BIO181

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181

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Biology

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Apr 3, 2024

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BIO181 Transcription Translation Worksheet If any questions are unanswered, you will receive a score of 0. The extension questions are required . Name: Class Time: Part 1: Transcription ______________________________________________________________________________________ READ THIS : DNA is often referred to as a genetic blueprint. In the same way that blueprints contain the instructions for construction of a building, the DNA found inside the nuclei of cells contains the instructions for assembling a living organism. The DNA blueprint carries its instructions in the form of genes. In most cases the genes direct the production of a polypeptide, from which other more complex proteins, such as enzymes or hormones, may be constructed. These polypeptides and other molecules run the organism’s metabolism and, in multicellular organisms, dictate what each cell’s job is. So, what is the language of these instructions and how are they read and decoded by the cellular organelles? This activity will focus on the decoding of genes in eukaryotes. ______________________________________________________________________________________ Note - These are the same images. The image on the left is provided as a Google drawing. To edit it, draw on it, add circles or labels to it, double click the image while in Google Docs. Add in what you need to, then hit “Save and Close”. The image on the right is clearer but cannot be edited. 1. Refer to Model 1. a. What is the base-pair rule for a DNA strand matching an RNA strand? For DNA, Adenine pairs with Thymine and Guanine pairs with Cytosine. For RNA, Adenine pairs with Uracil and Guanine pairs with Cytosine. b. Compare this base-pair rule with that of two DNA strands. C ompared to two DNA strands, the base-pair rule states that for DNA, DNA also pairs T with A because DNA does not have Uracil like RNA. In DNA, Thymine bonds to Adenine and Adenine bonds to Thymine. In RNA, Thymine bonds to Adenine, but Adenine bonds to Uracil. 1

2. Which strand of the DNA contains the “blueprint” for the pre-mRNA? The template strands of the DNA contains the “blueprint” for the pre-mRNA. 3. Consider Model 1 a. In which direction is the DNA template strand read? The DNA template strand is read from 3’ to 5’. b. The DNA template strand and pre-mRNA strand are anti-parallel. With this in mind label the 3ʹ and 5ʹ ends of the pre-mRNA strand in Model 1. The left end is the 3’ end, and the right end is the 5’ end. c. In which direction is the pre-mRNA molecule constructed? On which end are new bases added? The pre-mRNA molecule is constructed from 5’ to 3’. _______________________________________________________________________________________ READ THIS: In eukaryotes the enzyme RNA polymerase joins with several transcription factor proteins at the promoter, which is a special sequence of base pairs on the DNA template strand that signals the beginning of a gene. The transcription factor proteins, along with the RNA polymerase, is called the transcription initiation complex . This moves along the DNA template strand at about 40 base pairs per second producing pre-mRNA. When the RNA polymerase reaches the terminator sequence of base pairs on the DNA template strand, it completes the production of pre-mRNA and releases it into the nucleoplasm. _______________________________________________________________________________________ 4. Where on the DNA strand does the transcription initiation complex form? The transcription initiation complex forms on the promotor of the DNA strand. 5. Imagine the diversity of functions in which your cells participate. a. Do you think that the cells in your big toe contain the same DNA as the neurons in your brain? Yes, I think that the cells in my big toe contain the same DNA as the neurons in my brain because every cell in the human body has the same DNA. 2

b. Will an individual cell transcribe every, or almost every, gene in its genome? Explain. No, an individual cell cannot transcribe every, or almost every, gene in its genome because some of the products are not needed for that specific action. Note - These are the same images. The image on the left is provided as a Google drawing. To edit it, draw on it, add circles or labels to it, double click the image while in Google Docs. Add in what you need to, then hit “Save and Close”. The image on the right is clearer but cannot be edited. 6. Compare the pre-mRNA to the mRNA leaving the nucleus in Model 1. a. What has been removed from the pre-mRNA to make it into mRNA? An intron was removed from the pre-mRNA to make it into mRNA. b. What has been added to the mRNA that was not present in the pre-mRNA, and where on the mRNA strand are the additional items located? A methyl cap was added to the mRNA that was not present in the pre-mRNA. The additional items are located on the poly (A) tail that was added to the 3’ end. The poly A tail and a 5’ cap are added. 7. Can mRNA diffuse through a membrane? Why or why not? mRNA cannot diffuse through a membrane because it has to have a transport protein. 3

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______________________________________________________________________________________ READ THIS: Introns are sections of pre-mRNA that are noncoding. That is, they don’t provide useful information for the production of the polypeptide being synthesized. There is evidence that suggests these introns allow certain sections of DNA to code for different polypeptides when different sections are removed. The removal of specific sections is triggered by a signal response in the cell. The portions of the pre-mRNA that remain are called exons . The methyl cap (sometimes called the GTP cap or 5ʹ cap) helps the mRNA molecule move through the nuclear pore and attach to a ribosome, its final destination. mRNA is a short lived molecule. Once in the cytoplasm the mRNA will be subject to exonucleases that immediately start removing individual nucleotides from the 3ʹ end of a nucleic acid. The individual mRNA nucleotides will then be free to be used again during the process of transcription. ______________________________________________________________________________________ 8. The human genome contains about 25,000 genes and yet produces about 100,000 different polypeptides. Propose an explanation of how this is possible. Introns allow certain sections of the DNA to code for different polypeptides when different sections are removed from the DNA strand. Different mRNA combinations make up different genes. 9. Using the information in Read This section , develop a hypothesis to explain the advantage of the poly-A tail added to the 3’ end of the mRNA. The poly A tail prevents the information carrying part of the mRNA from being removed by the exonucleases before the formation of a polypeptide can occur. It also protects the coding on the mRNA. 10.Different mRNA molecules can have poly-A tails of different lengths. Considering the purpose of adding the poly-A tail (from the previous question), why are some tails longer than others? Justify your answer. Some tails are longer than others because proteins that are needed for long periods of time will come from mRNA with long poly A tails. Proteins that are needed for brief periods of time will come from mRNA with shorter poly A tails. Some tails are longer so that they can last longer in the cytoplasm. 11.Summarize the steps of transcription. 1. Initiation: recognition sites, transcription factors, RNA polymerase unzips DNA and reads the transcription region and produces RNA strands. 2. Elongation: elongating DNA. 5

3. Termination: sequence of nucleotides at the end of the transcribed DNA and reads it to stop transcribing. RNA is released and RNA polymerase releases. 6

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Part 2: Translation _______________________________________________________________________________________ READ THIS : The message in your DNA of who you are and how your body works is carried out by cells through gene expression. In most cases this means synthesizing a specific protein to do a specific job. First, mRNA is transcribed from the DNA code. Then, the mRNA sequence is translated into a polypeptide sequence. _______________________________________________________________________________________ Note - These are the same images. The image on the left is provided as a Google drawing. To edit it, draw on it, add circles or labels to it, double click the image while in Google Docs. Add in what you need to, then hit “Save and Close”. The image on the right is clearer but cannot be edited. 12.Model 1 defines the code scientists have discovered that relates the nucleotide sequence of mRNA to the amino acid sequence of polypeptides. The language of mRNA is often described as a “triplet code.” Explain the significance of this reference. This reference is significant because three nucleotides are needed to code one amino acid. There are 3 code combinations that code for amino acids. 13.If an mRNA molecule had 300 nucleotides in the coding region of the strand, how many amino acids would be in the polypeptide that was synthesized? Show mathematical work to support your answer. 100 amino acids would be in the polypeptide that was synthesized. You divide 300 by 10 to get 100. 14.Consider the information in Model 1. a. Compare all of the codons for Proline. What are the similarities and differences? The codons for Proline all start with CC. The third base is different, however. They all start with CC, but the last base is different. b. Considering that mistakes can occur during transcription and DNA replication, what advantage is there for an organism to have multiple mRNA sequences code for the 7

same amino acid? The advantage for an organism to have multiple mRNA sequenced coding for the same amino acid is that if one codon were to make a mistake or fail, the mRNA would still translate properly. Even if the third base if wrong, you still have a good chance of getting the same amino acid. 15.Use the mRNA codon chart in Model 1 to complete the table below: 5’ or 3’? (for mRNA) 1 st position 2 nd position 3 rd position 4 th position 5 th position 6 th position 5’ or 3’ (for mRNA) DNA 3’ TAC CTT CGG ATG GTC ACT 5’ mRNA 5’ AUG GAA GCC UAC CAG UGA 3’ peptid e N- terminus Met (start) Glu Ala Tyr Gln Stop C-terminus 16.According to the table in Model 1, what amino acid is at the beginning of every polypeptide? Met (AUG) is at the beginning of every polypeptide. Methyanine is the start codon. 8

Note - These are the same images. The image on the left is provided as a Google drawing. To edit it, draw on it, add circles or labels to it, double click the image while in Google Docs. Add in what you need to, then hit “Save and Close”. The image on the right is clearer but cannot be edited. 17.Describe why the 3 stages of translation are named as they are. ( Hint: What do these terms mean in standard English? ) Initiation is when the process of translation begins. Elongation is when the DNA sequence gets longer. Termination is when the process of translation ends. 18.According to Model 2, when the mRNA leaves the nucleus, to which cellular complex does it attach? When the mRNA leaves the nucleus, it attaches to the small ribosomal unit. 19.The mRNA attaches to the complex at the sequence AUG. What is the significance of this sequence of nucleotides? The significance of this sequence of nucleotides is that AUG is the start codon. 20.Describe the movement of the ribosome as translation occurs. As translation occurs, the ribosome moves from 5’ to 3’. _______________________________________________________________________________________ READ THIS : The ribosome is a large complex of ribosomal RNA (rRNA) and proteins. It consists 9

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of two subunits. The smaller subunit binds to the mRNA strand and the larger subunit holds the tRNA molecules in place while the covalent peptide bond is formed between the amino acids. Several ribosomes can attach to an mRNA molecule simultaneously. This allows for many polypeptide chains to be synthesized at once. _______________________________________________________________________________________ 21.The tRNA molecules in a cell are short sequences of nucleotides (about 80 bases) that contain an anticodon and carry a specific amino acid. a. Find the tRNA in Model 2 that is carrying the Histidine (His). What sequence of nucleotides makes the anticodon on this tRNA molecule? CAU is the sequence of nucleotides that makes the anticodon on this tRNA molecule. b. What codon on mRNA would match this anticodon? Histidine would match this anticodon. CAC would match. c. Verify that the codon you wrote in part b codes to Histidine by looking at the table in Model 1. The codon I wrote in part B codes to Histidine because either CAC or CAU would match when you look at the table in Model 1. d. What anticodon would be found on a tRNA molecule carrying Glycine (Gly)? ( Note: There are several correct answers here.) CCG, CCA, and CCU could all be found on a tRNA molecule carrying Glycine. 22.The “t” in tRNAs is short for transfer. In a complete sentence, explain why this molecule is called transfer tRNA is called a transfer RNA because it transfers the correct amino acid in the correct sequence. 23.Describe two things that occur during termination as illustrated in Model 2. Two things that occur during termination are: 1. A release factor binds to the last codon and stops the process of adding more amino acids to the polypeptide. 2. There is a water molecule added to the end of the amino acid chain. Ribosomes and proteins are released and the process stops at the stop codon. 10

Part 3: Extension Questions- Must be complete to earn credit 24.Consider the hypothetical protein represented by triple letter amino acid designations: N terminus Met‐Asn‐His‐Glu‐Pro‐Ser‐Tyr C-terminus a. Propose one RNA sequence that could encode this sequence (be sure to include start and stop codons and indicate 5’ and 3’ ends). 5’ AUG-AAU-CAU-GAA-CCG-AGU-UAC-UAG 3’ b. Propose the Template DNA strand sequence that could be transcribed into your RNA sequence (be sure to include start and stop codons and indicate 5’ and 3’ ends). 3’ TAC-TTA-GTA-CTT-GGC-TCA-ATG-ATC 5’ 25.Here is the beginning of a protein‐encoding gene sequence. Assume the bottom strand is the template strand. 5’-ATGAAGTTTGGCACTTAA-3’ 3’-TACTTCAAACCGTGAATT-5’ a. Give the RNA transcript that would be transcribed from this DNA sequence. The RNA transcript that would be transcribed from this DNA sequence would be 5’ AUGAAGUUUGGCACUUAA 3’. b. Translate the mRNA sequence into a protein sequence. 5’ -Met -Lys-Phe-Gly-Thr- 3’ c. What would be the effect on the final protein product if a mutation caused the following single base‐pair insertion: 5’-ATGAAGATTTGGCACTTAA-3’ 3’-TACTTCTAAACCGTGAATT-5’ The effect on the final protein product if a mutation caused the following single base- pair insertion would be that the process will never terminate because there is no stop codon. d. What would be the effect on the final protein product if a mutation caused the following single base‐pair substitution: 12

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5’-ATGAAGTTCGGCACTTAA-3’ 3’-TACTTCAAGCCGTGAATT-5’ The effect on the final protein product if a mutation caused the single base-pair substitution would be the same as before because UUC and UUU code for the same amino acid. 5’ -AUG-AAG-UUC-GGC-ACU-UAA- 3’ e. What would be the effect on the final protein product if a mutation caused the following single base‐pair substitution: 5’-ATGAAGTTTCGCACTTAA-3’ 3’-TACTTCAAAGCGTGAATT-5’ The effect on the final protein product would be that the final protein has Arg instead of Gly. 5’ -AUG-AAG-UUU-CGC-ACU-UAA 3’ Met-Lys-Phe-Arg-Thr-Stop f. What would be the effect on the final protein product if a mutation caused the following single base‐pair substitution: 5’-ATGTAGTTTGGCACTTAA-3’ 3’-TACATCAAACCGTGAATT-5’ The effect would be that the process would stop. Since UAG is a stop codon, the process stops here. 5’ -AUG-UAG- 3’ Met-Stop g. Using your knowledge from Bio181, predict which of the above mutation(s) you would expect to be the most severe in terms of the overall effect on the person carrying such a mutation? Justify your answer. The last mutation would be the most severe in terms of the overall effect on the person carrying such a mutation because the last mutation, which is the single base 13

pair substitution because it substitutes one base for another which changes the amino acid. 26.A mutation is a change in the DNA sequence which can then propagate to the mRNA and protein. A silent mutation is one that does not affect protein structure. Write a code for an original DNA strand containing at least 12 bases, and then mutate the original DNA so that the final protein is unaffected. 5’ -ATGAAGTTTGGCACTTAA- 3’ 3’ -TACTTCAAACCGTGAATT- 5’ 5’ -ATGAAGTTCGGCACTTAA- 3’ 3’ -TACTTCAAGCCGTGAATT- 5’ The final protein stays the same because UUU and UUC code for the same amino acid. 14

BIO181

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