Protein Synthesis S7 4P Biology 23-24 PDF

S7 4P Biology 23-24 Topic 2: Protein Synthesis  P R OTEIN SY N THES IS Syllabus Links 1/20/2025 2  P R OTEIN SY N THES IS Lesson 1 – DNA Transcription Write Everything you can recal...

S7 4P Biology 23-24 Topic 2: Protein Synthesis  P R OTEIN SY N THES IS Syllabus Links 1/20/2025 2  P R OTEIN SY N THES IS Lesson 1 – DNA Transcription Write Everything you can recall about DNA & proteins on the WB. 1/20/2025 3  Assessments Formative Graded Questions (Worth 1/3 of a formative grade) Application Challenge Summative Topic Test! 1/20/2025 4  P R OTEIN SY N THES IS Learning Outcomes: To Be able to explain the process of transcription. 1/20/2025 5  P R OTEIN SY N THES IS What use is DNA? Recall from S6… The whole purpose of DNA is to produce proteins. The types of proteins produced are controlled by the sequence of bases in your DNA. What are the 4 bases in DNA & 1 other base in RNA? Adenine, Thymine, Guanine, & Cytosine Uracil replaces thymine in RNA. 1/20/2025 6  The first step of producing a protein P R OTEIN SY N THES IS is TRANSCRIPTION 1/20/2025 7  TR AN S CR IP TION Steps in Transcription… Initiation Elongation Termination Transcription produces an RNA molecule known as messengerRNA (or mRNA) that leaves the nucleus. 1/20/2025 8  TR AN S CR IP TION Initiation Initiation. RNA polymerase binds to a sequence of DNA called the promoter, found near the beginning of a gene. Each gene (or group of co-transcribed genes, in bacteria) has its own promoter. Once bound, RNA polymerase separates the DNA strands, providing the single-stranded template needed for transcription. 1/20/2025 9  TR AN S CR IP TION Elongation Elongation. One strand of DNA, the template strand, acts as a template for RNA polymerase. As it "reads" this template one base at a time, the polymerase builds an RNA molecule out of complementary nucleotides, making a chain that grows from 5' to 3'. The RNA transcript carries the same information as the non-template (coding) strand of DNA, but it contains the base uracil (U) instead of thymine (T). 1/20/2025 10  TR AN S CR IP TION Termination Termination. Sequences called terminators signal that the RNA transcript is complete. Once they are transcribed, they cause the transcript to be released from the RNA polymerase. An example of a termination mechanism involving formation of a hairpin in the RNA is shown below. 1/20/2025 11 Second Explanation. If you understand, use the factsheet to look ahead towards TR AN S CR IP TION EX PL ANA TION 2 Translation. 1/20/2025 12  13 TR AN S CR IP TION EX PL ANA TION 3 1/20/2025  PRE-MRNA Eukaryotic Pre-mRNA In bacteria, RNA transcripts can act as messenger RNAs (mRNAs) right away. In eukaryotes, the transcript of a protein-coding gene is called a pre- mRNA and must go through extra processing before it can direct translation. Eukaryotic pre-mRNAs must have their ends modified, by addition of a 5' cap (at the beginning) and 3' poly-A tail (at the end). Many eukaryotic pre-mRNAs undergo splicing. In this process, parts of the pre-mRNA (called introns) are chopped out, and the remaining pieces (called exons) are stuck back together. Why is this useful? 1/20/2025 14  TRAN SLATION STARTER Q Lesson 2 – Translation! An analysis of DNA from several cells is performed and the results are presented in figure 4 (A, T, C, G, expressed in %). Origin of DNA A C T G 1. Explain what are A, C, T, G in DNA. (Kn 0.5) Sea urchin 33,2 16,7 33,3 16,8 2.Compare the percentage values of A, C, T, G for cells from different organisms. (An 1) 3.Use your knowledge to explain your observations of question ii. (Kn 0.5 & Ap 0.5) Wheat 27,3 22,7 27,2 22,8 4.Compare the percentage values of A, C, T, G for the two types of human cells. (An 1) Human cells (from the same 5.Use your knowledge to explain your observations of question iv. (kn 0.5 & Ap 0.5) person) 6.Explain what could be the results with cells from an other human being. (Kn 1 Wc 1) Skin cell 30,5 19,9 29,8 19,8 Intestinal cell 30,5 19,9 29,8 19,8 1/20/2025 15  TRAN SLATION Translation 1/20/2025 16  MRNA mRNA Triplet codons tRNA Anticodons Amino Acids Polypeptide chain – bonds? Protein Structures 1/20/2025 17  18 P R OTEIN SY N THES IS 1/20/2025  What is translation? TRAN SLATION Note the structure of tRNA 1/20/2025 19  TRNA Note the structure of tRNA 1/20/2025 20  TRNA 1/20/2025 21  Translation – Step by Step Initiation The ribosome consists of two subunits. mRNA gets sandwiched between these two subunits. Elongation Termination 1/20/2025 22  Initiation In initiation, the ribosome assembles around the mRNA to be read and the first tRNA (carrying the amino acid methionine, which matches the start codon, AUG). This setup, called the initiation complex, is needed in order for translation to get started. 1/20/2025 23  Elongation ELONGATION Elongation is the stage where the amino acid chain gets longer. In elongation, the mRNA is read one codon at a time, and the amino acid matching each codon is added to a growing protein chain. Each time a new codon is exposed: A matching tRNA binds to the codon The existing amino acid chain (polypeptide) is linked onto the amino acid of the tRNA via a chemical reaction The mRNA is shifted one codon over in the ribosome, exposing a new codon for reading 1/20/2025 24  Termination TER M IN ATI O N Termination is the stage in which the finished polypeptide chain is released. It begins when a stop codon (UAG, UAA, or UGA) enters the ribosome, triggering a series of events that separate the chain from its tRNA and allow it to drift out of the ribosome. After termination, the polypeptide may still need to fold into the right 3D shape, undergo processing (such as the removal of amino acids), get shipped to the right place in the cell, or combine with other polypeptides before it can do its job as a functional protein. 1/20/2025 25  Lesson 3 – Proteins & Codon Wheels Explain what the diagrams show! 1/20/2025 26  What is this showing? Challenge: How does a ribosome know when to ‘stop’ reading a gene? 27 Practice! What is the sequence of amino acids? 1. GAUGAUGAUCCUAGCUAGC 2. AUGGGCAGAUCAGAGUAUA 3. AUGGAGCUAGCAUAGUGA 4. UAAGGUCAUGAGUAGACG 5. Which of the above codes has a start codon? 6. Which has a stop codon?  Codon Practice Questions on page 181-182 (New green textbook) 1/20/2025 29  Lesson 5 – Mutations Revision! 1/20/2025 30  Task Using the diagram below, and your textbooks, describe and explain the effect of the following mutations: Silent Hint: Use the following keywords: Missense Point mutation Transcription Nonsense Frame shift Indel Tertiary structure Expanding triple nucleotide repeats Challenge: Explain why not all mutations are harmful. A. Point mutations The genetic code consists of nucleotide base triplets within the DNA. During transcription of a gene, this code is copied to a length of mRNA as codons complementary to the base triplets on the template strand of the length of DNA. The sequence of codons on the mRNA is therefore a copy of the sequence of base triplets on the gene (coding strand of the DNA). There are three types of point mutation: 1. silent 2. missense 3. nonsense. 1. Silent mutations All amino acids involved in protein synthesis, apart from methionine, have more than one base triplet code. This reduces the effect of point mutations, as they do not always cause a change to the sequence of amino acids in a protein. This is often called the 'redundancy' or 'degeneracy' of the genetic code. A point mutation involving a change to the base triplet. where that triplet still codes for the same amino acid, is a silent mutation. The primary structure of the protein, and therefore the secondary and tertiary structure, is not altered (see Figure 2). 2. Missense mutations A change to the base triplet sequence that leads to a change in the amino acid sequence in a protein is a missense mutation. Within a gene, such a point mutation may have a significant effect on the protein produced. The alteration to the primary structure leads to a change to the tertiary structure of the protein altering its shape and preventing it from carrying out its usual function. Missense mutations Sickle cell anaemia results from a missense mutation on the sixth base triplet of the gene for the β - polypeptide chains of haemoglobin: the amino acid valine, instead of glutamic acid, is inserted at this position. This results in deoxygenated haemoglobin crystallising within erythrocytes, causing them to become sickle shaped, blocking capillaries and depriving tissues of oxygen. 3. Nonsense mutations A point mutation may alter a base triplet, so that it becomes a termination (stop) triplet. This particularly disruptive point mutation results in a truncated protein that will not function. This abnormal protein will most likely he degraded within the cell The genetic disease Duchenne muscular dystrophy is the result of a nonsense mutation. 3. Nonsense mutations B. Indel mutations Both insertions and deletions cause a frameshift. 1. Insertions and deletions If nucleotide base pairs, not is multiples of three, are inserted in the gene or deleted from the gene, because the code is non-overlapping and read in groups of three bases, all the subsequent base triplets are altered. This is a frameshift. When the mRNA from such a mutated gene is translated, the amino acid sequence after the frameshift is severely disrupted. The primary sequence of the protein, and subsequently the tertiary structure, is much altered_ Consequently. the protein cannot carry out its normal function. If the protein is very abnormal., it will be rapidly degraded within the cell. Insertions and deletions Some forms of thalassaemia, a haemoglobin disorder, result from frameshifts due to deletions of nucleotide bases. Insertions or deletions of a triplet of base pairs result in the addition or loss of an amino acid, and not in a frameshift. 2. Expanding triple nucleotide repeats Some genes contain a repeating triplet such as -CAG CAG CAG-. In an expanding triple nucleotide repeat, the number of CAG triplets increases at meiosis and again from generation to generation. Huntington disease results from an expanding triple nucleotide repeat. If the number of repeating CAG sequences goes above a certain critical number, then the person with that genotype will develop the symptoms of Huntington disease later in life There is more about the inheritance pattern of this disease in topic 6.2.2.  Not all mutations are harmful Many mutations are beneficial and have helped to drive evolution through natural selection. Different alleles of a particular gene are produced via mutation. The mutation that gave rise to blue eyes arose in the human population 6000-8000 years ago. Such a mutation may be harmful in areas where the sunlight intensity is high as the lack of iris pigmentation could lead to lens cataracts. However. in more temperate zones, it could enable people to see better in less bright light Early humans in Africa would have had black skin, the high concentrations of melanin protecting them from sunburn and skin cancer. When humans migrated to temperate regions. a paler skin would be an advantage. enabling vitamin D to be made with a lower intensity of sunlight. In such areas, people with fairer skin would have an advantage and be selected, as vitamin D not only protects us from rickets, it protects us from heart disease and cancer Some mutations appear to be neutral, being neither beneficial nor harmful, such as those that in humans. cause: inability to smell certain flowers. including freesias and honeysuckle differently shaped ear lobes. KEY DEFINITIONS – Point Mutations https://learn.genetics.utah.edu/content/basics/firefly/  Questions 1. Explain how the degenerate nature of the genetic code reduces the effects of point mutations. 2. The following is a sequence of DNA bases on a piece of coding strand DNA: ATG TTT CCT GTT AAA TAC CAT CAG CAG CGC TAG CAC Below are six possible mutations to the piece of DNA: I) ATG TTT CCT GTT AAA TAA CAT CAG CAG CGC TAG CAC II) ATG TTT CCT AIT AAA TAC CAT CAG CAG CGC TAG CAC III) ATT TTT CCT GTT AAA TAC CAT GAG CAG CGC TAG CAC IV) ATG TTC CTG TTA AAT ACC ATC AGC AGC GCT AGC AC- V) ATG TTT CCT GTT AAA TAC CAT CAG CAG CAG CAG CGC TAG CAC VI)ATG TTT CCT GTT AAA TAC CAT CAG CAG CGC TGG CAC In each case identify the type of mutation and explain the effect it will have on the translated protein. You will  Questions 3. For the normal sequence of DNA base triplets shown in question 2, write down: a) the sequence of bases on the corresponding DNA template strand b) the sequence of RNA bases an the piece of mRNA transcribed from this gene. 4. Sometimes during translation, a tRNA molecule may combine with the wrong amino acid. How may this affect the protein being assembled at a ribosome? 1/20/2025 52