Replication, Transcription and Translation | MCDB 1041, Assignments of Genetics

Download Replication, Transcription and Translation | MCDB 1041 and more Assignments Genetics in PDF only on Docsity! Replication, Transcription and Translation Objectives:  Draw how replication works, and know what the main players are  Be able to explain how directionality of DNA has consequences for cellular processes  Draw transcription, and know what the main players are  Explain how mutations can affect proteins ________________________________________________________________________ Replication Below is a simple diagram of DNA. 1. Draw the direction of replication on both strands, with arrows. 2. If this is a cartoon of a chromosome from a cell that has not yet undergone replication, draw, inside this cartoon, what the strands of DNA look like within the chromosome. If this is a replicated chromosome (such as you would see in mitosis or meiosis), draw, inside the cartoon, what the strands of DNA would look like within each of the sister chromosomes. Draw the original strands of DNA with a solid line and the newly replicated strands of DNA with a dotted line. 5’ 3’ 5’ 3’ 5’ 3’ 5’ 5’ 5’3’ 3’ 3’ Transcription Earlier this semester, you learned about Amy Roloff who has achondroplasia. Molly is Amy’s daughter and she is average height. Achondroplasia occurs because of a mutation in a gene called Fibroblast Growth Factor Receptor 3 (FGFR3). The gene codes for a protein that binds to a growth factor—this binding process is required for normal growth. A portion of Molly’s and Amy’s FGFR3 DNA sequence is shown below: 1. Circle the DNA base pairs that are different in Molly and Amy’s FGFR3 gene. To find out why a single base pair change in DNA can affect height in humans, we need to understand both transcription (information passes from DNA to RNA) and translation (information passes from RNA to protein). One strand of DNA is known as the template strand and the other strand is known as the coding strand. RNA is transcribed from the template strand. The RNA is complementary and anti-parallel to the template strand: 2. In the picture below, draw in the strand of RNA as a line with a 5’ and 3’ end that would be produced from this DNA. 3. What information in the diagram tells you which strand is the coding strand? Label each strand as coding or template. The transcript is made from the template strand. 4. The sequence below is a piece of the template DNA strand from the middle of the FGFR3 gene discussed earlier. What strand of RNA is produced from this template? (write your strand of RNA with 5’ on the left and 3’ on the right) 5’ AGCTACGGGGTG 3’ 5’CACCCCGUAGCU3’ mRNA 5. What is the sequence of the coding DNA strand? (label 5’ and 3’ ends) 5’CACCCCGTAGCT3’ 5’ http://tlc.discovery.com/fansites/lp bw/bios/molly.html AGCTACGGGGTG TCGATGCCCCAC http://tlc.discovery.com/fansites/lp bw/bios/amy.html Molly AGCTACAGGGTG TCGATGTCCCAC Amy 3’ 3’ 5’ 5’ 3’ template coding Transcription starts here D. Allele D, Identified in Chinese patients. 5-AAG GAG AAA GTA AGG AAC TTT GCT GCC ACA-3 3-TTC CTC TTT CAT TCC TTG AAA CGA CGG TGT-5 5-AAG GAG AAA GTT AGG AAC TTT GCT GCC ACA-3 3- TTC CTC TTT CAA TCC TTG AAA CGA CGG TGT-5 Extra Practice Problems Here is a DNA sequence (coding strand): 5’ GCGCCAGCCTTAA 3’ A normal protein made from this sequence includes a proline. A mutant protein is the same length but there is a histidine instead of a proline. Write the coding strand of the mutant DNA sequence: (Note: the first codon may NOT begin with the 5’ G, and the AUG start codon is further upstream and not included in this sequence) Wild-type: 5’ G CGC CAG CCT TAA 3’ Mutant: 5’ G CGC CAG CAT TAA 3’ normal Allele D Kind of mutation: silent What affect does the mutation have on the transcription? none What affect does the mutation have on the function of the protein? none Name:________________________ Mutation Activity Questions 1. What is the mRNA sequence transcribed from this DNA sequence? A. 5’ AGCC 3’ B. 5’ CCGA 3’ C. 5’ UCGG 3’ D. 5’ GGCU 3’ Answer: D 2. You are studying mRNA that has the repeating sequence: 5'-UGUGUGUGUGUGU…-3’. UGU encodes for the amino acid cysteine and GUG encodes for the amino acid valine. What kind of amino acid(s) would you expect to find when the mRNA is translated into protein? A. one protein consisting of repeats of cysteine B. one protein consisting of repeats of valine C. one protein consisting of alternating repeats of cysteine and valine Answer: C 3. When looking at a replicated chromosome at the beginning of meiosis or mitosis, which of the following is true? A. One sister chromosome contains a double helix composed of two original DNA strands, and the other sister chromosome contains a double helix composed of two new DNA strands. B. Each sister chromosome contains a double helix composed of one original DNA strand and one new copy. C. Each sister chromosome is a combination of old and new strands of DNA interspersed together. Answer: B 4. One form of deafness called CMT is caused by the absence of a single amino acid in the middle of the PMP22 protein. You sequence the PMP22 gene and gene products in a normal person and a person with CMT. When you are comparing the two sequences, you would expect to find that the person with CMT has: A. An insertion B. A deletion C. A new termination codon D. Either b) or c) E. without additional information, any of the above are possible Answer: B 5. The following DNA sequence corresponds to the two extreme ends of the coding strand: 5’ ATG-GAA//CAG-TGA 3’ Each codon is separated by a dash and the middle of the gene is represented by “//”. Referring to the codon table below, which of the following sequence changes is likely to interfere most with the function of this gene? A. 5’ ATG-GAA//CAA-TGA 3’ B. 5’ ATG-TAA//CAG-TGA 3’ C. 5’ ATG-GAG//CAG-TGA 3’ D. (a.) and (b.) are equally likely to interfere with the function of this gene E. (b.) and (c.) are equally likely to interfere with the function of this gene Answer: B 5ÕAGCC 3Õ 3ÕTCGG 5Õ Template strand Coding strand eng Pur, SVE Sve Sve Sy eu & lif g22 : nh SESE ES 5 E888 E\Es 3 3|32 ae a|s 3 315 3|2 3133 #) é 3313 3 3 BEEE é # BBSB/E8S8 3 BES8 #\& B5iss = £ sak 2282|223): arg FUL 3 SSeS