involving sex linkage. – Haemophilia A ( when blood does not clot so continuous bleeding. Protein ( factor VIII) coded for on gene on X chromosome, so is a male has the allele they will get it as only have one X chromosome. Parental phenotypesCarrier motherNormal father Parental genotypesXHXhXHXh XHY GametesX H X h X H Y Male Gametes Female Gametes XHXH Y XHXH X H X H – Normal F X H Y – Normal M XhXh X H X h – carrier F X h Y – Carrier M 2) to solve problems involving co-dominance. -sickle-cell anaemia (all individuals have the same mutation –B-polypeptide of haemoglobin at position 6 so valine produced instead of glutamic acid. Build up in capillaries and can stop blood flow and organs can become damaged. -Normal genotype – H A H A -Sickle-Cell – H S H S -Symptomless heterozygotes - H A S A Parent GenotypesHAHSHAHS HAHSHAHS GametesH A H S F (DOWN) M (ACROSS)HAHA HSHS HAH A H A (Normal)H A H S (Carrier) HSH A H S (carrier)H S H S ( Sufferer) E.g. Roan Cattle -Coat colour has two alleles ( C R = red, C W = white) -Homozygous C R C R have red coats, C W C W have white coats -Heterozygous C R C W have red and white hairs and have roan coats. If two homozygous then all roan (F1), and if two roan (F2) then 50 % roan and 25%white red.

10. 1) Interactions between gene loci There are cases where different genes on different loci interact at affect one genotype characteristic. One gene masks the expression of another gene – EPISTASIS. They may do this by working antagonistically(masking each other) or together in a complementary fashion. ANTAGONOSTICALLY: homozygous presence of recessive allele prevents expression on another allele at second locus. First locus are epistatic to the second, which are hypostatic. Epistasis is not inherited and reduces the phenotype variation. - Recessive epistasis: e.g. colour of Salvia flowers. 2 gene loci, A/a and B/b on different chromosomes are involved. pure breed pink-flowers have genotype AAbb, crossed with pure breed white-flowers have genotype aaBB, all F1 generation have AaBb(purple flowers). Interbreeding F1 had a ratio of 9:3:4. Homozygous aa is epistatic to both alleles of B/b. Neither B (purple) or b (pink) can be expressed if no common dominant allele, A, present. -Dominant epistasis: e.g. fruit colour of summer squash. -This occurs when a dominant allele at one gene mask the expression of the alleles at the second gene locus. Two gene loci invloved – D/d and E/e. One D = white fruit regardless of E/e. Homozygous dd, one E = yellow and 2 ee = green. 2 white, double heterozygous (DdEe) crossed, F2 is 12:3:1 Bateson and punnett (square) cross 2 white-flowers sweet peas ( ccRR x CCRR). – All F1 had purple flowers – F2 had purple/white 9:7 – Explanation = one dominant allele for both gene loci (C-R-) present for purple flowers as all others (ccR-,C-rr) are white – Because Homozygous recessive condition at EITHER locus masks expression of dominant allele at other – COMPLEMENTARY FASHION AS one gene codes for intermediate colourless pigment and second locus codes for enzyme that converts to final purple colour. colourless purple C/c R/r

11. 1) Determine sex? Human’s males = HETEROGAMETIC (XY) but in animal kingdom birds, butterflies and moths FEMALES ARE HETEROGAMETIC. In some animal (turtles), sex of offspring determined by temperature that eggs are incubated. Higher- = FEMALE. Oysters can change sex during life span. PLYMOUTH ROCK CHICKENS – - new hatched have to be sexed so farmer keep females as they are egg layers, hard because males don't have penis. – Dominant sex-linked allele, B, normal black feathers with white bar. New hatched chicks with B have barred feathers (white spot on black head), b have non barred and black head. – To produce pheneotype that can easily be sexed: males = XX and females = XY Female with white spot = X B Y Male with black head = X b X b (characteristic is recessive). Ratio of 1:1, and all females have black head and all males are barred so easy to tell the sex. In pale brindled moth there is a rare form black one. They are all female and when mated with a normal allele all the females are black and males are normal. this is because the allele for the black is dominant when on the Y chromosome, meaning that the males cannot get it. (XY B ) Parental phenoMale with black head Female with white spot Parental genoXbXbXbXb XBYXBY GametesX b X B Y Male gametes Female gametesXbXb XBXB X B X b (male barred) YX b Y (female black head)

12. 1)Interactions between gene loci 2 Coat colour in mice are either agouti (bands), black or albino. Gene for agouti has 2 alleles A/a. a is a mutation (when in homozygous produces black). Gene B/b on separate loci controls formation of pigment. Genotype B- can produce pigment, but bb cannot and are albino. When several agouti (AaBb) crossed, total is 9:3:4 (ag, bl, al) There are four comb shapes in domestic chickens. 2 gene loci, P/p and R/r interact to affect this. Effect of P/p depends on which R/r alleles are present in genotype. PPrr (pea-combed)bread with ppRR (rose-combed), progeny have PpRr (walnut combed). Walnut combed x walnut combed = 4 phenotypes : walnut combed, rose comb, pea comb, single comb in classic mendellion 9:3:3:1 for F2. colourless agoutiblack B/bA/a m(across) F (down) gametes PRPrpRPr PRPPRR walnutPPRr WalnutPpRR WalnutPPRr Walnut PrPPRr walnutPPrr peaPpRr walnutPPrr pea pRPpRR walnutPpRr walnutppRR rosePpRr walnut prPpRr walnutPprr peappRr rosePprr single GenotypePhenotype P-R-Walnut comb ppR-Rose Comb P-rrPea comb PprrSingle comb

13. 1) Chi-squared (x 2 )test If we think two characteristics are determined by 2 unlinked genes, expect a 9:3:3:1 ratio in F2. with large sample, not always exactly to ratio so must use a statistical test to see if significantly close, and if not model which it is based must be wrong. Chi-squared test the null hypothesis (says that no significance different between observed and expected and it is all chance). If no significance difference then we accept the null hypothesis,. RrGg pea plants grew 288 in F2 and produced phenotypes: 169 yellow/round, 54 green/round, 51 yellow/wrinkled 14 green/wrinkled. Close to 9:3:3:1 (16) expected should be yellow/round - 9/16 x 288= 162, yellow/wrinkled -3/16 x288 = 54, green/round – 3/16 x288 = 54, green/wrinkled – 1/6 x 288 =18. So calculate values of X 2: – X 2 = the sum of (E) ((observed no, O, - expected no, E)) 2 / expected no, E. – when coming up with this 3 factors taken into account. They were 1) difference may be +/- so they are squared to stop – cancelling out +. 2) dividing by E takes into account size of numbers. 3) sum takes into account number of comparisons being made. – Look up the calculated value of chi-squared in distribution table. If calculated value is smaller than critical then we accept null hypothesis but if not test is wrong. – Number of classes are the number of categories and the degree of freedom is this by -1.

14. 1)(Dis)continuous variation Discontinuous variation - the qualitative differences between PHENOTYPES, clearly distinguishable categories such as male/female, blood group. Continuous variation – the quantitative differenced between PHENOTYPES. No distinct categories such as height/mass of humans, cob length in maize or milk yield in cattle. Both types of variation can be result of more than one gene, but in dis, if more than one gene involved and they act in epistaic way and if only one gene its called monogenic. E.g. Cystic fibrosis, as two faulty alleles for CFTR gene, which codes for CF trans-membrane regulatory protein – chloride channel in membranes of lungs. DISCONT – – Different allele at same locus have large effect on phenotype – Different gene loci have quite different effects on phenotype E.g. Co-dominance, dominance and recessive patterns of inheritance. CONT- – Traits exhibiting cont controlled by 2 or more genes. – Each gene provide additive component to phenotype – Different alleles at each gene locus have small effect – Large number of genes have combined effect Known as polygenes and characteristic controlled knows as polygenic. They genes are unlinked, and they are on different chromosomes. – E.g. Controlling length of corn cobs. Alleles of each gene add 1cm(recessive) or 2cm(dominant) to cob length. So genotype AABBCC have cob length on 12cm, but aabbcc have cob length of 6cm. Although the genetic potential is there it is not always reached because of environmental factors that limit the expression of the genes. E.g. If inherited genetic potential to intelligence may not be expressed fully unless guide of school and home, good nutrition. Despite natural selection or artificial selection, variation is crucial to population. This is because when an environment changes those who are well adapted will survive and reproduce, passing on their advantages to their offspring. This is the basis of evolution by natural selection.

15. 1) Population genetics Gene Pool – the set of information carried by a population and can be measured using the HARDY-WEINBERG PRINCIPLE. Factors such as migration, selection, genetic drift and mutation can alter the amount of info in a gene pool. When studying Darwin’s evolution, scientists noted that they need to consider the frequency of alleles in pop, and not just offspring from individual meetings. In population genetics, focus is on the genetic structure of populations and they measure changes in alleles and genotype frequency down generations. The phenotype is observed in individuals and to measure the frequency of an allele must know: mechanism of inheritance for particular trait and how many different alleles of gene for trait is in population. Co-dominace traits, frequency of heterozygous phenotype same as genotype. – e.g. MN blood group, gene L has 2 alleles: LM and LN. – Each allele controls production of spec antigen on surface of RBC. Individual may be (pheno) M (geno LMLM or MM), type N (LNLN or NN) or type MN(LMLN or MN). Because they are co-domiant, determine frequency of alleles in population. BUT, if one allele is recessive, the heterozygote shows same phenotype as homozygote dominant. So frequency cannot be directly determined. Hardy-Weinberg principle is a mathematical model to calculate allele frequency in both recessive and dominant populations. Makes 4 assumptions: 1)population is very large (eliminates sampling error), 2) mating within population is random, 3)no selective advantage for any phenotype, 4) no mutation, migration or genetic drift. WORKED EXAMPLE FOR CF. Heterozygotes CFcf are symptomless carriers, and those affected are cfcf. In population of 2000, 1 person has CF. We want to know how many in pop are carriers. P = freq of dominant CF Q=freq of recessive cf Q 2 -= freq of genotype cfcf P 2 = freq of genotype CFCF PQ= freq of genotype CFcf, Dom from dad and rec from mum. Also cfCF also PQ. So freq of all heterozygotes is PQ+PQ=2PQ So for random mating freq of all genoypes is P 2 +2PQ + Q 2. Within pop freq of alleles P & Q add up to 1 (100%) Within pop freq of genotypes P 2, 2PQ,Q 2 = 1 (100%) So... Q 2 = 1 in 2000 so Q 2 = 0.0005 Q is square root so = 0.022 If P = Q = 1 then P = 1 - 0.022 so 0.978 The freq of carriers in 2PQ so 2 x 0.978 x 0.022 So 2PQ = 0.043 So this means that 4.3 people in 100 are carriers. 2000 x 4.3/100 = 86

16. 1) Roles of genes/environment in genetics. Many populations will reach their carrying capacity and remain stable, so sometimes not all of the young survive as if they did they population would continue to expand. Environmental factors limit grown of population such as the amount of space available, availability of food, light, minerals, water, predication and infection. These offer environment resistance. Some are abiotic ( NON-LIVING) and some are biotic (LIVING). Because of variation some members are better adapted and can out-compete others. E.g the predator who can run faster or has sharper teeth had advantage in the struggle for existence. Greater chance of surviving and passing these alleles onto offspring. In prey such as rabbits Selection Pressure allows the animal to escape and live to a reproductive age. For example mice with a agouti coat can blend in more so that allele is passed on more than white/black. Natural selection keeps things they way they are – Stabilising selection. If new phenotype does arise unlikely to survive as will not have the same advantage. But, if environment does change then selection pressure will change – e.g. Climate change and ground covered in snow, those with white fur will have an advantage. Frequency of these alleles in gene pool will increase. This directional selection leads to evolutionary change and it is a evolutionary force of natural selection. Large populations will be separated into sub-groups by various isolating mechanisms : 1) geographical barriers, such as river. 2)seasonal (temporal) barriers, such as climate change through year. 3) reproductive mechanisms, no longer physically able to mate – incompatible genitals. – Leaves two sub-populations that are then isolated from each other and is each case a different allele will be eliminated/increased within each one. Eventually the sub-species will not be able to inter-breed and will be a different species. Small populations may occur with natural disasters such as disease pandemic or volcano eruption that splits a species into two different areas and each one my evolve into an animal with more advangtage that can reach reproductive age and pass on the more favourable alleles.

17. 1)What is a species? Biological species concept is a group of similar organisms that can interbreed and produce fertile offspring. Concept is problematic when classifying living organisms that do no reproduce sexually, also some members of same species look different from each other and some isolated populations may appear to be very different from each other. Phylogenetic species concept is a group of organisms with a similar morphology(shape), physiology( biochemistry), embryology (stages of development) and behaviours, and all occupy the same niche. with improved methods of DNA sequencing can compare base sequences (haplotypes) on chromosomes of spec organisms. Analysis of this is carried out, and the differences, caused by base substitutions, are expressed as % divergence. % divergence = (number on substitutions / number of base pairs analysed) x 100 Any group of organisms with haplotypes that are more sim that those in any other group are called a clade. Use of molecular systematic (analysis) is a cladistic approach to classification. Assumes that classification corresponds to Phylogenetic descent and all valid taxa (groups) much by monophyletic. clade is a taxonomic group comprising single ancestral organism and all its descendants. Cos of this its described as a monophyletic group. Cladistics is hierarchical classification of species, based on evolutionary ancestry. Different from taxonomic classification system as: 1)Focuses on evolution, rather than similarities between species. 2)Places great importance on using objective and quantitative analysis 3)Used DNA and RNA sequencing 4)Uses computer programmes and data from nucleic acid sequencing to generate dendrograms/ cladograms to represent tree of life cos manual creation would be hard when dealing with large numbers of species. 5)Makes no distinction between extinct and extant species, both may be included in cladogram. Cladistics is different from Linnaean classification as doesn’t use kingdom, phylum, class and regards tree very complex. Not helpful to used a fixed number of levels. Cladistic approach often confirmed Linnaean and sometimes led to them being re-classified. Helped biologist understand the evolutionary relationships between species.

18. 1)Natural/artificial selection Artificial selection is when humans select the organisms with the useful characteristics and breed these together to get the best. TO PRODUCE THE MODERN DAY DAIRY COW – Main breeds of dairy cattle with high milk yields are : Holstein- Friesian, Guernsey, Jersey and Milky Shorthorn. – Repeatedly selected cows with high milk yields and bred them over generation. This is down by 1) measuring each cow’s milk yield and recorded 2) Progeny of bulls tested to find which daughter has highest milk yield 3) Good quality bulls kept as semen can be collected and artificially impregnate females. 4) Some elite cows given hormones so produce many eggs 5) Eggs fertilised in vitro (glass) and implanted into surrogate mothers. 6)Embryos can be cloned to give identical embryos. TO PRODUCE BREAD WHEAT, TRITICUM AESTIVUM – most wild wheat are diploid with 14, so haploid is 7. can undergo polyploidy (nuclei contain more than one diploid set of chromosomes). Modern bread is hexaploid, so 42 chromosomes in nucleus of each cell. Contains 3 genomes, AUAUBBDD Genome AUAU from wild wheat, T.Urartu Genome BB from wild emmer wheat, T. Turgidum Genome DD from wild goat grass, Ar. Tauschii. Using Linnaean all wheats that can interbreed are classified as being same species, but more recent genetic classification been used. Both are valid, but should not be confused. Winter wheats grown in UK as climate not as extreme, wheat has soft grains with low protein content for biscuits but not bread. Spring wheat (colder winters) harder, higher protein content and can be used for bread. Characterises that farmers focus on when breeding wheat are those such as high resistance to fungal infections, increased yield, high protein content, straw stiffness and resistance to lodging (stems bending over in wind).