Crick's Central Dogma states that there is a unidirectional flow of genetic information DNA ---> RNA -----> Protein Explain why this may not always be the case. (Describe a scenario in which movement may occur in the opposite direction).
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Crick's Central Dogma states that there is a unidirectional flow of genetic information
DNA ---> RNA -----> Protein
Explain why this may not always be the case. (Describe a scenario in which movement may occur in the opposite direction).
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- As you should recall, DNA, when not being actively transcribed, has a double helical structure. This portion of the DNA has had the two strands separated in preparation of transcribing for a needed protein. The following is one of the two complimentary strands of DNA: 3' - AACCAGTGGTATGGTGCGATGATCGATTCGAGGCTAAAATACGGATTCGTACGTAGGCACT - 5' Q: Based on written convention, i.e. the 3'-5' orientation, is this the coding strand or the template strand? ______________________________ Q: Assuming this strand extends from base #1 to #61 (going left to right), interpret the correctly transcribed mRNA and translated polypeptide for bases 24 - 47: mRNA: ___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___- polypeptide chain: ________--________--________--________--________--________--________--________What is the meaning of the statement “The code is redundant but not ambiguous?”.You would like to add a nuclear localization sequence (NLS) of Lys-Lys-Lys-Arg-Lys to a protein that is usually found in the cytoplasm of a yeast cell. To accomplish this, you introduce the nucleotide sequence encoding the NLS into the gene that encodes the cytoplasmic protein of interest. a. What is the size of the nucleotide insert that will encode the NLS? Briefly explain. 5' 3' b. Below is a diagram of the gene encoding the cytoplasmic protein of interest in the yeast genome. If your goal is to put the NLS at the carboxyl (C) terminus of the protein, at which location (A-E) should the NLS be inserted? Briefly explain. A TATAA ATATT promoter +1 B ATG TAC D TAA ATT stop codon E 3' 5'
- Often geneticists look for suppressors to find interactive proteins. Which of the type(s) of suppressors you put for part a will help to identify interacting proteins, and which type(s) will not? What are two (or one, if we don’t get a chance to talk about two of them in class) other techniques (not necessarily “genetic” techniques, but at least, lab techniques) that help to identify identifying proteins?Complete the protein synthesis for the partial DNA sequence for a normal FGFR3 gene (TOP) and mutated FGFR3 gene (BOTTOM). Remember, when filling in mRNA, use capital letters only. When filling in amino acids, use three letters, with the first letter capitalized. If you do not use this format, your answer may be marked wrong. DNA CCG TTC GGG GAA ССС MRNA Amino Acid DNA CCG TTC GGG GAA TCC MRNA Amino AcidThe code for a fully functional protein is actually coming from an mRNA transcript that has undergone post-transcriptional processing which is essentially way too different from the original code in the DNA template. Given: Val-His-Leu-Thr-Pro-Glu-Glu (Protein with known amino acid sequence) Requirement: Original DNA code. Itemize the steps you would take to get to know the original DNA code of the protein in focus.
- The first 15 bases of the original coding informational strand of DNA (which continues after what is shown) are 5-ATGAAACCCGGGTTT(...)-3'. Which of the following mutations to this original strand would be silent and which would likely only have a small effect on the coded protein? (Again, only the first 15 bases of the DNA strand are shown, even though they continue on. HINT: To solve this, you might write out the template DNA strand, the mRNA strand that would be made, and the resulting amino acids for the start of the protein.) Mutation #1: 5-ATGAAAGCCCGGGTT(...)-3' Mutation #2: 5-ATGAACCCCGGGTTT(...)-3' Mutation #3: 5-ATGAACCCGGGTTTA(...)-3' Mutation #4: 5-ATGAAGCCCGGGTTT(...)-3' Mutation #5: 5-ATGTAACCCGGGTTT(...)-3'Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.The sequences of DNA bases below represent parts of the genes responsible for the production of one type of protein, an enzyme, produced by Botana curus and Species X, Y, and Z Under each DNA sequence, write the complementary messenger RNA base sequences that each of these gene fragments would produce. Note: Unlike during DNA replication, in the production of messenger RNA, the DNA base “A” specifies the RNA base “U.”. Use the universal genetic code table provided (see Universal Code attachment) to translate the messenger RNA base sequences into sequences of amino acids in the protein produced by each species. Write the sequences of amino acids under the messenger RNA sequences.
- Which of the following are elongation factors involved in the attachment of new aminoacyl-tRNAs? EF-Ts EF-Tu EF-p EF-GAccording to wobble rules, what codons should be recognized by the following anticodons? What amino acid residues do these correspond to?(a) 5′ ¬ICC ¬ 3′ (b) 5′ ¬GCU ¬3′(c) By binding one L-tryptophan molecule/monomer, the trp repressor binds to DNA to suppress syn- thesis of L-tryptophan in E. coli. Below is the amino acid sequence of the helix – (reverse) turn – helix region of the trp repressor that binds to DNA compared to the sequence of the corresponding DNA binding motif of the Prl protein, a different type of repressor protein. A diagram of the trp repressor dimer is also shown. reverse turn trp helix 4 70 Trp -Gly-Glu-Met-Ser-Gln-Arg-Glu-Leu-Lys-Asn-Glu-Leu-Gly-Ala-Gly- Ile- Prl -Ser-Glu-Glu-Ala-Lys-Glu-Glu-Leu-Ala-Lys-Lys-Cys-Gly-Ile-Thr- Val- Pri heilix trp helix 5 80 90 Trp Ala-Thr-Ile-Thr-Arg-Gly-Ser sgn-Ser-Leu-Lys-Ala-Ala- Prl Ser-Gln-Val-Ser-Asn-Trp-Phe-Gly-Asn-Lys-Arg-Ile-Arg- Prl helix