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MCB 142 second midterm: Molecular Genetics, Exams of Genetics

University of California - BerkeleyGenetics

A midterm exam for MCB 142, a course on Molecular Genetics. The exam consists of ten short answers and five problems, with a total of 100 points. The exam covers topics such as DNA replication, human genome, restriction enzymes, start codon, and translation. The exam format is described in detail, and students are reminded to write their name, ID, and TA's name on every page. The document also includes a section for graders to score each page of the exam.

Typology: Exams

2021/2022

Uploaded on 05/11/2023

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Name _________________________________________ ID ______________________
GSI Name ------------------------------------------
1
MCB 142 second midterm: Molecular Genetics
Please write your name, ID, and TA’s name on the top of ALL 8 pages of this exam.
Please put all your answers in the spaces below.
Please use pen to write your answers. If you use pencil, you cannot request a re-grade.
You may not use any calculator, cellular phone, internet connection, playstation, or other
electronic device during this exam.
Please double check that you have written your name, ID, and TA’s name on top of
every page in the exam.
Exam format: There are ten short answers (4-7 points each, total of 45 points), and five
problems (10-15 points each, total of 60 points). The whole exam is worth 100 points.
You have 90 minutes to complete this exam. Please pace yourself, and don’t get bogged
down in any one problem (especially the short answers, where typically you will either
know the right answer, or you won’t). Good luck.
For Grader’s Use Only
Page 2 score (out of 24) ______
Page 3 score (out of 21) ______
Page 4 score (out of 15) ______
Page 5 score (out of 10) ______
Page 6 score (out of 10) ______
Page 7 score (out of 10) ______
Page 8 score (out of 10) ______
TOTAL SCORE (out of 100) _______
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GSI Name ------------------------------------------ MCB 142 second midterm: Molecular Genetics Please write your name, ID, and TA’s name on the top of ALL 8 pages of this exam. Please put all your answers in the spaces below. Please use pen to write your answers. If you use pencil, you cannot request a re-grade. You may not use any calculator, cellular phone, internet connection, playstation, or other electronic device during this exam. Please double check that you have written your name, ID, and TA’s name on top of every page in the exam. Exam format: There are ten short answers (4-7 points each, total of 45 points), and five problems (10-15 points each, total of 60 points). The whole exam is worth 100 points. You have 90 minutes to complete this exam. Please pace yourself, and don’t get bogged down in any one problem (especially the short answers, where typically you will either know the right answer, or you won’t). Good luck. For Grader’s Use Only Page 2 score (out of 24) ______ Page 3 score (out of 21) ______ Page 4 score (out of 15) ______ Page 5 score (out of 10) ______ Page 6 score (out of 10) ______ Page 7 score (out of 10) ______ Page 8 score (out of 10) ______ TOTAL SCORE (out of 100) _______

GSI Name ------------------------------------------

  1. Replication (4 points). DNA replication is “semiconservative.” This means (a) one strand watches Fox News, the other CNN (b) in a given double helix, one strand is inherited directly from the parental helix and the other strand is newly synthesized as its complement (c) both strands are inherited from the parent, but DNA repair removes half of the sequence randomly, keeping (“conserving”) the other half. (d) If a (previously unlabeled) bacteria is grown in the presence of labeled nucleotide precursors, half of all future descendents will have labeled DNA ANSWER: B.
  2. Human genome (4 points). True or False: “Exons make up only a small fraction (<5%) of the human genome.” ANSWER: TRUE
  3. Restriction enzymes (4 points). Restriction enzyme recognition sites are often palindromic because (a) DNA is an antiparallel double helix, so the two strands are the equivalent (b) In a recognition site the number of A’s is the same as the number of T’s, and the number of G’s is the same as the number of C’s. (c) Restriction enzymes typically bind to DNA as dimers (d) Processing an mRNA requires the spliceosome, which “restricts” the genomic contribution to a gene by removing introns ANSWER: C
  4. Start codon (4 points). True or False: “Every mRNA begins with the “start codon” AUG.” ANSWER: FALSE. Only the coding sequence starts with AUG – there is typically additional 5’ untranslated (UTR) sequence
  5. Translation (4 points). Which of the following element(s) are essential for the process of translation from a mature mRNA? For full credit, be sure to identify ALL elements. (a) RNA polymerase (b) DNA polymerase (c) Spliceosome (d) 5’ capping enzyme (e) Ribosome (f) Reverse transcriptase (g) Transfer RNA

GSI Name ------------------------------------------

  1. Replication bubble (7 points). On the diagram of a bidirectional replication bubble below (a) draw arrows to indicate the 5’-to-3’ directionality of all newly synthesized strands ANSWER: You’re given strand direction on the templates; strand direction on the newly synthesized strands are, as always, antiparallel. (b) label each strands as “leading” or “lagging” ANSWER: “leading” strands point into the fork; they get extended as the fork moves “forward”. Lagging strands point away from the fork.

lagging

lagging

leading

leading

GSI Name ------------------------------------------

  1. Restriction digests II (15 points). A linear DNA molecule is subjected to complete restriction digestion by (1) EcoR1 alone, (2) BamHI alone, and (3) both enzymes together. The fragments are run on a gel. Results are shown below. (a) [5 points] How long is the original DNA molecule? ANSWER: all lanes adds up to 10+5+2 = 9+8 = 9+5+2+1 = 17 kb (b) [5 points] How many EcoR1 recognition sites does it have? ANSWER: Since the DNA is given as initially linear, three fragments are generated by two cuts. (c) [5 points] Does the longest EcoR1 fragment contain a BamH1 restriction site? ANSWER: Yes. The longest EcoRI fragment is 10 kb. Since there is no 10 kb fragment in the combined EcoRI/BamHI digest, the 10 kb EcoRI fragment must be cut further by BamH1. EcoR1 BamH1 both 10 kb 9 kb 5 kb 8 kb 2 kb 1 kb

GSI Name ------------------------------------------

  1. Transcription and translation (10 points). The partial sequence

5’ TCTAGCCTGAACTAATGC 3’

3’ AGATCGGACTTGATTACG 5’

is found in the middle of the protein-coding region of a gene from a bacterial genome. That is, the stop codon of the gene is found outside of the sequence shown. (a) Which strand is the mRNA-like strand? Please explain your logic. ANSWER: There are six possible reading frames. As noted, since this is the middle of a protein-coding region, the correct reading frame should not have a stop codon (UGA, UAA, UAG) If top strand were the mRNA-like strand, there are three possible mRNA reading frames as indicated:

5’ UCU AGC CUG AAC UAA UGC 3’

5’ U CUA GCC UGA ACU AAU GC 3’

5’ UC UAG CCU GAA CUA AUG C 3’

but all three frames have a stop codon (underlined). So none of them is the mRNA-like strand. If the bottom strand is mRNA-like, then the three reading frames are (reading as always from 5’ to 3’)

5’ GCA UUA GUU CAG GCU AGA 3’

5’ G CAU UAG UUC AGG CUA GA 3’

5’ GC AUU AGU UCA GGC UAG A 3’

and as noted the bottom two have stop codons. So by elimination the correct answer must be that the bottom strand is mRNA-like.

GSI Name ------------------------------------------ (b) What is the amino acid sequence of the peptide encoded by this (partial) gene? Indicate which is the amino terminal end of the peptide. ANSWER: mRNA is translated from 5’ to 3’, producing a corresponding peptide from the amino terminal to the carboxy terminal.

5’ GCA UUA GUU CAG GCU AGA 3’

Amino Ala Leu Val Gln Ala Ser Carboxy

Note: The genetic code table is found below

GSI Name ------------------------------------------

  • (1,8)
  • (2)
  • (3,4,7) The interpretation is that deletion 5 is likely a deletion in the gene corresponding to complementation group (3,4,7), since it behaves in the same way as these point mutatios. Deletion 6 is somewhat more complicated – it is not complemented by the genes corresponding to BOTH (1,8) AND (3,4,7). It is therefore a deletion that involves BOTH of these genes. This explains why it appears in two group in part (a).

GSI Name ------------------------------------------

  1. Human variation [10 points]. A polymorphic microsatellite locus is PCR amplified in four individuals from the same family, and the resulting products are run on a gel. The alleles are labeled “1”, “2” , “3”, and “4” from longest to shortest. (a) (5 points). From which parent did “Kid1” inherit allele 2? ANSWER: Mom is heterozygous, carring alleles 1 and 4; similarly, Dad has alleles 2 and 3. Kid1 has alleles 2 and 4, and so got the “2” allele from Dad. (b) (5 points). Kid1 and Kid2 don’t share any alleles. Can we infer that Kid2 is not related to Kid1? Why or why not? ANSWER: Kid1 and Kid2 have genotypes at this locus that are consistent with Mom and Dad being the parents of both. So we can’t rule out that they’re related. But its also possible that they had different parents – only one locus is not enough to

Mom Dad Kid1 Kid