Gene Expression Notes PDF

Gene expression As Biology ( WBI11) Jan 2025 Gene expression Class work : 1. State definition of mutation 2. Why mutation cause changes in new cells 3. State types of mutations 4. State types of chromo...

Gene expression As Biology ( WBI11) Jan 2025 Gene expression Class work : 1. State definition of mutation 2. Why mutation cause changes in new cells 3. State types of mutations 4. State types of chromosmal mutation 5. Explain why most mutations have no observable effect 6. Explain how change in DNA sequence lead to change in enzymatic activity 7. features of sickle cell anaemia 8. Gene / dominant / recessive /test cross/ codominance / sex linked  > - Gene mutation Random change in base sequence of DNA during DNA replication ….thus producing new allele..leading to protein of different function / shape. Where errors are being copied during replication Causes of mutation : Exposure to mutagens : When wrong base is being inserted chemicals as tobacco smoke and mustard gas Triplet code……DNA Physical such U.V rays , x rays Codon…..mRNA Types of mutation Anticodon ….tRNA Point mutation Frame shift ( substitution ) - Affects only one triplet Deletion Insertion Change in one single base of DNA code One nucleotide is One extra missed out..so Change nucleotide is Has one of three effects in one single base of inserted ….so the DNA code entire base entire base Silent mutation Non sense Mis sense sequence is sequence is altered altered..coding for an after mutation..so The base The base substitution can The base entirely different substitution is substitution can be all amino acids be non sense. Where the protein … where all silent..where the mis sense mutation amino acids are after mutation are altered codon altered codon where the altered different after affected ,code for code for same corresponds to a stop codon , code for mutation …so 3D shape different protein amino codon , so new poly another amino acid of protein is changed acid..genetic peptide might be shorter The to Closer to the start , the greater the effect code degenerate CCA…proline TTC….phenylaninine AGC….serine CCT…..arginine Shift reading backwards one TTG…stop codon AGT….serine place Shift forward one place  r ab lg iha.N Dr Dr.NIhal Gabr 151 6 Errors in DNA replication can give rise to mutations. The diagram shows the bases in a length of DNA. C Length of DNA A T G C T C A T T T A C C A T C G A Base number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 The table shows the genetic code for the amino acids. Genetic Amino Genetic Amino Genetic Genetic Amino acid Amino acid code acid code acid code code AAA CAA GAA TAC Lysine Glutamine Glutamic acid Tyrosine AAG CAG GAG TAT TCA AAC CAT GAC TCC Asparagine Histidine Aspartic acid Serine AAT CAC GAT TCG TCT ACA CCA GCA ACC CCC GCC Threonine Proline Alanine TGG Tryptophan ACG CCG GCG ACT CCT GCT CGA GGA AGA CGC GGC TGC Arginine Arginine Glycine Cysteine AGG CGG GGG TGT CGT GGT CTA GTA AGC CTC GTC TTA Serine Leucine Valine Leucine AGT CTG GTG TTG CTT GTT ATA TTC ATC Isoleucine Phenylalanine TTT ATT - ATG Methionine The genetic codes TAA, TAG and TGA are stop codons. (a) State the sequence of the first four amino acids coded for by this length of DNA. (1) Methionine, leucine , isoleucine, tyrosine.................................................................................................................................................................................................................................................................................... 14 *P62792RA01428*  (b) A change in a single base can cause a change in the amino acid sequence produced in protein synthesis. (i) Name the type of each mutation described below. (2) Base number 3 becomes cytosine (C).................................................Substitution.............................................................................................................................. Deletion Base number 6 becomes number 5 in the sequence........................................................................................................................................ Insertion Base number 9 becomes number 10 in the sequence.................................................................................................................................... *(ii) Explain the possible effects of these three types of mutation on the amino acid sequence coded for by this length of DNA. Use the information in the table to support your answer. (6) Substation ( point mutation ).................................................................................................................................................................................................................................................................................... Where there is a change in a single base of one triple code.................................................................................................................................................................................................................................................................................... This may result in one of three possibilities.................................................................................................................................................................................................................................................................................... Silent mutation.................................................................................................................................................................................................................................................................................... Where the altered codon will code for the same amino acid due to the fact that genetic code are degenerate ✓........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ Example: if base 6 was replaced by adenine then CTA still coding for leucine. Non sense.................................................................................................................................................................................................................................................................................... Where the altered codon will be stop codon , terminate translation early , so.................................................................................................................................................................................................................................................................................... shorter polypeptide........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ Example if base 12 was replaced by Adenine Mis sense.................................................................................................................................................................................................................................................................................... Where the altered codon code for a new amino acid.................................................................................................................................................................................................................................................................................... Example if base 3 replaced by cytosine , then this will code for isleucine instead of methionine..................................................................................................................................................................................................................................................................................... Deletion.................................................................................................................................................................................................................................................................................... Where one base is removed , causing the entire sequence to be altered (shift the reading.................................................................................................................................................................................................................................................................................... frame backward one place)..................................................................................................................................................................................................................................................................................... So all amino acids after mutation will be altered. (Total for Question 6 = 9 marks) Example: removal of base 4 cause sequence to become methionine serine phenylalanine threonine. 15 *P62792RA01528* Turn over Where one base is added twice , causing the entire sequence to be altered (shift the reading frame forward one place). So all amino acids after mutation will be altered. Example: adding T between base 9 and 10 sequence becomes leucine proline serine  B. Chromosomal mutation Whole Chromosomal chromosome mutation mutations Change in the position of entire The loss or duplication of whole chromosome genes within a chromosome. during meiosis. Example: down syndrome which is caused by a whole chromosome mutation at chromosome 21 - Gene mutation : Read only change in nucleotide( base) sequence of DNA , so a new allele is formed. This can happen by deletion , substitution, addition. r This change lead to change in the transcribed mRNA (mRNA with altered codons). ab In case of substitution : a new amino acid with different R group may be incorporated into the growing chain at the ribosome during translation. Causing a change in primary structure of protein( amino acid sequence on polypeptide). lg Change in three dimensional shape of protein. So different protein with altered function or totally un functional protein may be produced. iha Also mutation can lead to cancer characterised by uncontrolled cell division to form a mass of cells (functionless) known as a tumor..N 1 DNA- 2 mRNA- 3 Polypeptide- One triplet is One codon is One amino acid is different different different Dr (substitution) 4 Function of protein- polypeptide forms a different shape. Amino acid Phe replaced by Leu so bind cant be firmed to form shape of active site 5 Phenotype- enzyme can’t function because shape of active - site is changed. Lack of enzyme causes a genetic disease. Individual is not healthy. Dr.NIhal Gabr 152  Explain why most mutation has no observable effect ?\ 1. Genetic code is degenerate 2. DNA repair mechanism 3. Occur in non coding DNA 4. One allele might be affected Explain how a change in DNA sequence might lead to loss of enzyme activity ? 1. Mutation takes place where there is a change in the base sequence of gene , so new allele is formed 2. Results in change in mRNA codons 3. So different tRNA with different antocodons will be involved , as the tRNA will carry different ( incorrect ) amino acid to the ribosome 4. So incorrect amino acid might be incorporated into the growing polypeptide chain , so change in sequnec of amino acids meaning changing the primary structure 5. Polypeptide folding will be different so change in tertiary structure ( 3D) 6. Active site will have a different shape 7. So substrate can no longer bind to active site 8. So ESCs Features of sickle cell disease Haemoglobin carrying capacity is reduced / haemoglobib carry less oxygen. RBCs become sickle shaped so smaller surface area so less gas exchange by allowing less diffusion of oxygen from RBCs to body cells Blood vessels get blocked so reducing amount of oxygen being transported to body cells so less aerobic respiration So heart muscles become deprived of oxygen , so heart attack/ angina  Features of sickle cell disease 1. RBCs become sickled shaped 2. Less / smaller surface area 3. Haemoglobin less soluble 4. So less able to bind to oxygen 5. So less oxygen diffuse from RBCs in blood to body cells 6. / less O2 transported to body cells 7. So less aerobic respiration..less energy ….anaemia 8. As well as heart attack if heart muscle is deprived of oxygen 9. RBCs stick together and block capillaries OR Haemoglobin carrying capacity is reduced / haemoglobib carry less oxygen. RBCs become sickle shaped so smaller surface area so less gas exchange by allowing less diffusion of oxygen from RBCs to body cells Blood vessels get blocked so reducing amount of oxygen being transported to body cells so less aerobic respiration So heart muscles become deprived of oxygen , so heart attack/ angina  Second: 2C.2 :pattern of inheritance: Gene: A length of DNA coding for specific protein, determining specific characteristics. Allele: Alternative forms of same gene. May be: Dominant Recessive r Allele expressed in phenotype only ab Allele expressed in phenotype when individual is homozygous for whether the individual is that recessive trait.( both alleles homozygous or heterozygous for coding for recessive trait) that allele. lg Homozygote : Heterozygote : iha An individual An individual when both alleles where the 2 coding for a alleles coding for.N particular a particular characteristic are characteristic are identical different Dr Genotype: Genetic make up of an organism with respect to a particular feature OR combination of /pair of / two / all alleles present in an organism of a particular trait. Phenotype: All characteristics of an organism Determined by an interaction between genes(genotype) and environment (observable features) Dr.NIhal Gabr 154 True breeding: A homozygous organism which will always produce the same offspring when crossed with another true breeding organism for the same characteristic. which means the parents must be both dominant or both recessive. Monohybrid cross A genetic cross where only one gene for one characteristic is considered. r ab lg Test cross A test made to find out the genotype of an individual with dominant phenotype iha for a particular gene by crossing it with one known to have the homozygous recessive genotype for the same gene. To reveal the parental genotype ( being homozygous dominant or heterozygous..N Dr Dr.NIhal Gabr 155  Codominance: When pair of alleles are equally dominant , so in heterozygous where both alleles at a gene locus are fully expressed in the phenotype. Example blood groups I A and I B are codominant. This means both alleles are expressed and produce their proteins (antigens) which act together without mixing. A B So person with genotype I I will have both antigens, antigen A and antigen B on the surface of their erythrocytes and will have blood group AB. 1 2 r ab Parental phenotype : mother group A Parental phenotype : mother group B lg Father group O Father group A iha 3 Exam hint: When asked to explain why a certain organism is used for.N scientific experiments: 1. May have short life cycle. 2. Organism may be cheap Dr Parental phenotype : mother group A and easy to keep. Father group B 3. It may produce many offsprings Pair of alleles are equally dominant , so in heterozygous , where both alleles at a gene locus are fully expressed in the phenotype So the phenotype is intermediate of both A id dominant over O O is recessive B is dominant over O A and B are exmaple of codominance Dr.NIhal Gabr 156  Sampling error: The theoretical ratios of phenotypes that are predicted by a genetic cross are usually seen (approximately) in real genetic experiments , however, the numbers are never precise. This is may be due to: 1. Reproduction is a result of chance. Where combination of alleles in each gamete is completely random and so is the joining of particular gamete. Yet the theoretical diagrams we draw doesn’t show this. 2. Some offsprings die before they can be sampled. Example some seeds don’t germinate and some embryos miscarry. 3. Inefficient sampling techniques, example its very easy to allow few Drosophila to escape. Solution r 1. So sampling error must be taken into account specially when you are using smaller sample ab 2. use fast growing plants , Drosophila, and certain fungi and bacteria are so useful as they all produce large number of offsprings kn a short time. Why use drosophila in lg scientific exp ? Genetic pedigree: 1. Cheap and easy to keep Dominant feature..one allele Dd Dd D 2. Produce many offsprings / fast Recessive feature …2 alleles iha 3. Short life cycle Dd dd dd dd d dd dd dd dd dd dd.N dd dd Dr Recessive Learning tip: Always remember to look for the alleles shows individuals that show the recessive - low phenotype because - these are the only frequency Solution ones where you can be sure of the genotype- they are double recessive. Dr.NIhal Gabr 157 Sex linked diseases Genetic disease controlled by faulty allele ( resulting from a mutated allele), carried on a sex chromosome (x-chromosome) , making it more common in males than in females and is inherited on X- chromosome to the next generation from the parent. Recessive sex linked diseases, are more common in males than females, as males have only one X chromosome and thus can’t be heterozygous carriers, so if they inherit a recessive allele on X chromosome from the mother they are diseased. Example 1.Red green color blindness The normal (dominant) allele of the gene causes a protein to be produced that forms the pigment in the cones of the eye that detects green light. The mutant (recessive) allele does not cause this pigment to be formed ( b ) mutant recessive allele have difficulty seeing the difference between red & green.  2. Haemophilia condition caused by gene mutation which affects production of a certain protein that is important in clotting. The disease is recessive, h People with hemophilia tend to bleed internally Carrier female Normal male H XY h XX X X Y H h H X HH HhH H XX, XX, XY , XY Normal female , carrier female , normal male , haemophilic male Cystic fibrosis Faulty recessive allele Due to mutation in gene coding for CFTR protein on AUTOSOME Affect production of mucus by epithelial cells CFTR gene is a large gene ( higher risk of mutation ) where mutation can lead to abnormal CFTR protein and cause stick mucus 1. Normal CFRT gene 2. So code for a normal CFTR channel protein so can function properly 3. CL- ions leave the cells through CFTR proteins into the mucus. 4. Sodium channels inhibited by CFTR so Na+ ions remain outside sell in mucus 5. So NACL causes the mucus to be hypertonic 6. So water move out of the cells by osmosis 7. Resulting in water thinner mucus 8. Cilia can beat 9. Cilia can move mucus away from airways / bronchi/ bronchioles. 1. Mutation in the CFTR gene ( faulty allele ) leads to change in primary structure of CFTR protein 2. So CFTR cant function properly 3. CL- ions build up inside the cell 4. Water doesn’t move out of the cells by osmosis / or water moves out from mucus and enter the c by osmosis. 5. Resulting in thick sticky mucus 6. Cilia can’t beat 7. Cilia cant move mucus away as its TOO THICK 8. Reducing air flow to alveoli so reducing rate of gas exchange Where this will result in reduction in concentration gradient/ diffusion of gases ( as less oxygen is reaching alveoli ) 9. Mucus trap dust and bacteria , so can cause infection The effect on respiratory system 1. Thick sticky mucus 2. Narrowing air ways …trachea , bronchi and bronchioles 3. Reduce air flow into alveoli 4. Concentration gradient gradient between blood and alveoli will decrease 5. Reduce gas exchange 6. So less OXYGEN carried by blood 7. Less aerobic respiration 8. Less ATP 9. Lack of energy 10. Coughing sticky mucus 11. Pathogens reach the lungs increasing susceptibility to lung infection The effect on digestive system 1. Block to the pancreatic duct 2. No enzymes ( amylase / trypsin / lipase ) 3. Not reaching duodenum / small intestine 4. so digestion would be less efficient 5. Enzymes trapped in pancreas will start digesting cells of pancreas …diabetes 6. Block villi by excessive mucus so reduce surface area for absorption so less absorption of nutrients The effect on reproductive system : In women : weak chance of being fertile.Block the oviduct ……REDUCE CHANCE OF EGG MOVEMENT , decrease chance of fertilisation.Block the cervix , so decrease chance for the sperm to reach the egg. Implantation impaired In men : A) lack of vas deference / sperm duct..which is needed to carry sperm out of the testes into semen ( no sperm leaving the testes ) B) vas deference is present yet blocked by thick sticky mucus so less / no sperm leaving the testes The effect on sweat glands The effect on sweat glands With normal CFTR 1. CFTR mutation so less CL- ions 1. CFTR allow Cl- ions to move into the will move into the cells epithelial cells 2. So more salts (. Chloride and 2. So less Cl- ions are lost in sweat sodium ) are lost in sweat 3. Reducing water loss and preventing 3. So more water will leave epithelial dehydration cells by osmosis. Increasing chance of dehydration. Number of cases decreasing : 1. more Genetic screening 2. more carriers idesntified 3. More coupls choose IVF 4. More couples choosing not have child 5. Frequency of the allele decrease in population so there will be a decrease in CF phenotype 6. More migration / chnage in population diversity  Explanation video part 1 Genetic screening IPedigree analysis 1. Genetic testing to family 2. If one partner in a. Couple is known to be a carrier …..so its advisable to make the other partner tested. 3. If the other partner is aCarrier-has faulty allele …..they have to take a decision regarding having a child A) either not to have a child B) prenatal testing screening then take a decision. Amniocentesis Chorionic villus sampling 1. By taking 20 cm3 from the amniotic fluid using 1.taking sample from placenta 8 a needle at week 16 of pregnancy …..sample fetal to 12 weeks of pregnancy epithelial cells and blood cells 2. DNA anaylsis 2. Culture for 2 -3 weeks then screen / DNA analysed Advantages : 1. Carried earlier Advantages : ' 2. Results are available faster Less miscarriage risk 3. Larger sample. Disadvantages : Disadvantages: 1. 16 weeks is too late to take a decision terminating pregnancy ( traumatic) 1, 0.5 to 1 % spontaneous abortion. 2. Sex linked characteristics cant be 2. 0.5 to 1% spontaneous abortion detected 3. Expenxive , so not offfered to all pregnant woman.  Other genetic defects might be found Genetic testing Helping parents to decide A) prenatal testing screening then take a decision. B) preimplantation genetic diagnosis Amniocentesis 1. Where a parent already has a family history or a child affected by a genetic disease. Chorionic villus sampling 2. So they can carry pre implantation genetic diagnosis based on technique of IVF. Explanation video part 2 3. Where the sperm and the egg are fertilised outside the body 4. After few cell divisions , a single cell is removed from each embryo 5. Genetic make up is checked and only those embryos free of the problem alleles are placed into the mother’s uterus to implant and grow 6. This removes the faulty allele from the gene pool 7. In case of genetic disorders found only in boys , only female embryos would be implanted. Disadvantage : 1. IVF requires the female to take high doses of hormones which may be carcinogenic 2. Not all people can afford to pay for IVF Advantages of screening embryo : Allow the parents to choose and take a decision Reducing social pressure Other genetic disorders may be found. Removal of mutant allele from the family tree Problems with genetic testing : Because still individuals could live a healthy life as only a Social Ethical genetic predisposition ). 1. Risk false positive and negative result ( inaccurate results) 1. Social stigma of having a disabled child 2. Healthy fetus may be aborted if false positive result 2. Cost implications to health service or 3. May result miscarriage / spontaneous abortion individuals. 4. Ethical concern about the potential life ( killing un ethical ) 3. Social pressure 5. Who has the right to choose / decide if test should be done and 4. Religion issues. terminating life of a fetus who has the right to live. What is the step taken if the parent know they are having an abnormal child ? 1. They go and meet trained experts..( genetic counselors ) provide help A) by awareness of the disease and how to prepare and manage B) help in decision making according to ethical , social and religion issues.  To investigate effect of a drug 1. Get large sample of people 2. Divide into two groups 3. Group A …drug 4. Group B …placebo 5. C.v..all should be same age , gender , fitness , not taking any other drugs ,..for validity 6. dependent..measure the effect of drug after 30 days repeat over 1 year 7. Use a statistical test analysis 8. Calculate the SD and draw error bars 9. To know if there is a significant difference or not Check lost for topic 6 unit 1 1. Types of mutations 2. Describe three type of substitution mutation A) silent B) non sense C ) mis sense 3. Describe the two types of mutations causing frame shift A) deletion…shift reading backward one place B) insertion ….shift reading forward backward one place 4. Why errors occur during replication 5. Why mutation have no observable effect 6. Explain how a change in DNA sequence might lead to loss of enzyme activity 7. Sickle cell anaemia as an example of genetic disorder 8. Definitions of A) gene , b) allele, c) dominant , d) recessive , e) homozygote , f) heterozygote , g) genotype , f) phenotype I) true breeding , j) test cross, k) codominance , 9. Sampling error reasons ( random) 10 why use drosophila in scientific experiments. 11. Uses of genetic pedigree 12, autosomes, homogametic, heterogametic. 13. Sex linked diseases.( haemophilia and color blindness) 14. Cystic fibrosis A) normal gene …. B) in case of mutant allele 15. Effect of cystic fibrosis on A) respiratory sy stem B) digestive system C) reproductive system D) sweat glands 16. Genetic testing A) when to carry genetic testing B) types of genetic testing 1. Prenatal screening ( amniocentesis and chorionic villus sampling ) 2. Preimplantation genetic diagnosis For each know the steps , advantages , vs disadvantages 17. Problems with genetic testing ( ethical and social) 18. What can parents do if they know they are having an abnormal child?  Biology Topic 2: Membranes, proteins, DNA, and Gene Expression 16 October 2020 Code: WBI11/01 Paper 1 2C.2. Patterns of inheritance r ab lG ha Ni & Bb Bb n wa ag bb bb.N Dr Direct Combination of alleles Dr.Nagwan Gabr& Dr. Nihal Gabr 374 Biology Topic 2: Membranes, proteins, DNA, and Gene Expression Direct 50% or 0.5 Direct Parents : genotype Bb Bb Gametes B, b B, b x r ab Offsprings genotype Bb, Bb, BB, bb lG Ratio 3 orange : 1 white ha Ni & Skills n wa 1…..10000….in wild 1x200 =0.03 ag 200……..6000. ….in captivity. 6000.N 6000….1 Dr 200……x Dr.Nagwan Gabr& Dr. Nihal Gabr 375  18 June 2021. WBI11. June12021 Paper 4 Phenylthiocarbamide (PTC) is a chemical that has a very bitter taste to some individuals (tasters). The ability to taste PTC is determined by a gene that codes for a bitter-taste receptor on the tongue. Parents The pedigree diagram shows some of the tasters and non-tasters in a family. Off springs showing both taster and non taster Key So parents show dominant feature male taster TTE 1 2 female taster XY XX male non-taster Recessive female non-taster It T T XX XY T3 4tt 5 6 7 XX XY XX T T……taster XX t …..no taster - XY 8 9 (a) Complete the diagram to show the following information: (1) individual 7 as a male taster individual 8 as a male non-taster individual 9 as a female taster. Female 6 is non taster so she must have inherited 2 faulty ( recessive ) alleles one from each parent Yet her father is taster so if it was sex linked so she would have inherited X chromosome with normal allele and become carrier ( taster). 10 *P65812A01032*  (b) Describe the difference between each of the following pairs of terms, using the information in the pedigree diagram to illustrate your answer. (i) Gene and allele (2) Gene is A sequence of bases that code for a polypeptide where.................................................................................................................................................................................................................................................................................... according to this example the gene was for bitter taste receptor........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ Allele is a version of a gene , according to this example ,we had.................................................................................................................................................................................................................................................................................... allele for tasting and another allele for non tasting........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ (ii) Genotype and phenotype (2) Genotype is the genetic make up including a.................................................................................................................................................................................................................................................................................... combination of alleles , according to this example we.................................................................................................................................................................................................................................................................................... have TT, Tt, tt..................................................................................................................................................................................................................................................................................... Phenotype is observable characteristic , according to.................................................................................................................................................................................................................................................................................... this example being a taster or a non taster......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... (c) Explain which is the dominant allele. Use the information in the pedigree diagram to support your answer. (2) Those who can tase PCT.................................................................................................................................................................................................................................................................................... Parents 1 and 2 were tasters and they had children with both.................................................................................................................................................................................................................................................................................... phenotypes being taster and non taster................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. 11 *P65812A01132* Turn over  (d) Explain why this gene is unlikely to be located on the X chromosome. Skills Use the information in the pedigree diagram to support your answer. (2) Female 6 is non taster so she must have inherited 2.................................................................................................................................................................................................................................................................................... faulty ( recessive ) alleles one from each parent.................................................................................................................................................................................................................................................................................... Yet her father is taster so if it was inherited on X.................................................................................................................................................................................................................................................................................... chromosome so she would have inherited X.................................................................................................................................................................................................................................................................................... chromosome with normal allele and become carrier.................................................................................................................................................................................................................................................................................... ( taster)..................................................................................................................................................................................................................................................................................... (Total for Question 4 = 9 marks) 12 *P65812A01232*  20 June 2022. WBI11. Paper 1 8 Mutations can give rise to cancer or genetic disorders. (a) Cancer is one of the main causes of death in the world. In 2018, in the UK, there were 541 589 deaths in total and 166 800 of these were due to cancer. Calculate the ratio of deaths caused by cancer to deaths not caused by cancer. Give your answer to two decimal places. (2) Cancer =166800 Not cancer = 541589-166800= …….. 374789 166800 374789 0.45...................... Answer........................................ (b) Scientists have identified many mutations in cancer cells and are trying to identify the mutations that are significant. The chart shows the proportion of cancers caused by some types of mutation. other types of mutation type 2 substitution mutation 25% type 1 substitution 50% mutation 10% type 3 substitution 15% mutation (i) Name two other types of mutation. (1) 1............................................................................Insertion................................................................................................................................................................................................... 2............................................................................Deletion................................................................................................................................................................................................... 24 *P70960A02428*   (ii) Estimate the percentage of cancers caused by type 3 substitution mutations. (1) 65% Triplets Universal Result in Silent mutation Dont overlap as each amino can have Degenerate more than one codon Answer.............................................................. % (iii) A type 1 substitution mutation in a gene alters the DNA and mRNA but does not affect the protein synthesised from the mutated gene. Triplet code Codon Explain how a substitution mutation can alter the DNA and mRNA but not the protein. (4) Substitution mutation involves substitution of one base in one.................................................................................................................................................................................................................................................................................... triplet code.................................................................................................................................................................................................................................................................................... Where genetic code is degenerate.................................................................................................................................................................................................................................................................................... More than one codon code for the same amino acid.................................................................................................................................................................................................................................................................................... So silent mutation where the altered triplet code will code for the same amino acid.................................................................................................................................................................................................................................................................................... So protein will have same sequence of amino acids with same type.................................................................................................................................................................................................................................................................................... so same structure......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... 25  *P70960A02528* Turn over  (c) Phenylketonuria (PKU) is a genetic disorder. Recessive faulty allele This disorder is inherited in a similar way to cystic fibrosis. The pedigree diagram shows the phenotypes of individuals in one family affected by PKU. 1 2 Bb Bb Key: female unaffected by PKU BB/ Bb male unaffected by PKU 3 4 5 6 BB/ Bb female with PKU bb bb male with PKU 7 8 9 BB/ Bb BB/ Bb BB/ Bb (i) What is the probability of couple 1 and 2 having a fourth child that is a male affected by PKU? (1) A 0.000 Male child is 50% X B 0.125 Bb x Bb C 0.250 BB, Bb, Bb, bb D 0.500 75% : 25% Diseased 50x 25 =1250 Prove ….is dominant feature Parents 1, 2 are showing unaffected Yet they have offsprings with both phenotypes ( affected and un affected ) 26 *P70960A02628*   *(ii) Discuss the extent to which this pedigree diagram can be used to identify the genotypes of all the members of this family. (6) For Parents1 and 2 must be heterozygous carriers to the disease.................................................................................................................................................................................................................................................................................... As they are both have no PKU yet some of there kids( 3 and.................................................................................................................................................................................................................................................................................... 4) have PKU..................................................................................................................................................................................................................................................................................... For Offsprings 3 and 4 must be homozygous recessive.................................................................................................................................................................................................................................................................................... As they both have PKU.................................................................................................................................................................................................................................................................................... So they must have inherited 2 recessive alleles one from each carrier.................................................................................................................................................................................................................................................................................... parent.................................................................................................................................................................................................................................................................................... Against.................................................................................................................................................................................................................................................................................... So individuals 5 and 6 we cant be sure if they are homozygous dominant or.................................................................................................................................................................................................................................................................................... heterozygous carriers..................................................................................................................................................................................................................................................................................... Because they both dont have PKU.................................................................................................................................................................................................................................................................................... And both genotypes will code for the phenotype of no PKU.................................................................................................................................................................................................................................................................................... Individual 6 in not a family member and we dont know anything about her.................................................................................................................................................................................................................................................................................... family and parents..................................................................................................................................................................................................................................................................................... Against So individuals 7,8 and 9 we cant be sure if they are homozygous.................................................................................................................................................................................................................................................................................... dominant or heterozygous carriers..................................................................................................................................................................................................................................................................................... Because they both dont have PKU.................................................................................................................................................................................................................................................................................... And both genotypes will code for the phenotype of no PKU.................................................................................................................................................................................................................................................................................... No offsprings to help suggest their genotypes (Total for Question 8 = 15 marks) TOTAL FOR PAPER = 80 MARKS Comment Discuss ….describe with evidence ( for and against ) Evaluate 27  *P70960A02728*  22 Jan 2022. WBI11. Paper 1 8 Sickle cell disease is caused by a gene mutation affecting the β-globin chain of haemoglobin. (a) The mutation occurs in the seventh triplet code of this gene. This mutation results in the amino acid Glu being replaced with the amino acid Val. The table shows the sequence of bases in the first part of the DNA in a person who does not have sickle cell disease. It also shows the corresponding sequence of amino acids in the β-globin chain. Position of first second third fourth fifth sixth seventh eighth ninth triplet code DNA AUG GUG CAC CUG ACU CCU GAG GAG AAG β-globin chain (start) Val His Leu Thr Pro Glu Glu Lys (i) Give the seventh triplet code in the gene for the β-globin chain in a person who has sickle cell disease. (1) GUG.................................................................................................................................................................................................................................................................................... (ii) Name the type of mutation that causes sickle cell disease. (1) Substitution.................................................................................................................................................................................................................................................................................... (iii) The amino acid Glu is hydrophilic (polar) and the amino acid Val is hydrophobic (non-polar). Related topic 1 Suggest why this mutation causes haemoglobin molecules to stick together. (3) Different amino acids have different R groups with different.................................................................................................................................................................................................................................................................................... properties that form different bonding........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ R groups of glu polar repell other R groups on other.................................................................................................................................................................................................................................................................................... haemoglobin molecules …in case of normal haemoglobin.................................................................................................................................................................................................................................................................................... R groups facing outside will be hydrophobic so hydrophobic.................................................................................................................................................................................................................................................................................... interaction between hydrophobic R groups of Val............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ Repel from water 25  *P69498A02528* Turn over  (b) In 2020, about 140 million babies were born in the world. About 305 800 babies are born with sickle cell disease each year. Estimate the ratio of babies born with the disease to babies not born with the disease. (2) Not born with 140 x 10 6 305800 139 694 200 disease 305800/ 139 694 200 Accept ….1: 458 ① / -Q 2 x 10 : 1 Answer.............................................................. (c) The red blood cells in a person with sickle cell disease are sickle shaped and less elastic. They also have a shorter lifespan than healthy red blood cells. These sickle shaped red blood cells cannot carry as much oxygen as healthy red blood cells and they get stuck in the capillaries. The graph shows oxygen dissociation curves for groups of people with sickle cell disease and those without the disease. The shaded areas represent the range of values for each group of people. 100 people without sickle cell disease 75 PO2 Percentage of Person with and haemoglobin saturated with 50 person without oxygen (%) + results people with 25 sickle cell disease 0 0 5 10 12546777 Partial pressure of oxygen / kPa w 26 *P69498A02628*   (i) A p50 value is the partial pressure of oxygen that results in 50% saturation