Hw1 Gen320fall07answer
- 1. Genetics 320 REVISED HW 1. Due Sept5, Wed. 11AM. TEXT EXPLANATIONS MUST BE TYPED 1. Complete the following chemical structure. I AM ASKING FOR A C-C DINUCLEOTIDE HYDROGEN-BONDED TO A G-G DINUCLEOTIDE. THE DOTTED LINES INDICATE THE AXIS OF EACH DINUCLEOTIDE. Insert all relevant atoms to form 2 G-C base pairs, with phosphodiester backbone connecting them. Put dashed line showing hydrogen bonds involved in base pairing. Put a box around atoms that are changed when cytosine is deaminated to uracil. Put a triangle around the 3’hydroxyl groups where polymerase could add a base during DNA replication. Indicate the 3’ ends of each strand: write “3’ end” DNA is negatively charged when it is isolated from the cell. Where does the negative charge come from? Oxygen on the phosphate O-P=O O C-G C-G H H H CH2 O-P=O O O CH2 OH N N N H O H H H H CH2 O CH2 O OH 5’end 3’end N N 3’end 3’end O-P=O O N N N H O O-P=O O N O N N N H O N O N N N
- 2. 2. Indicate which mutation has a greater consequence to a gene’s function from each of the following pairs of mutations. BRIEFLY discuss your reasoning. : Missense mutation near the 5’ end of the ORF compared to a nonsense near the 5’ end of the ORF. Nonsense near the 5’end of the ORF; a nonsense will stop translation, so no protein 3’ of mutation is made, while a missense only changes 1 amino acid. Nonsense mutation near the middle of the ORF compared to a nonsense near the 3’ end of the ORF. Nonsense near the middle of the ORF; no protein made 3’ of mutation. Frameshift near the 5’ end of the ORF compared to near the 3’ end of the ORF. Frameshift near 5’end…reading frame altered, so amino acids 3’ of mutation different, ending in a stop codon most likely.
- 3. 3. Consider the domain structure and the phenotypes of mutants in the gene shown below. The three domains are shown (amino acids 10-100 for the rectangle, 120-150 for the triangle, and 180-210 for the circle). The gene has two functions, one in memory and one in hearing. The gene has no introns. Cells (wt or “wildtype”=normal; mut1, mut2) are homozygous for each allele. Amino 1 10 100 120 150 180 210 250 Acid # Cell Wt mut1 mut2 Cell Phenotypes Hearing memory Mut2 is a deletion that removes the triangle domain joining the two DNA regions shown. Is this deletion likely an in frame or out of frame deletion. Explain. IN FRAME; the oval domain is still made in this mutant, so correct frame of codons read by ribosome only if deletion inframe--out of frame ribosome reads other amino acids . What is a likely molecular consequence and phenotype of a frameshift mutation in the triangle domain? Explain. Hearing and memory loss, because correct amino acids not read by ribosome for either triangle nor oval domains. Oval required for memory, triangle for hearing (from mut 1 and mut 2). What is a likely molecular consequence and phenotype of a missense mutation between codons 150 and 180 in which an alanine is mutated to a glycine. Explain. No phenotype=wildtype, because alanine to glycine minor R group change, and because 150 -180 is a ‘linker” region with no specific domain nor function, most likely. What is a likely molecular consequence and phenotype of a nonsense mutation at codon 100? Hearing and memory mutant of short protein of 100 amino acids. NOTE: This graded. 1pt each answer
- 4. 4. Describe the fate of “the p1 protein” for each of the mutants listed. In the p1 protein the specific amino acids 1,2,3 and 6 are essential for function; amino acids 4 and 5 are “linker” and can be any amino acid. p1 synthesis requires DNA damage and the p53 enhancer (see below). Describe a likely consequence of each mutation; is a transcript made , is the transcript translated, how long is the translated protein, and the overall amount of functional protein in the cell. Bases in the TATA box that are essential for RNA polymerase binding are capitalized. Capital letters in the ORF are to help you identify the amino acids essential for function. Consult Genetic Code Table. You can answer with More, Less, Unchanged compared to wt. p53 binding site TATA box +1 12 TATAccg ggccaattgcg A TGGGCATA cccctg CCCTGA wild type YES YES 6 YES a. Deletion in the p53 binding site Less Less still 6 Less b. Mutation of TATA to TATtgca Less Less still 6 Less c. Base 12; A to C YES NO 0 NO d. ATA to ATT YES YES 6 YES e. ATA to TTC YES YES 6 NO f. cccc to tagc YES YES 3 NO g. Insertion c after g at +2 YES YES 6 YES h. ATG to TGA YES NO 0 NO Transcript Translated #AA Function p53 biology: p53 is a transcriptional activator, or enhancer. Without it, the RNA polymerase cannot transcribe the p1 gene very well. Note to class: c thru h was graded. 1pt each line.
- 5. 5. Describe briefly how a gene family arises. The gene family can consist of two genes with related but slightly different functions. First there was one gene. An error led to a gene duplication, so there were two identical genes. Then mutation occurs in one of them to alter function. 6. Discuss briefly the relevance of the idea that ATP is not limiting for many processes in higher eucaryotic cells for the presence of introns. A student pointed out that to transcribe 250kb, then delete most of the RNA to make a mRNA of 6 kb seems ‘wasteful” in terms of ATP..indeed, each nucleotide polymerized costs to make the NTP. We surmise, then that ATP is not limiting, otherwise introns would not be so huge and “wasteful” of the cells energy.