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A2- Biology Student Guide Unit 5 Control in Cells and Organisms. 3.5.6- Genetic control of protein structure and function In this unit you will learn what populations are and how we can study them. In looking at the factors that affect population size, consideration is given to abiotic factors, competition and predation. Name: Teacher: Form:
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A2- Biology Student GuideUnit 5 Control in Cells and Organisms.3.5.6- Genetic control of protein structure and function In this unit you will learn what populations are and how we can study them.In looking at the factors that affect population size, consideration is given to abiotic factors, competition and predation.
Name:
Teacher:
Form:
Effective Learning 1-Excellent; 2-Good; 3-Needs to improve; 4-Cause for concern
Assessment:Written test Date of test ___________________
Target Attainment Grade:How will I achieve this?
Evaluation:Self assessed learning grade ____Comments – (What can I do differently next time I do this?)
Teacher assessed learning grade ____Comments:
Independent Learning• Do you work outside of formal learning sessions? • Are you able to demonstrate independent learning?
Questioning • Do you interrogate material/resources • Ask your LT / peers sensible, in depth questions • Question sources of information
Planning • Do you read additional material in preparation for the session • Plan your work in advance • Use assessment criteria to plan your tasks
Review/Amend • Do you look back at notes/tasks and review them regularly • Amend tasks having spoke/discussed with others or read more widely • Amend/review after advice from LTs • Active listening • Listen to understand • Be proactive and contribute in sessions/activities • Listen to other students in the group
Collaboration • Work effectively with others in the group • Take part in group work to ensure that everyone is involved • Take a range of roles within group work to encourage effective collaboration
SPECIFICATION
Assessable Learning Outcomes Covered
How did I review this
The genetic code
The genetic code as base triplets in mRNA which code for specific amino acids..
The genetic code is universal, non-overlapping and degenerate.
The structure of molecules of messenger RNA (mRNA) and transfer RNA (tRNA)
Candidates should be able to compare the structure and composition of DNA,mRNA and tRNAPolypeptide Synthesis
Transcription as the production of mRNA from DNA. The role of RNA polymerase.The splicing of pre-mRNA to form mRNA in eukaryotic cells.
Translation as the production of polypeptides from the sequence of codons carriedby mRNA.
The role of ribosomes and tRNA.
Candidates should be able to• show understanding of how the base sequences of nucleic acids relate to theamino acid sequence of polypeptides, when provided with suitable data
• interpret data from experimental work investigating the role of nucleic acids.Recall of specific codons and the
Gene MutationGene mutations might arise during DNA replication. The deletion and substitution of bases.Gene mutations occur spontaneously. The mutation rate is increased by mutagenicagents.
Some mutations result in a different amino acid sequence in the encodedpolypeptide. Due to the degenerate nature of the genetic code, not all mutations result in a change to the amino acid sequence of the encoded polypeptide.
The rate of cell division is controlled by proto-oncogenes that stimulate cell division and tumour suppressor genes that slow cell division. A mutated proto-oncogene, called an oncogene, stimulates cells to divide too quickly. A mutated tumour suppressor gene is inactivated, allowing the rate of cell division to increase.
3.5.7 Gene Expression is controlled by a number of featuresMost of the cells DNA is not translated Totipotent cells are cells that can mature into any body cell.During development, totipotent cells translate only part of their DNA, resulting in cellspecialisation.In mature plants, many cells remain totipotent. They have the ability to develop in
vitro into whole plants or into plant organs when given the correct conditionsTotipotent cells occur only for a limited time in mammalian embryos. Multipotentcells are found in mature mammals. They can divide to form only a limited number ofdifferent cell types.Totipotent and multipotent stem cells can be used in treating some geneticdisorders.
Candidates should be able to• interpret data relating to tissue culture of plants from samples of totipotent cells• evaluate the use of stem cells in treating human disorders.Regulation of transcription and translationTranscription of target genes is stimulated only when specific transcriptional factorsmove from the cytoplasm into the nucleus.
The effect of oestrogen on gene transcription.Small interfering RNA (siRNA) as a short, double-strand of RNA that interferes withthe expression of a specific gene.
Candidates should be able to• interpret data provided from investigations into gene expression• interpret information relating to the use of oncogenes and tumour suppressorgenes in the prevention, treatment and cure of cancer• evaluate the effect on diagnosis and treatment of disorders
Key word DefinitionRNA
Uracil
Pre mRNA
Universal code
Degenerate code
Universal code
Codon
Anticodon
Transcription
Complementary
Template strand
Coding strand
RNA polymerase
Transcription Factor
Splicing
Nuclear pore
Translation
Ribosome
Peptide bond
Amino acid
Mutation
Nonsense mutation
Misense Mutation
Silent mutation
Tumour suppressor gene
Proto oncogene
Stem cell
Totipotent
Pluripotent
Multipotent
Gene expression
Transcriptional Regulation
Translational Regulational (post transcriptional)
Protein DNA complex
DNA binding domain
Si RNA
Practice Questions:You should complete these questions and then self assess your answers using the Mar scheme provided on the MLE.- Insert any missed answers in a different colour.
Q1. (a) The table shows the mRNA codons for some amino acids.
Codon Amino acid
CUA Leucine
GUC Valine
ACG Threonine
UGC Cysteine
GCU Alanine
AGU Serine
(i) Give the DNA sequence coding for cysteine.
.............................................................................................................(1)
(ii) Name the amino acid coded by the tRNA anticodon UCA.
.............................................................................................................(1)
(b) A particular gene is 562 base-pairs long. However, the resulting mRNA is only 441 nucleotides long. Explain this difference.
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......................................................................................................................(1)
(c) Tetracycline binds to bacterial ribosomes. This is shown in the diagram.
Protein synthesis in bacteria is similar to that in eukaryotic cells. Explain how tetracycline stops protein synthesis.
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(Total 5 marks)
Q2.(a) The genetic code is described as being degenerate. What does this mean?
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.(1)
(b) What is a codon?
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(c) (i) What is the role of RNA polymerase during transcription?
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(ii) mRNA can be converted to cDNA.
Name the enzyme used in this process.
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(d) The diagram shows the base sequence on DNA where a restriction endonuclease cuts DNA.
Use evidence from the diagram to explain what is meant by a palindromic recognition sequence on DNA.
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(1)(Total 6 marks)
Q3. Figure 1 shows part of a gene that is being transcribed.
Figure 1
(a) Name enzyme X.
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(b) (i) Oestrogen is a hormone that affects transcription. It forms a complex with a receptor in the cytoplasm of target cells. Explain how an activated oestrogen receptor affects the target cell.
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(ii) Oestrogen only affects target cells. Explain why oestrogen does not affect other cells in the body.
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(c) Some breast tumours are stimulated to grow by oestrogen. Tamoxifen is used to treat these breast tumours. In the liver, tamoxifen is converted into an active substance called endoxifen. Figure 2 shows a molecule of oestrogen and a molecule of endoxifen.
Figure 2
Use Figure 2 to suggest how endoxifen reduces the growth rate of these breast tumours.
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(Total 6 marks)
Q4. (a) Transcriptional factors are important in the synthesis of particular proteins.Describe how.
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(b) The flowchart shows how small interfering RNA (siRNA) affects the expression of a particular target gene.
(i) The siRNA-protein complex attaches to an mRNA molecule coding for a particular protein (step 2). Explain what causes the siRNA to attach only to one sort of mRNA molecule.
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.............................................................................................................(1)
(ii) Describe and explain how expression of the target gene is affected by siRNA.
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(iii) Scientists have suggested that siRNA may be useful in treating some diseases.Suggest why siRNA may be useful in treating disease.
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(Total 7 mark
Q5.Human immunodeficiency virus (HIV) particles have a specific protein on their surface.This protein binds to a receptor on the plasma membrane of a human cell and allows HIV to enter. This HIV protein is found on the surface of human cells after they have become infected with HIV.
Scientists made siRNA to inhibit expression of a specific HIV gene inside a human cell. They attached this siRNA to a carrier molecule. The flow chart shows what happens when this carrier molecule reaches a human cell infected with HIV.
(a) When siRNA binds to mRNA, name the complementary base pairs holding the siRNA and mRNA together. One of the bases is named for you.
............................................................with.....................................................
.....................Adenine.........................with.....................................................(1)
(b) This siRNA would only affect gene expression in cells infected with HIV.
Suggest two reasons why.
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(c) The carrier molecule on its own may be able to prevent the infection of cells by HIV.
Explain how.
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(Total 7 marks)
Q6. The black mamba is a poisonous snake. Its poison contains a toxin.
The table shows the base sequence of mRNA that codes for the first two amino acids of this toxin.
Base sequence of anticodon on tRNA
Base sequence of mRNA A C G A U G
Base sequence of DNA
Complete the table to show
(a) (i) the base sequence of the anticodon on the first tRNA molecule that would bind tothis mRNA sequence
(1)
(ii) the base sequence of the DNA from which this mRNA was transcribed.
(1)
(b) The length of the section of DNA that codes for the complete toxin is longer than the mRNA used for translation. Explain why.
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.......................................................................................................................(1)
(c) A mutation in the base sequence of the DNA that codes for the toxin would change the base sequence of the mRNA.
Explain how a change in the base sequence of the mRNA could lead to a change in the tertiary structure of the toxin.
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(d) The black mamba’s toxin kills prey by preventing their breathing. It does this by inhibiting the enzyme acetylcholinesterase at neuromuscular junctions. Explain how this prevents breathing.
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(Extra space) .................................................................................................
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(Total 7 marks)
Q7. The diagram shows part of a pre-mRNA molecule.
(a) (i) Name the two substances that make up part X.
................................................... and .................................................(1)(ii) Give the sequence of bases on the DNA strand from which this pre-mRNA has been transcribed.
.............................................................................................................(1)
(b) (i) Give one way in which the structure of an mRNA molecule is different from the structure of a tRNA molecule.
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.............................................................................................................(1)
(ii) Explain the difference between pre-mRNA and mRNA.
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(c) The table shows the percentage of different bases in two pre-mRNA molecules.The molecules were transcribed from the DNA in different parts of a chromosome.
Part of chromosome
Percentage of base
A G C U
Middle 38 20 24
End 31 22 26
(i) Complete the table by writing the percentage of uracil (U) in the appropriate boxes.(1)
(ii) Explain why the percentages of bases from the middle part of the chromosome and the end part are different.
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(Total 7 marks)
