Thymine Cytosine Adenine Guanine Nucleobases of DNA Base pair DOODL Helix of sugarphosphates DNA Deoxyribonucleic acid Nucleobases RNA Ribonucleic Acid Uracil Cytosine Adenine Guanine Nucleobases of RNA
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- pen+ GEAR Normal DNA Normal RNA Amino Acids Mutant DNA Mutant RNA: Name: FRAMESHIFT MUTATIONS occur when a base is added (or removed) from 8. Determine the amino acid chain coded for by the following sequence. Supp another A is added after the first codon. Complete the chart showing the amir normal and mutant sequence. Amino Acids Investigation: DNA. Proteins. and Mutatic TGG AGT CGA GGT TGG AAG TCG AGG T Why are frameshift mutations likely to cause more problems than a p 9. Cystic fibrosis is a disease that causes mucus to build up in th misshapen protein in the cell membrane which interferes with the popociated with the disorder3’ – ACCTCTTACTTTTATATATAGGGAAGACTAATTGTC – 5’ Transcribe the template strand be sure to use the 5’ and 3’ directions appropriately on the mRNA, feel free to rewrite the DNA strand if needed to make it easier to interpret. Make sure to label the mRNA with a "5'cap" and place 10 A's to form the poly-A tail.Name: Date: 2. The sequence of a fragment of one strand of DNA is AATTGCATATACGGGAAATACGACCGG. Transcribe this s sequence into MRNA. er bns eldst eboo oi ebitqeqylog erlt to noihiog eri qu elsm bluow tsri abios onime Jlaw as ye s 1ot noitem atelomet AHG 3. The following MRNA ştrand is being used to asemble a polyp
- Basisse triplet nommer/ Base triplet number 1 2 3 4 5 6 7 Menslike DNA-volgorde / Human DNA sequence ATG TGT CCA TTA ACG TGC ACA Name the codon that is formed from base triplet number 2 on the DNA sequence. Write down the names of the amino acids coded for by base triplets 6 and 7. Write down the full names Draw the structure of the dipeptide Thr-Cys at pH7. Clearly indicate the following: the amino terminus (N) the carboxyl terminus (C) a peptide bond an α-carbon atom If a mutation changes base triplet 1 from ATG to ATA, why will this not change the protein formed? E.The following is a sequence of base triplets in DNA F.If guanine, found in the first base…The original DNA sequence TACACCTTGGCGACT I need the mRNA sequence and the amino acid sequence And also the mutation typeCOMPLEMENTARY DNA SEQUENCE OF GACGGCTTAAGATGC
- 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' GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT 3' 3' CACGATCGCCCTTACTCGACCCTATGATCATCCCGA 5' Template Strand: 9. Using the template strand, transcribe the DNA above, Be sure you write your sequence 5 - 5 a indicate the 5' and 3' ends of any nucleic acid molecule(s). 10. Use the codon chart below to translate your mRNA into an amino acid sequence. Begin at the first codon. Third First position (5' end) Second position position (3'end) UGU Cys UAU Tyr Cc UGC Cys UGA Stop UGG Trp UCU Ser -Y UAC Tyr UAA Stop UAG Stop UUU Phe - F UUC Phe UUA Leu UUG Leu FL UCC Ser -- UCA Ser UCG Ser CGU Arg CGC Arg ER CGA Arg CGG Arg CCU Pro CAU His CUU Leu CUC Leu -- CAC His CAA Gln CAG Gln CCC Pro -P A - CUA Leu CUG Leu CCA Pro CCG Pro AAU Asn AAC Asn AGU Ser AGC Ser AGA Arg ACU Thr AUU lle AUC lle AUA lle AUG Met M ACC Thr -T ACA Thr ACG Thr A. AAA Lys K AAG Lys -R AGG Arg A. GAU Asp -D GAC Asp GGU Gly GGC Gly GCU Ala GUU Val GUC Val GCC Ala A -G GGA Gly GGG Gly A -V GUA Val GUG Val GCA Ala GCG Ala GAA Glu -E…Cynt 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
- DNA 5' ATGGCTTCTCAATACTGCTTTGTTTTGGTT 3' template strand 3' TACCGAAGAGTTATGACGAAACAAAACCAA 5' coding strand Write down the sequence of nucleotides in a fragment of an m-RNA molecule that will be produced based on the information in the DNA fragment above (start with 5' and end with 3'). If you separate codons in MRNA with blank spaces, it will be easier to do the next step. MRNA: 5' Using a three-letter code for amino acids write the sequence of the first ten amino acids of the protein pectate lyase (refer to the table of 64 codons from a lecture or a textbook).5-ccuaaucg-34 3'-acctgcctataccggattagetetgatectaagcatgtc-5 The diagram above shows an RNA primer hydrogen-bonded to a DNA template. Which letter indicates the site where DNA polymerase would add nucleotides to this structure? OA OB D Any of these is possible CUsing Figures 8.7 and 8.9 as a guide, draw a dinucleotide composed of C and A. Next to this, draw the complementary dinucleotide in an antiparallel fashion. Connect the dinucleotides with the appropriate hydrogen bonds. FIGURE 8.9 The two polynucleotide chains in DNA run in opposite directions. The left strand runs 5 to 3, and the right strand runs 3 to 5. The base sequences in each strand are complementary. An A in one strand pairs with a T in the other strand, and a C in one strand is paired with a G in the opposite strand. FIGURE 8.7 Nucleotides can be joined together to form chains caled polynucleotides. Polynucleotides are polar molecules with a 5 end (at the phosphate group) and a 3 end (at the sugar group). An RNA polynucleotide is shown at the left, and a DNA polynucleotide is shown at the right.
