16. Inheritance
A section of Biology, 9700
Listing 10 of 199 questions
There are a number of mutations affecting the production of fetal haemoglobin, HbF, and normal adult haemoglobin, HbA. • The HbA allele codes for the normal β-globin polypeptide of haemoglobin. • The HbS allele, caused by a base substitution mutation, codes for an abnormal β-globin polypeptide. • The base substitution results in the amino acid glutamine, which has a polar R group, to be replaced by valine, which has a non-polar R group, in the polypeptide. The abnormal haemoglobin molecules (HbS) form fibres in low partial pressures of oxygen (pO2). The fibres cause red blood cells to become sickle shaped and the cells can block blood capillaries. Individuals with adult haemoglobin molecules that are all abnormal (HbS) have sickle cell anaemia. This is a painful chronic condition that can be life-threatening. Explain why this mutation causes the HbS to form fibres. Fetal haemoglobin, HbF, is produced by the fetus until just before birth, when adult haemoglobin begins to be made. By the age of six months, adult haemoglobin has replaced most of the HbF. This change occurs when the genes coding for HbF are switched off and the genes coding for adult haemoglobin are switched on. • A base substitution, British-198, causes fetal haemoglobin to continue to be produced. • Normally by the age of six months, the concentration of HbF reduces to less than 1% of total haemoglobin. • With the British-198 mutation, the concentration of HbF may be as high as 20% of total haemoglobin in an adult. • HbF has a higher affinity for oxygen at low pO2 than adult haemoglobin. Individuals who have both sickle cell anaemia and British-198 mutation have reduced symptoms of sickle cell anaemia. Suggest why having the British-198 mutation reduces the symptoms of sickle cell anaemia. In adults with the British-198 mutation, the gene coding for a fetal haemoglobin polypeptide remains switched on. This is due to the presence of a protein that controls gene expression. State the term that is used to describe a protein that controls gene expression. Gel electrophoresis can be carried out to test individuals for the different versions of haemoglobin: HbA, HbS and HbF. • A buffer with alkaline pH is used to make all haemoglobin molecules negatively charged. • HbS molecules have an additional positive charge compared to HbA. Describe and explain how gel electrophoresis is used to diagnose sickle cell anaemia. Four individuals had their haemoglobin analysed by gel electrophoresis. One of the individuals was heterozygous for the HbA and HbS alleles and had a condition known as sickle cell trait (SCT). Some of the results are shown in . In , lane 1 and lane 5 are complete. lane 1 individual with normal phenotype one-month- old baby with normal phenotype individual with SCT individual with sickle cell anaemia reference lane 2 lane 3 lane 4 lane 5 HbA HbF HbS Predict the results for the individuals analysed, by adding bands to lanes 2, 3 and 4 on .
9700_w20_qp_43
THEORY
2020
Paper 4, Variant 3
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The lac operon is a section of DNA present in the genome of the bacterium Escherichia coli. The structural genes of the operon are only fully expressed when E. coli is exposed to high lactose concentrations. is a diagram showing the lac operon and a nearby region of the E. coli genome. I lacA transcription transcription Name sections 1, 2, 3 and 4 shown in . Certain enzymes need to be present inside E. coli for it to be able to take up and use lactose. When the genes of the lac operon are expressed, the enzymes β-galactosidase and lactose permease are produced in large quantities. Outline the functions of β-galactosidase and lactose permease. β-galactosidase lactose permease Gene I is located a short distance away from the lac operon. The product of gene I, a repressor protein, is a constitutive protein. State what is meant by the term constitutive protein. Explain why the structural genes of the lac operon are not expressed when lactose is absent. A strain of E. coli has been produced with a mutation in gene I. Expression of this gene results in a non-functional repressor protein. Suggest a negative effect that this mutation will have on this strain of E. coli.
9700_s18_qp_41
THEORY
2018
Paper 4, Variant 1
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Yeast cells respond to changes in glucose concentration in their environment by using transcription factors to switch off genes. When glucose is present: • Mig1 transcription factors bind to the promoters of five genes • Mig1 binding to the promoters stops transcription of these genes. The genes that are repressed by Mig1 code for five enzymes that allow yeast cells to metabolise the sugar galactose when glucose is absent. Complete Table 5.1 to show three chemical differences between a transcription factor, such as Mig1, and a promoter. Table 5.1 transcription factor promoter difference 1 difference 2 difference 3 Mig1 binds to promoter sites with these features: • 17 base pairs long • includes a region of five repeating adenine-thymine pairs • includes a region of six repeating cytosine-guanine pairs. Promoter sites to which Mig1 binds are known as Mig1-binding promoter sites. Bioinformatic techniques were used to analyse the yeast genome to look for sections of DNA that match these features. The information obtained for four chromosomes is shown in Table 5.2. Table 5.2 yeast chromosome chromosome size / base pairs number of Mig1-binding promoter sites per chromosome A 230 018 B 813 184 C 316 620 D 1 531 933 Explain why bioinformatic techniques were used to obtain the information in Table 5.2. Identify, with a reason, the yeast cell chromosome that is most likely to include genes that code for enzymes that metabolise galactose. Mig1 binds to 27 promoters on these four chromosomes. Yeast cells also have other chromosomes where Mig1 binds to additional promoters. Five different enzymes, coded by five genes, must be made for yeast cells to metabolise galactose. Suggest reasons why an individual diploid yeast cell has a larger number of Mig1-binding promoter sites than the expected number of ten. The repression of genes involved in galactose metabolism in yeast is similar to events at the lac operon in the bacterium Escherichia coli. Explain how E. coli represses the production of proteins needed to metabolise lactose sugar.
9700_s20_qp_42
THEORY
2020
Paper 4, Variant 2
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Gene expression in a cell is controlled. When a gene is expressed (switched on), the gene is transcribed. When a gene is not expressed (switched off), the gene is not transcribed. Environmental changes can cause some genes to be switched on or switched off. An example of control of gene expression in prokaryotes is regulation in the lac operon. The lac operon is a length of DNA that is made up of different parts. shows a simple diagram representing the lacI gene and the lac operon. lacI Outline the main features of the lac operon. Explain the role of the lacI gene in the regulation of the lac operon. The lac operon codes for inducible enzymes. Repressible operons code for repressible enzymes. Suggest and explain why it is an advantage to a prokaryote to have a repressible operon.
9700_s23_qp_43
THEORY
2023
Paper 4, Variant 3
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The lac operon is a section of DNA present in the genome of Escherichia coli. The structural genes of the lac operon are only fully expressed when the bacteria are exposed to high lactose concentrations. is a diagram showing the lac operon and a nearby region of the E. coli genome. I P transcription transcription O lacZ lacY lacA shows how the lac operon consists of structural genes and regulatory sequences. Use to identify two structural genes. Complete Table 6.1 to name each structural gene and its product. Table 6.1 structural gene name of gene product Gene I is transcribed all the time to produce its protein. This is constitutive expression. Explain why some genes show constitutive expression. Describe the effect of the product of gene I on the functioning of the lac operon. If E. coli is put into a nutrient medium containing lactose, some new enzymes are synthesised. These are described as inducible enzymes. Explain what is meant by an inducible enzyme. The structural genes of the lac operon are not expressed when lactose is absent. Suggest one reason why this is beneficial to E. coli.
9700_w18_qp_42
THEORY
2018
Paper 4, Variant 2
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The lac operon of prokaryotes contains a group of structural genes that are under the same control and are transcribed together. Another operon found in prokaryotes is the trp operon. summarises the structure and control of the trp operon. trpR promoter operator attenuator (regulates the extent of transcription) structural genes are transcribed RNA polymerase binds to promoter inactive repressor active repressor tryptophan trpE trpD trpC trpB trpA trpR promoter operator attenuator no transcription of structural genes trpE trpD trpC trpB trpA Describe the differences in structure and control between the lac operon and the trp operon. Suggest why structural genes in operons are transcribed together. trpA is an example of a structural gene and trpR is an example of a regulatory gene. Describe the differences between the functions of structural genes and regulatory genes. trpA codes for the enzyme tryptophan synthase. Tryptophan synthase catalyses the formation of the amino acid tryptophan. Explain why tryptophan synthase is an example of a repressible enzyme. Control of gene expression in eukaryotes is more complex than in prokaryotes. In plants, the control of gene expression can involve plant hormones, such as gibberellin, and proteins known as JAZ and MYC. Describe how gibberellin activates genes in plant cells. Transcription of some plant genes is prevented when JAZ proteins bind to other proteins known as MYC. When JAZ proteins are broken down, MYC proteins are free to bind to DNA. This allows transcription to begin. State the term that is used to describe proteins such as MYC proteins.
9700_w22_qp_42
THEORY
2022
Paper 4, Variant 2
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The bacterium, Escherichia coli, can use glucose or disaccharides, such as lactose, in its metabolism. Lactose needs to be hydrolysed by the enzyme β-galactosidase to form glucose and galactose, which can then be used by E. coli. The production of β-galactosidase is controlled by a length of DNA called the lac operon. shows the lac operon when lactose is absent. P O lacZ lacY lacA RNA polymerase repressor On , draw the positions of RNA polymerase and the repressor molecule when lactose is present. P O lacZ lacY lacA The protein coded for by lacY is not involved in the control of gene expression. Name the type of gene represented by lacY. Name the protein product coded for by lacY and state the precise role of this protein. In an investigation into the growth of E. coli, a sample of the bacterium was grown in a medium that contained limited concentrations of glucose and lactose. The population size of E. coli was measured at regular intervals. shows the population growth curve obtained for this investigation. E. coli population size time Describe and suggest explanations for the population growth curve shown in .
9700_s22_qp_42
THEORY
2022
Paper 4, Variant 2
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Questions Discovered
199