Name_BS50 Exam 3 Key (Fall 2005) Page 2 of 5

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1 Name_BS50 Exam 3 Key (Fall 2005) Page 2 of 5 Question 1. (14 points) Several Hfr strains derived from the same F + strain were crossed separately to an F - strain, giving the results indicated in the table below. Draw a chromosome map for the genes indicated. Include the location and orientation of the F factors. Define all symbols used Time elapsed before cells carrying markers were recovered Hfr strain 5min 10min 15min 20min 25min 30min 1 pro gal lac 2 pur ala leu lac 3 pro thy thr 4 pur ile thr Fertility Factor Origin (first to enter) Terminus (last to enter) Thr pro Hfr 3 Hfr 1 Thy 15 min gal lac (14 pts) Genes (order and mapping times): 8 pts; no deduction if a single mapping time in error; -1 pt. if mapping times listed, but not indicated on map. Hfr s: 6 pts; positions (2 pts); relative orientations (2 pts); details (2 pts). Position of Hfr s must be indicated as single sites: -1 pt if drawn as an extents. Details for Hfr s (any 2 of the following): absolute orientation; use of conventional symbol to indicate; precise localization (esp. for Hfr4 adjacent to ala). For aborted attempts, 1 pt if indicated as circular; if drawn as 4 linears, graded on the basis of one (2 pts for genes; 1 pt for Hfr). Question 2. (22 points). You are studying a fairly common human trait and have discovered a linked restriction fragment length polymorphism (RFLP). Use the pedigree from the following family along with the Southern blot analyzing inheritance of the RFLP to answer the following questions. Ile Hfr 2 5 min pur 5 min Hfr 4 ala leu

2 Name_BS50 Exam 3 Key (Fall 2005) Page 3 of 5 a) Just by looking at the pedigree (and ignoring the DNA data), what mode(s) of inheritance could account for the pedigree? Circle all possible answers. autosomal dominant autosomal recessive X-linked dominant X-linked recessive (4 pts) 1 pt for each correctly chosen or not b) Considering the DNA data as well, what is the mode of inheritance for this trait? (4 pts) X-linked recessive c) Give the genotypes of the following individuals are completely as possible. Define all symbols used. R 7 = 7kb RFLP A = dominant no trait R 6 = 6kb RFLP a = recessive trait R = 4kb+3kb RFLP Y = Y chromosome Individual A: R 6 a/ R 7 A Individual B: R a/ Y Individual E: R 7 A/ R a (8 pts) 2 pts for each (1 pt for each chromosome); 2 pts for correct nomenclature; -3 pts if correct linkage not indicated. d) Individual E marries a phenotypically normal male. Their son is found to have only a 4 kb and 3 kb band when analyzed with the same probe used above. Assuming the RFLP is actually 10 cm from the trait for this gene, what is the probability that their son will have this trait? Show your reasoning. (6 pts) 90 % (3 pts if no further explanation); son carries 4+3 RFLP marker from mother (3 pts; OK if implicit, deducted if misunderstood) and Y from father (1 pt; OK if implicit); 10 % chance that a recombination event occurred between the RFLP marker and the disease allele from mother (2 pts; must explicitly mention recombination, cross-over, or linkage, or show in a diagram). -2 pts if assumed E has 90% chance of carrying the disease allele; -2 to 3 pts for other extraneous factors. If answer to b/c incorrect, could receive up to 5 pts if consistent. Question 3. (18 points) a) The following lists a number of genotypes for the Lac operon (including partial diploids). In each case, say whether the genotype listed will show activity of β-galactosidase (lacz) and lactose permease (lacy) when the inducer (allolactose) is added and in the absence of inducer: Genotype β-galactosidase (lacz) lactose permease (lacy) Not induced Induced Not induced Induced I + P + O + Z + Y I s P + O + Z + Y

3 Name_BS50 Exam 3 Key (Fall 2005) Page 4 of 5 I + P + O c Z + Y I + P + O + Z Y + / I + P + O c Z + Y I + P O c Z Y + / I + P + O + Z + Y I + P O + Z + Y / I P + O + Z Y (15 pts) -1 pt for each incorrect; >3 incorrect scaled (eg: -11pts. given 7/15). b) For the lac operon, what would be the phenotype of a mutation that prevents CAP from binding camp. (3 pts) Genes of the lac operon would never be transcribed at high level; or basal level of transcription only, even in presence of inducer. 1-2 pts if indicate no activity, depending upon detail of explanation. Question 4. (18 points) A plant has been discovered that has two different true breeding strains, each one homozygous for a different allele of the gene "Height". One strain is homozygous for the dominant, wild-type allele H and the plants are very tall. The other strain is homozygous for the recessive, mutant h allele and the plants are all very short. Below are the DNA coding regions for the two different alleles. Both alleles have three exons, although Exon 2 of the h allele is longer and contains an in-frame stop codon (UAA). Exon 3 can encode a DNA binding motif (a homeodomain). The presence of a homeodomain is characteristic of a transcriptional regulator. A) Which allele produces a longer processed mrna? Which allele produces a longer protein? Explain. (Assume all in-frame stop codons are shown.) (8 pts) h produces longer mrna (2 pts) because exon 2 is longer (2 pts); H produces longer protein (2 pts), because h has a premature stop (2 pts). B) Based on the information given above, what do you think Height protein does in the plant? Why is the H allele dominant? (5 pts) Functions as a transcription factor (1 pt); regulates genes that affect height or growth (2 pts); is dominant because one copy of the gene is sufficient (2 pts). Partial credit for explanation of dominance: trans-acting (1 pt); produces functional product and h does not (1 pt). If additional incorrect explanation offered, -1 pt. Compensatory credit: described as haplo-sufficient (1 pt); the h allele product has no deleterious effect

4 Name_BS50 Exam 3 Key (Fall 2005) Page 5 of 5 (1 pt); if detailed explanation of transcriptional regulation, but did not address phenotype (1 pt). C) Another mutant of the H allele is discovered. It has all the Height coding DNA, but no trace of Height mrna. Suggest what the molecular nature of this mutation might be, and predict its phenotype. (5 pts) Mutation in the promoter (2 pts); phenotype (3 pts): short (2 pts), recessive (1 pt), OR equivalent to h/h (3 pts). Compensatory credit: if detailed explanation of promoter/polymerase, but did not address phenotype (1 pt). Alternative explanations, such as mutation that affects local chromatin structure, accepted if adequately explained. Question 5. (22 points) A fragment of mouse DNA digested with EcoR1 carries the gene M. This DNA fragment, which is 8 kb long, is inserted into a bacterial plasmid at an EcoR1 site. The recombinant plasmid is then digested with different restriction enzymes. The size of these fragments after digestion are described below: EcoR1: 6kb, 8kb BamH1: 3kb, 11kb EcoR1 & BamH1: 1kb, 2kb, 4kb, 7kb Pst1: 14kb Pst1 & EcoR1: 4kb, 6kb A) Draw a restriction map of this circular plasmid. PstI 3 kb 4 kb BamHI 1 kb EcoRI Gene M 4 kb EcoRI 2 kb (8 pts) Sites in correct relative order (4 pts); correct map distances (4 pts). B) If a Southern blot is prepared from EcoR1 digested DNA, BamH1 digested DNA, and EcoR1 & BamH1 digested DNA, which fragments will hybridize to a probe containing only plasmid DNA? EcoR1: _6kb BamH1: _3kb, 11kb BamHI

5 Name_BS50 Exam 3 Key (Fall 2005) Page 6 of 5 EcoR1 & BamH1: _2kb, 4kb (6 pts; 2 pts each). If one of two bands indicated, 1pt; no credit if extra bands indicated. C) What fragment sizes do you expect from a Pst1, BamH1 double digest? (8 pts) 3kb (2 pts) and 8kb (2 pts) Two different 3kb fragments are produced (not required for full credit) D) You wish to express the mouse protein in bacteria. What problem(s) might you encounter, and how could you solve it? (4 pts) If the mouse gene has introns (1 pt) they would not be spliced out (1 pt); can avoid this problem by using a cdna of the mouse gene (2 pts). OR Prokaryotic promoters and transcription differ from eukaryotic (2 pts); include a bacterial promoter in construct (2 pts). If missed the point, but described transformation process (1 pt). Question 6. (6 points) Histone acetyltransferase (HAT) is the enzyme responsible for adding acetyl groups to histones (particularly H4). What would be the consequence of having a low activity HAT enzyme? (6 pts) Would result in more tightly packed nucleosomes (4 pts) and thus lower levels of transcription (2 pts). Mention of nucleosomes required for full credit. Partial credit for more condensed chromatin structure (3 pts for first component); more heterochromatin or chromatin remodeling (2 pts for first component).