3' A TAT --IIL 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 DNA C GU UGA UG G MRNA TRNA Amino Acid Met
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Q: 1. T DNA strand NUCLEUS MRNA CU CYTOPLASM 6. 5. Lysine 7. AAGUUU UGUUCA AA 4. 2.
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Q: TAC TGA TCG ACC CCC ATA ATG AAAAIC MRNA TRNA AA
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- pcc300ATAAADATATAOOTTAA 1. Use the genetic code table and the information in the diagram below to determine the amino acids that would make up the portion of the polypeptide shown. Include information for a key as well. DNA template 3' G CATA ACAGAGGATT-5' al bnsua AMAm pniwollot erfT E transcription s yd bnsita ebitgeqylog s sidmeaze of beae RNA strandUU UAOUOUU A-emoaodin 5'-CGUA AUUGUC UCCUUA- 3' J J JL erit o elinW (s) translation bluow terdt aspnso sigootiwsone polypeptide viemetis ns ebivo19 (d) ent ot etslanT Key:BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation LeadpleEukaryotic Genetic Sequence: 5'-TAC CAT GAT CCC TAT - 3' 1. What would be the newly synthesized DNA strand and explain how the strand will be replicated. Where in the cell would this occur? 2. What would be the synthesized mRNA strand, and how is it transcribed from the original DNA strand, and then converted from a pre-mRNA strand to a mature mRNA? Where in the cell does this occur? 3. What would be the anti-codons for the tRNA. What are the amino acids generated based on the RNA. How are these amino acids translated into protein and where in the cell does this happen?
- In which strand is the promoter? 5' TCATGAGATA GCCATGCATA TCTAGTATGT AGGCATCTGA GTTTATATCT CA 3’ 3' AGTACTCTAT CGGTACGTAT AGATCTATCA TCCGTAGACT CAAATATAGA GT 5'www D le C 3⁰ A B Indicate True (T) or False (F) for the following statements. Only use the letter (T/F) in the space provided 1. The name of this process is best known as Rho dependent termination 2. The enzyme C called DNA polymerase incorporates ribonucleotides into B called the mRNA False 3. The DNA region A contains inverted palindrome sequences which results in formation of a stem-loops structure 4. During this process, the structure D called terminating hairpin forms and increases the enzyme affinity which terminates transcriptionCynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-3" It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X G
- First Letter A G U с 22. Using the provided "Genetic Code-Reference" answer the following question. Based on the following DNA template strand, write out the amino acid chain produced. 23. Consider the following mRNA base codon sequence 5'-AUC-GAA-3' and the provided "Genetic Code-Reference". Genetic Code-Reference UUU UUC UUA UUG CUU CUC CUA CUG (mutated or silent) (mutated or silent) b. Briefly explain your reasoning for each. (be sure to include both parts) AULLY AUU a. Label which of the following would result in a mutated amino acid sequence or a silent mutation. (May help to first determine the original amino acid sequence, then compare to mutations) U Phe mRNA codon sequence: anticodon sequence: amino acid sequence: Leu Leu 5'-AUA-GAA-3' Val 5'- AUC-GAC-3' AUC Ille AUA AUAJ *AUG Met/Start GUU GUC GUA GUG UCU UCC UCA UCG) CCU) CCC ccc CCA 000 CCG ACU ACU ACC ACA ACG, C GCU GCC GCA GCG Second Letter Ser Pro Thr 3'-CAA-GTC-TGT-5' Ala UAU UAC) Tyr Туг A **UAA Stop UAG Stop CAU] CACJ…5' - ATG GGG CCC GTT TTC AAT ATG CAG GTC CAT CCG TAC GTA CAG GCC GGA ATT TGA - 3' There are two introns in this DNA sequence. Remember introns start with GT and end with AG. (a) How many base pairs are in intron 1 and intron 2.Telomerase supplies its own template RNA molecule as shown in Figure 3 below: B AAUCCCAAU TTAGGGTTAGGGTTAGGGTTAGGGTTAGGGTTAGGG-W' JAATCCCAATCCCAATCCCAA-X' Figure 3 (i) Label the ends (5' and 3') on the DNA and RNA strand at position X, Y and Z, in the Figure 3. (ii) Draw and explain the two loop structures at the end of telomere.
- 48 Second letter If any single nucleotide is deleted from the DNA sequence shown below, what type of mutation is this? UUU U UUC UUA UCU Phe UCC UAU UGU Cys ANTISENSE 5' GGACCCTAT3' UAC Tyr UGC UAA Stop UGA Stop UAG Stop UGG Trp Ser UCA Leu UUGL" UCG CUU CU CAU) His CAC) CGU CGC CGA Arg CUC C Leu Pro CAA1 Gin CAG) CUA CCA CUG CG CGG AAU AAC Asn AGC AAA AAG Lys AGG Arg AUU ACU AGU Ser AUC lle ACC Thr AUA ACA AGA AUG Met ACG GCU GCC GUU GAU] GGU GUC Val GAC Asp GGC Ala Gly GUA GCA GAA GGA Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a FRAMESHIFT SILENT NONSENSE MISSENSE Third letter First letterDNA STRAND IS 3’ TAC-AGC-ACT-CAG-TCA 5’a. what is the non-template/sense/coding strand?b. What is the arrangement of the m-RNA?c. What's the chain arrangement of the amino acids that will be made according to the order of the RNA?d. If on the non-coding strand of DNA there is suddenly one T base that sneaks into the 4th sequence (from the left), or causes a mutation, then how will the RNA be formed?e. What's the chain arrangement of the amino acids produced by this mutation?-Write down the complementary DNA sequence. TACCTAGCG CACATGTAGGTGGGCAAAGTT -Write down the complementary mRNA sequence for each of the following DNA sequence. A: TACCTAGCGCACATGTAGGTGGGCAAAGTT B: TAC ATG GTT ACA GTC TAT TAG ATG CTA TTT ACT TAG -If the first G changes to A what kind of mutation will happen? Show the change in amino acid sequence. This is base substitutions involve the replacement of one nucleotide with another. And it changes one amino acid coding, producing a missense mutation TAC CTA GCA CAC ATGTAGGTGGGCAAAGTT TAC CTA ACACACATGTAGGTGGGCAAAGTT