BIO 325 Final

Human DNA replication occurs at a rate of 2,500 bp/minute per fork. The largest human chromosome 1 has 250 MBP (megabasepair = 10^6 bp) of DNA. If chromosome 1 has 100 origins of replication that initiate simultaneously, how long, in hours, will it take to complete replication of chromosome 1?

(250 x 10^6 bp / 2,500 bp/min) x (1/2) = 50,000 minutes per origin = 833 hours per origin / 100 origin = 8.3 hours

Although they are rare, mistakes do occur during DNA replication as DNA polymerase adds a wrong nucleotide to the growing chain. In vivo (cells), the error rate of DNA polymerase is one mistake in every 10^9 bases copied. How many new mutations per division will accumulate in the human genome, which is 6 GBP (gigabasepair =10^9 bp)? What DNA repair mechanism would correct an error made by DNA polymerase that failed to proofread?

(6 x 10^9 nt) x (1 mutation / 10^9 nt) = 6 new mutations per division Mismatch Repair (MMR)

Histone modification is also involved in epigenetic regulation of the LCT gene. Name one histone modifying enzyme recruited by the activator, and describe its effect on the histone tails and the chromatin structure.

- HAT acetylates histone tails and opens up chromatin or - HMT methylates histone tails and opens up chromatin or -HDM demethylates histone tails and opens up chromatin

In humans, the ability to digest lactose, the sugar in milk, declines after weaning because of decreasing levels of the enzyme lactase encoded by the LCT gene; this condition is knows as the lactase non persistent (LNP) trait. However, some individuals maintain high levels of lactase and are able to digest lactase into adulthood; this condition is known as the lactase-persistent (LP) trait. Although the change in LCT gene expression is not completely understood, it involves both transcriptional and post transcriptional regulation. What method would you employ to distinguish between the three genotypes at the -13910 locus: TT, CT and CC?

- PCR followed by sequencing or - Whole genome or gene sequencing and comparison to RefSeq or - DNA or SNP microarray to identify the genotype at the locus

MYOC gene provides instructions for producing a protein called myocillin. Although the exact function of myocillin in the eye is unknown, the mutant allele makes a form of protein that folds incorrectly and then accumulates abnormally in tiny canals through which eye fluid normally drains into the bloodstream, thereby causing glaucoma. One mutant allele of MYOC is sufficient to cause juvenile glaucoma, but both mutant alleles of CYP are required to cause juvenile glaucoma. In a test cross with a dihybrid, what is the expected proportion of offspring with juvenile glaucoma?

1/4 CcMm glaucoma 1/4 ccMm glaucoma 1/4 Ccmm normal 1/4 ccmm glaucoma 3/4 Juvenile Glaucoma

As the RNA primer is removed from the end of a new strand, DNA becomes shorter by the primer's length, and the chromosome becomes shorter and shorter over successive rounds of replication. Telomerase counteracts the loss of DNA by extending telomeres found at both ends of linear DNA. At which end of template DNA, 5' or 3', does telomerase function? And what type of chemical bond is formed during the process?

3' end phosphodiester bond is formed

If the palindromic restriction site used for cloning is 6 bp (base pairs) in length, and only the first three bases are known in the sequence 5' GCC___ 3', what are the last three bases of the sequence? What type of chemical bond is broken by restriction enzymes?

5' GGC 3' phosphodiester bond

BRCA1 mRNA, which is 7,388 nucleotides (nt) in length, produces BRCA1 protein with 1863 amino acids. What is the length of the UTRs (untranslated regions) in nt?

7,388 - (1863 x 3) = 1799 nt (or 1796 nt excluding the stop codon)

In corn, three autosomal genes are on the same chromosome. a and b are 20 m.u. apart, b and c are 5 m.u. apart, and a and c are 25 m.u. apart. A plant with a+ b c+ / a b+ c genotype is crossed to a b c / a b c plant. The wild type allele (denoted by +) is dominant to the mutant allele for each gene. What proportion of the progeny will have the wild type phenotype (a+ b+ c+)? A. 0.5% B. 1% C. 5% D. 20% E. 25%

A. 0.5%

Hachimoji DNA is an artificial DNA made of eight building blocks. IN addition to the four nucleotides found in standard DNA (A, T, G, C), hajimoji DNA contains four additional nucleotides that base pair through hydrogen bonding: P pairs with Z, and B pairs with S. If a molecule of hachimoji DNA has 10% A, 10% P and 20% B, what is the proportion of G bases? A. 10% B. 15% C. 20% D. 30% E. 60%

A. 10%

Diploid oranges have 18 chromosomes. Diploid bananas have 22 chromosomes. What would be the somatic chromosome number of a trisomic variant of oranges and bananas, respectively? A. 19; 23 B. 21; 25 C. 27; 33 D. 54; 66

A. 19; 23

You perform a cross between a black bodied male fly with normal wings and a brown bodied female fly with miniature wings. The F1 generation consisted of brown-bodied males and females, but all males had miniature wings and all females had normal wings. When F1 males and females were crossed F2 flies belonged to four different phenotypic categories in both male and female progeny in the following ratio: brown body with normal wings, black body with normal wings, brown body with miniature wings, and black body with miniature wings in 3:1:3:1 ratio. Which gene is autosomal, and which gene is X-linked? A. Body color- autosomal; wing morphology- X linked B. Body color- X linked; wing morphology- autosomal C. Both genes are autosomal D. Both genes are X linked

A. Body color- autosomal; wing morphology- X linked

You perform a cross between a black bodied male fly with normal wings and a brown bodied female fly with miniature wings. The F1 generation consisted of brown-bodied males and females, but all males had miniature wings and all females had normal wings. When F1 males and females were crossed F2 flies belonged to four different phenotypic categories in both male and female progeny in the following ratio: brown body with normal wings, black body with normal wings, brown body with miniature wings, and black body with miniature wings in 3:1:3:1 ratio. What is the dominant allele of each gene? A. Brown body and normal wings B. Brown body and miniature wings C. Black body and normal wings D. Black body and miniature wings

A. Brown body and normal wings

Fragile X syndrome, one of the most common forms of inherited mental retardation, is caused by a rare, X linked dominant allele. A normal man whose mother is also normal mates with his affected first cousin whose father is normal. The mothers of this man and his mate are sisters. Who passed on the disease allele to the affected woman who married her first cousin? A. Grandmother B. Grandfather C. Both grandparents D. Either grandparent

A. Grandmother

Angelina Jolie's mother and maternal grandmother died of ovarian cancer. Her maternal aunt died of breast cancer. Angelina Jolie opted to have a blood test to see if she had a BRCA1 mutation and found that one copy of the gene was wild-type, and the other copy had a nonsense mutation that would cause production of a nonfunctional protein. What is the genotype of the non-tumor cells of her mother, maternal aunt, and maternal grandmother? A. Heterozygous B. Homozygous mutant C. Homozygous Wild Type

A. Heterozygous

Which of the following descriptions of a reciprocal translation is true? A. It can be caused by double stranded breaks or aberrant crossing over B. Fertility of homozygotes is reduced C. Translocation chromosomes in heterozygotes pair normally during meiosis D. Homozygotes usually do not survive E. Translocation would change the total number of genes

A. It can be caused by double stranded breaks or aberrant crossing over

What would happen if the pol gene (with reverse transcriptase and endonuclease activity) of a retrotransposon was nonfunctional but the long terminal repeats were wild type? A. It would only move in the pretense of a related retrotransposon that provides a functional pol enzyme B. It would be defective in transposition even in the presence of a functional pol enzyme C. It could allow mobilization of other related retrotransposons D. It could move on its own

A. It would only move in the pretense of a related retrotransposon that provides a functional pol enzyme

In cats, the dominant big O allele of the X linked orange gene is required to produce orange fur; the recessive little o allele of this gene yields black fur. Tortoiseshell cats have coats with patches of orange fur alternating with patches of black fur. In a cross between a female tortoiseshell cat and a male black cat, which of the following events could produce an XXY male tortoiseshell cat? A. Nondisjunction in meiosis 1 of either the female or male cat B. Nondisjunction in meiosis 2 of either the female or male cat C. Nondisjunction in either meiosis 1 or meiosis 2 of the female cat D. Nondisjunction in either meiosis 1 or meiosis 2 of the male cat

A. Nondisjunction in meiosis 1 of either the female or male cat

Which chromosome undergoes nondisjunction or loss to produce a transformed daughter cell? A. RB+ Chromosome B. RB- Chromosome

A. RB+ Chromosome

One of the following sequences was obtained from a cloned piece of genome that includes parts of two exons of a gene. The other sequence was obtained from the corresponding part of a cDNA clone. Nucleotide sequences in lower case are unique to sequence A. Sequence A: 5' CACCTGTTGAAGCAAGgtaagaatgaagcattggagcatactgttctttttccttttcctatcttaaacatacattttttaaatgtgcagGAAGAAGCTCCATGGGCACTGGTCTCAGTGGTGGGAAACGTCCTAG 3' Sequence B: 5' CACCTGTTGAAGCAAGGAAGAAGCTCCATGGGCACTGGTCTCAGTGGTGGGAAACGTCCTAG 3' Which sequence is derived from genomic DNA? A. Sequence A B. Sequence B

A. Sequence A

Angelina Jolie's mother and maternal grandmother died of ovarian cancer. Her maternal aunt died of breast cancer. Angelina Jolie opted to have a blood test to see if she had a BRCA1 mutation and found that one copy of the gene was wild-type, and the other copy had a nonsense mutation that would cause production of a nonfunctional protein. What is the probability that Angelina Jolie's biological children are carriers of her BRCA1 mutation? (BRCA1 is a gene on chromosome 17.) A. They have 50% chance of being carriers. B. They are all carriers. C. None of them can be a carrier.

A. They have 50% chance of being carriers.

Messleson and Stahl relied on equilibrium density gradient centrifugation to resolve the DNA containing 14N from the DNA containing 15N. They started off with DNA in media containing the heavy isotope, and then switched to media containing the lighter isotope for DNA replication to take place. After three rounds of semiconservative replication, how many bands of DNA would you see and which band would contain more DNA molecules? A. Two bands (one intermediate, one light) - the light band contains more DNA molecules. B. Two bands (one intermediate, one light) - the intermediate band contains more DNA molecules C. Two bands (one intermediate, one heavy) - the heavy band contains more DNA molecules D. Two bands (one intermediate, one heavy) - the intermediate band contains more DNA molecules

A. Two bands (one intermediate, one light) - the light band contains more DNA molecules.

If you want to introduce a missense mutation in the beta-globin gene (GAG>GTG) to make a mouse model of sickle cell anemia, would you use donor DNA for CRISPR-mediated gene editing? A. Yes B. No

A. Yes

Huntington disease is a dominant disease caused by expansion of the trinucleotide repeat region of the Htt gene that results in the production of a Huntingtin protein with an expanded number of glutamines. An animal model with most features of this syndrome could be created by A. randomly inserting a transgene containing a disease-causing mutant allele of the Htt gene to a mouse or primate genome. B. knocking in a wild-type copy of the Htt gene to a mouse or primate genome. C. knocking out both copies of the wild-type Htt gene from a mouse or primate genome D. randomly inserting a transgene containing a wild-type allele of the Htt gene to a mouse or primate genome. E. knocking out one copy of the wild-type Htt gene from a mouse or primate genome.

A. randomly inserting a transgene containing a disease-causing mutant allele of the Htt gene to a mouse or primate genome.

Five to ten percent of all breast cancers, which can affect both females and males (to a lesser extent), have inactivating mutations in genes BRCA1 or BRCA2, which are important players in DNA double-strand break repair. Spectral karyotyping (a chromosome "painting" technique) of BRCA1- and BRCA2-mutant cells reveals gross chromosomal rearrangements such as translocations and deletions, as well as fusions. Is the BRCA gene a tumor suppressor or a proto-oncogene? A. tumor supressor B. Proto-oncogene

A. tumor supressor

See the following list of eukaryotic elements for a gene that has a single intron: Intron; 3' UTR; Promoter; Stop codon; Exon 1; Start Codon; Exon 2; 5' Cap; Poly-A tail; 5' UTR How many of these element(s) is/are absent from both the mRNA and primary transcript? A. 0 B. 1 C. 2 D. 3 E. 4

B. 1

See the following list of eukaryotic elements for a gene that has a single intron: Intron; 3' UTR; Promoter; Stop codon; Exon 1; Start Codon; Exon 2; 5' Cap; Poly-A tail; 5' UTR How many of these element(s) is/are present in the primary transcript but absent from the mRNA? A. 0 B. 1 C. 2 D. 3 E. 4

B. 1

Sequence B: 5' CACCTGTTGAAGCAAGGAAGAAGCTCCATGGGCACTGGTCTCAGTGGTGGGAAACGTCCTAG 3' How many possible open reading frames that extend through the sequence B, which is the RNA like strand, exist? A. 0 B. 1 C. 2 D. 3

B. 1

The XG locus on the human X chromosome has two alleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells, while the recessive XG+ allele does not allow the antigen to appear. The XG locus is 10 m.u. from the STS locus. STS+ allele produces normal activity of the enzyme serious sulfates (STS), while the recessive STS allele results in lack of this enzyme activity. A man with neither XG antigen nor STS enzyme has a normal daughter with both XG antigen and STS enzyme. The daughter is expecting a child. If the child is a son, what is the probability that he will have either the XG antigen or the STS enzyme (but not both)? A. 5% B. 10% C. 45% D. 50% E. 90%

B. 10%

Inherited mutations in BRCA1 or BRCA2 increase the risk of several cancers in addition to breast and ovarian cancer, and inherited breast cancer is associated with younger age of onset compared to non-inherited, sporadic cancer. Which of the following events in a normal cell from an individual inheriting a mutant allele of BRCA1 or BRCA2 could cause the descendants of that cell to grow into a tumor? (Select all that apply.) A. Mitotic recombination that results in gene conversion of the mutant allele to the wild-type allele B. A point mutation in the allele inherited from the normal parent C. A deletion of the gene copy inherited from the normal parent D. A deletion of the gene copy inherited from the affected parent E. Mitotic nondisjunction or loss of the chromosome inherited from the affected parent F. A second point mutation in the allele inherited from the affected parent

B. A point mutation in the allele inherited from the normal parent C. A deletion of the gene copy inherited from the normal parent

Most breast and ovarian cancers are sporadic (not inherited), but some are the result of inherited predisposition, principally due to mutations in genes BRCA1 and BRCA2, which are important players in DNA double-strand break repair. Individuals with breast cancer can undergo genetic testing to find if s/he inherited one of the mutant BRCA alleles or not. Which of these results would support a sporadic origin of breast cancer? (Assume that there is no additional cancer.) A. Both blood cells and tumor cells are homozygous wild-type. B. Blood cells are homozygous wild-type and tumor cells are homozygous mutant. C. Blood cells are heterozygous and tumor cells are homozygous mutant. D. Both blood cells and tumor cells are homozygous mutant. E. Blood cells are homozygous wild-type and tumor cells are heterozygous.

B. Blood cells are homozygous wild-type and tumor cells are homozygous mutant.

How could you make a genomic library useful for sequencing an entire genome? A. Fragment the genomic DNA prepared from a single cell by partial restriction enzyme B. Fragment the genomic DNA prepared from many cells by partial restriction enzyme C. Fragment the genomic DNA prepared from many cells by complete restriction enzyme D. Either A or B E. Either B or C

B. Fragment the genomic DNA prepared from many cells by partial restriction enzyme

You want to test a hydroxylating agent that adds an -OH group for mutagenicity by doing the Ames test. Hydroxylamine adds -OH to cytosine, and hydroxylated C pairs with A instead of G. There are different types of his- auxotrophs that you can choose from in order to detect reversion to his+ prototrophs in response to a compound under test. What type of his- mutants do you need to use when testing hydroxyl amine for the Ames test? A. A-T mutant (G-C is wild type) B. G-C mutant (A-T is wild type) C. Insertion or deletion mutants D. Any type of point mutants

B. G-C mutant (A-T is wild type)

The human IGF2 gene is autosomal and maternally imprinted. Two alleles of this gene encode two different forms for the IGF2 protein. One allele encodes a larger 60K blood protein; the other allele encodes a smaller 50K blood protein. In the analysis of blood proteins from a coupled named Bill Sr. and Joan, you find only the 60K protein in Joan's blood and only the 50K protein in Bill's blood. They have 2 children, Jill and Bill Jr. Jill produces only the 50K protein and Bill Jr. produces only the 60K protein. What is the genotype of Bill Sr.? A. He is homozygous B. He is heterozygous for the 2 alleles C. It is inconclusive with this data.

B. He is heterozygous for the 2 alleles

p53 is a transcription factor whose function is lost in many cancers. MDM2 binds to the transcriptional activation domain of p53 to inhibit its function; MDM2 also covalently adds ubiquitin, a small protein that causes proteasome-mediated degradation, to p53. Is MDM2 a tumor suppressor or a proto-oncogene? A. tumor supressor B. Proto-oncogene

B. Proto-oncogene

Who is Dolly's genetic mom for her chromosomal DNA? A. Egg Donor B. Somatic Nuclear Donor C. Surrogate Mom

B. Somatic Nuclear Donor

The following linear DNA molecule contains two EcoRI restriction sites. EcoRI cuts between the first G and the second A in 5'G^AATTC3'. 5' AGATGAATTCGCTGAAGAACCAAGAATTCGATT 3' 3' TCTACTTAAGCGACTTCTTGGTTCTTAAGCTTA 5' The digested sequence above is then inserted into a plasmid vector, whose polylinker region is shown below. The entire vector contains one EcoRI and one BamHI site. BamHI cuts between the first G and the second G in 5'G^GATCC3' 5'...CGGATCCCCTAAGATGAATTCCGCGCGCATCGGC...3' 3'...GCCTAGGGGATTCTACTTAAGGCGCGCGTAGCCG...5' Will the EcoRI or BamHI digestion make sticky ends (with what overhangs) or blunt ends? A. Sticky ends with 3' overhangs B. Sticky ends with 5' overhangs C. Blunt ends

B. Sticky ends with 5' overhangs

Cantu syndrome is a rare condition characterized by excess hair growth (hypertrichosis), a distinctive facial appearance, heart defects, and several other abnormalities. Cantu syndrome results from mutations in the ABCC9 gene, which encodes a subunit of potassium ion channel. How is it possible for mutation in one gene to cause a number of distinct phenotypes in an individual? A. There are multiple mutant alleles of the ABCC9 gene, and they affect the function of the potassium channel differently- some mutations are more severe than other. B. The ABCC9 gene product functions in multiple tissues and in different biochemical processes. C. To make a functional potassium ion channel, it requires to have both the ABCC9 gene and another gene that encodes a different subunit. D. Environmental factors can influence whether a mutant genotype displays the expected phenotype

B. The ABCC9 gene product functions in multiple tissues and in different biochemical processes.

Diploid oranges have 18 chromosomes. Diploid bananas have 22 chromosomes. What fruit has the greatest chance to produce unbalanced gametes? A. Diploid bananas B. Triploid bananas C. Diploid oranges D. Triploid oranges

B. Triploid bananas

The following linear DNA molecule contains two EcoRI restriction sites. EcoRI cuts between the first G and the second A in 5'G^AATTC3'. 5' AGATGAATTCGCTGAAGAACCAAGAATTCGATT 3' 3' TCTACTTAAGCGACTTCTTGGTTCTTAAGCTTA 5' The digested sequence above is then inserted into a plasmid vector, whose polylinker region is shown below. The entire vector contains one EcoRI and one BamHI site. BamHI cuts between the first G and the second G in 5'G^GATCC3' 5'...CGGATCCCCTAAGATGAATTCCGCGCGCATCGGC...3' 3'...GCCTAGGGGATTCTACTTAAGGCGCGCGTAGCCG...5' How many DNA fragment(s) would be produced when you cut the recombinant DNA with EcoRI or BamHI? A. 1 EcoRI fragment; 1 BamHI fragment B. 2 EcoRI fragments; 2 BamHI fragments C. 2 EcoRI fragments; 1 BamHI fragment D. 1 EcoRI fragment; 2 BamHI fragments

C. 2 EcoRI fragments; 1 BamHI fragment

How may pairs of PCR primers are needed to amplify three SSR loci? A. 1 pair B. 2 pairs C. 3 pairs D. 6 pairs

C. 3 pairs

Sequence A: 5' CACCTGTTGAAGCAAGgtaagaatgaagcattggagcatactgttctttttccttttcctatcttaaacatacattttttaaatgtgcagGAAGAAGCTCCATGGGCACTGGTCTCAGTGGTGGGAAACGTCCTAG 3' Which of the following sets of primers could you use to amplify the entire target DNA sequence A shown above? (The sequence only shows one of the two strands of the DNA molecule) A. 5' CACCTGTTGA 3' and 5' AACGTCCTAG 3' B. 3' GTGGACAACT 5' and 5' AACGTCCTAG 3' C. 5' CACCTGTTGA 3' and 3' TTGCAGGATC 3' D. 3' GTGGACAACT 5' and 3' TTGCAGGATC 3'

C. 5' CACCTGTTGA 3' and 3' TTGCAGGATC 3'

In radishes, color and shape are each controlled by a single locus with two incompletely dominant alleles. Color may be red (R1R1), purple (R1R2), or white (R2R2), and shape can be long (L1L1), oval (L1L2), or round (L2L2). How many different phenotypes do you expect to see among the offspring of a cross between two plants heterozygous at both loci? A. 3 B. 6 C. 9 D. 12

C. 9

In corn, three autosomal genes are on the same chromosome. a and b are 20 m.u. apart, b and c are 5 m.u. apart, and a and c are 25 m.u. apart. A plant with a+ b c+ / a b+ c genotype is crossed to a b c / a b c plant. The wild type allele (denoted by +) is dominant to the mutant allele for each gene. What proportion of the progeny will be wild type for a and b but mutant for c (a+ b+ c)? A. 4% B. 5% C. 9.5% D. 19% E. 20%

C. 9.5%

p53 responds to diverse stress signals by activating transcription of genes whose protein products orchestrate specific cellular responses, including transient cell cycle arrest (through p21), DNA repair, and apoptosis. p21, one of the genes activated by p53, is a CDK inhibitor that blocks the activity of a particular CDK-cyclin complex, which phosphorylates Rb and allows the entry of a cell into the S phase past the checkpoint. Which of the following is most likely to occur from inactivation of p53? (Select two that apply.) A. Cells arrest in G1 phase due to decreased phosphorylation of Rb B. Cells undergo apoptosis due to accumulation of DNA damage C. Cells are defective in the G1-to-S checkpoint due to increased phosphorylation of Rb D. Cells have a high number of point mutations and fragments of chromosomal DNA lacking telomeres and a centromere

C. Cells are defective in the G1-to-S checkpoint due to increased phosphorylation of Rb D. Cells have a high number of point mutations and fragments of chromosomal DNA lacking telomeres and a centromere

Which one of the following events is unlikely to be associated with cancer? A. Point mutation that enhances the function of a proto-oncogene B. A chromosomal translocation with a breakpoint near a proto-oncogene, now fused to a strong enhancer that activates expression C. Deletion of a proto-oncogene D. Mitotic nondisjunction in a cell heterozygous for deletion of a tumor-suppressor gene E. Mitotic recombination in a cell heterozygous for deletion of a tumor-suppressor gene

C. Deletion of a proto-oncogene

Tobacco mosaic virus (TMV) that infects tobacco plants is made up of RNA and proteins. Which one is NOT an ideal and informative experiment designed to test whether RNA or protein acts as the hereditary material in TMV? A. Use 32P and 35S isotopes to selectively label RNA and proteins, respectively, and observe which isotope enters plant cells to program viral production. B. Mutate RNA only or protein only to test which mutation can be passed down to progeny via viral replication. C. Determine which molecule has greater potential for diversity and stability to serve as the genetic material. D. Reconstitute TMV in vitro by mixing purified proteins and RNA that originate from two different TMV variants and are distinguishable. The molecule that directs synthesis of the other is most likely the genetic material. E. Add RNase or protease enzyme to the TMV extract to test which enzyme inhibits transduction of tobacco plants

C. Determine which molecule has greater potential for diversity and stability to serve as the genetic material.

The human IGF2 gene is autosomal and maternally imprinted. Two alleles of this gene encode two different forms for the IGF2 protein. One allele encodes a larger 60K blood protein; the other allele encodes a smaller 50K blood protein. In the analysis of blood proteins from a coupled named Bill Sr. and Joan, you find only the 60K protein in Joan's blood and only the 50K protein in Bill's blood. They have 2 children, Jill and Bill Jr. Jill produces only the 50K protein and Bill Jr. produces only the 60K protein. If Bill Jr. is heterozygous, what must be the genotype of his sister Jill? A. She is homozygous B. She is heterozygous for the 2 alleles C. It is inconclusive with this data.

C. It is inconclusive with this data.

One problem associated with CRISPR/Cas9 technology is off-target effects. Part of the reason is that single base-pair mismatches between the target site and the sgRNA in the 20 bp DNA/RNA hybrid do not prevent Cas9 cleavage of the target site. What is the best strategy to use in order to specifically target the mutant Htt allele that causes Huntington disease without off-target effects? A. Design sgRNA that can base pair with several regions in the genome, including the Htt locus, to ensure efficient cleavage of the mutant allele. B. Use multiple sgRNAs, each of which has one mismatch with the target sequence at different nucleotide positions. C. Make sure that no other sequences in the genome are identical or too similar to the 20 bp sequence involved in base pairing with the sgRNA. D. Use mutant Cas 9 enzyme with decreased target specificity. E. Look for a unique target sequence in the Htt gene that lacks a PAM site nearby.

C. Make sure that no other sequences in the genome are identical or too similar to the 20 bp sequence involved in base pairing with the sgRNA.

Trisomy 21 results in Down Syndrome Which of these events cannot produce Down syndrome individuals or mosaics? A. Nondisjunction of chromosome 21 during first meiotic division B. Nondisjunction of chromosome 21 during second meiotic division C. Mitotic loss of chromosome 21 D. Mitotic nondisjunction of chromosome 21 E. All answer choices can produce Down syndrome individuals or mosaics

C. Mitotic loss of chromosome 21

The MAP kinase pathway is activated by ligand binding to a transmembrane receptor, which switches Ras on via GEF (Guanine nucleotide Exchange Factor) that counteracts GAP (GTPase Activating Protein) activity. Ras-GTP in turn activates the downstream signaling intermediates (kinases), which are activated by phosphorylation. Which drug CANNOT be used to target cancer cells that have elevated receptor activation? A. Small molecule inhibitors that bind the ATP-binding pocket of kinases B. Antibodies specific for ligand to prevent receptor binding C. Small molecule inhibitors that interfere with GAP activity D. Small molecule inhibitors that interfere with GEF activity E. Antibodies that block the ligand-binding domain of receptor

C. Small molecule inhibitors that interfere with GAP activity

As discussed in lecture, CRISPR/Cas9 system evolved in bacteria to provide anti-viral immunity. Bacterial genome has the CRISPR locus that produces short CRISPR RNAs, but it does not contain any PAM sites. However, the bacteriophage genome sequence targeted by CRISPR does. How is this fact crucial to keeping the bacterial genome intact and selectively destroying the phage genome? (Select two that apply.) A. The presence of the PAM site in the target sequence of the bacteriophage genome prevents the bacterial chromosome from being cleaved by Cas9. B. The absence of the PAM site in the bacterial CRISPR locus causes the phage chromosome to be cleaved by Cas9. C. The absence of the PAM site in the bacterial CRISPR locus prevents the bacterial chromosome from being cleaved by Cas9. D. The presence of the PAM site in the target sequence of the bacteriophage genome causes the phage chromosome to be cleaved by Cas9.

C. The absence of the PAM site in the bacterial CRISPR locus prevents the bacterial chromosome from being cleaved by Cas9. D. The presence of the PAM site in the target sequence of the bacteriophage genome causes the phage chromosome to be cleaved by Cas9.

What is the relationship between chromosomal instability and cell proliferation characteristic of tumorigenesis? (Select two that apply.) A. The higher the frequency of cell division, the smaller the chance for genomic instability because of activation of DNA repair. B. There is no relationship between chromosomal instability and cell proliferation rate. C. The greater the chromosomal instability, the higher the probability of increased cell proliferation due to driver mutations. D. The greater the chromosomal instability, the higher the probability of increased cell death due to apoptosis. E. The higher the frequency of cell division, the greater the chance for genomic instability because of more opportunities to create mutations.

C. The greater the chromosomal instability, the higher the probability of increased cell proliferation due to driver mutations. E. The higher the frequency of cell division, the greater the chance for genomic instability because of more opportunities to create mutations.

Messleson and Stahl relied on equilibrium density gradient centrifugation to resolve the DNA containing 14N from the DNA containing 15N. They started off with DNA in media containing the heavy isotope, and then switched to media containing the lighter isotope for DNA replication to take place. How would the result change if they started their experiment with media containing the light isotope and then switched to the media containing the heavier isotope? A. Two bands (one intermediate, one light) - the light band contains more DNA molecules. B. Two bands (one intermediate, one light) - the intermediate band contains more DNA molecules C. Two bands (one intermediate, one heavy) - the heavy band contains more DNA molecules D. Two bands (one intermediate, one heavy) - the intermediate band contains more DNA molecules

C. Two bands (one intermediate, one heavy) - the heavy band contains more DNA molecules

The XG locus on the human X chromosome has two alleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells, while the recessive XG+ allele does not allow the antigen to appear. The XG locus is 10 m.u. from the STS locus. STS+ allele produces normal activity of the enzyme serious sulfates (STS), while the recessive STS allele results in lack of this enzyme activity. A man with neither XG antigen nor STS enzyme has a normal daughter with both XG antigen and STS enzyme. The daughter is expecting a child. What is the daughter's genotype? A. XG+ STS+ / XG+ STS+ B. XG STS / XG STS C. XG+ STS+ / XG STS D. XG+ STS / XG STS+

C. XG+ STS+ / XG STS

The three genotypes (TT, CC and CT) also correlate with differential methylation of CpG dinucleotides in the vicinity. Considering the epigenetic control of the LCT gene, which genotype would display the highest DNA methylation levels in the region near -13910? And would the transcriptional activator promote or prevent the recruitment of the DNA methyltransferase enzyme that methylates CpG dinucleotides?

CC genotype activator would prevent the recruitment of the DNA methlytransferase

MYOC gene provides instructions for producing a protein called myocillin. Although the exact function of myocillin in the eye is unknown, the mutant allele makes a form of protein that folds incorrectly and then accumulates abnormally in tiny canals through which eye fluid normally drains into the bloodstream, thereby causing glaucoma. One mutant allele of MYOC is sufficient to cause juvenile glaucoma, but both mutant alleles of CYP are required to cause juvenile glaucoma. How can you explain the mode of inheritance for juvenile glaucoma caused by CYP or MYOC mutation based on the information on the mutant alleles? In other words, explain the phenotype of a heterozygote in relation to the molecular basis of mutation and the allele interaction for each gene individually

CYP heterozygotes are phenotypically normal because one functional copy makes enough enzyme to metabolize estradiol and prevent glaucoma. MYOC heterozygotes are phenotypically abnormal because, presumably, the mutant protein can accumulate even in the presence of the normal protein and cause glaucoma dominantly.

Juvenile glaucoma can arise due to mutations in one of two genes A and B. Either the dominant allele of gene A or the recessive allele of gene B can cause glaucoma. Having both the dominant allele of gene A and the recessive allele of gene B (A_bb) results in a phenotype with more severe than the phenotype caused by either one alone. A test cross is performed with double heterozygote AB/ab. If A and B assort independently, what is the proportion of test cross progeny for each of the three possible phenotypes: normal, glaucoma, and severe glaucoma?

Cross AB/ab x ab/ab 25% AB/ab glaucoma 25% aB/ab normal 25% Ab/ab severe glaucoma 25% ab/ab glaucoma 50% glaucoma; 25% normal; 25% severe glaucoma

Juvenile glaucoma can arise due to mutations in one of two genes A and B. Either the dominant allele of gene A or the recessive allele of gene B can cause glaucoma. Having both the dominant allele of gene A and the recessive allele of gene B (A_bb) results in a phenotype with more severe than the phenotype caused by either one alone. A test cross is performed with double heterozygote AB/ab. If A and B are 20 m.u. apart, what is the proportion of test cross progeny for each of the three phenotypes?

Cross AB/ab x ab/ab 40% AB/ab glaucoma 10% aB/ab normal 10% Ab/ab severe glaucoma 40% ab/ab glaucoma 80% glaucoma, 10% normal, 10% severe glaucoma

CYP gene encodes an enzyme that belongs to cytochrome P450 family and is involved in estradiol metabolism. Mutant CYP enzyme is defective in converting estradiol to 4-hydroxy estradiol. One wild type copy of CYP gene is sufficient to break down estradiol to 4-hydroxy estradiol. A different enzyme called aromatase produces estradiol from testosterone. Juvenile glaucoma is associated with increase in estradiol level. Assume normal phenotype for individuals unable to make estradiol. -------------(Aromatase)---------- Testosterone ----------> Estradiol (CYP) -------> 4-OH- Estradiol IN a dihybrid cross (a cross. between heterozygotes for both CYP and aromatase), what is the expected proportion of offspring with juvenile glaucoma?

Cross: AaCc x AaCc 9/16: A_C_ normal 3/16: aaC_ normal 3/16: A_cc glaucoma 1/16: aacc normal 3/16: Juvenile Glaucoma

Fragile X syndrome, one of the most common forms of inherited mental retardation, is caused by a rare, X linked dominant allele. A normal man whose mother is also normal mates with his affected first cousin whose father is normal. The mothers of this man and his mate are sisters. What is the probability that the couples first daughter will be affected? A. 1 B. 1/8 C. 1/4 D. 1/2 E. 0

D. 1/2

Cantu syndrome is a rare condition characterized by excess hair growth (hypertrichosis), a distinctive facial appearance, heart defects, and several other abnormalities. Cantu syndrome results from mutations in the ABCC9 gene, which encodes a subunit of potassium ion channel. Having one copy of the ABCC9 mutant allele results in Cantu syndrome. Having two copies of the mutant allele results in stillbirth (not born). If two people with Cantu syndrome have a child together, what is the probability that their child will also have the trait? (Note that only live progeny are considered) A. 0 B. 1/4 C. 1/2 D. 2/3 E. 3/4

D. 2/3

Homozygosity for extremely rare mutations in a human gene called SCN9A causes complete insensitivity to pain and a total lack of the sense of smell. The gene encodes a 1977-amino acid protein. Different point mutations in the SCN9A gene were identified in three families with the condition. IN family one, a G to A substation in Exon 15 yields a 898-amino acid protein. In family two, deletions of a single base results in a 766-amino acid protein. In family 3, a C to G substitution in exon 10 yields a 458 amino acid protein. What codon is mutated in family 1? A. 5' UAC 3' B. 5' UCA 3' C. 5' UAA 3' D. 5' UGG 3'

D. 5' UGG 3'

A modified uridine xo5U in the wobble position can base pair with A, G, U, or C. What codon(s) can base pair with the tRNA with anticodon 5'xo5UAC3' ? A. 5'AUG3' B. 5'UUG3' C. 5'CUG3' D. 5'GUG3' E. All of the above

D. 5'GUG3'

Trisomy 21 results in Down Syndrome What would be the number of FISH signals for a gene located on chromosome 21 in a somatic cell from a Down syndrome individual? Assume the cell is in mitotic metaphase A. 2 B. 3 C. 4 D. 6

D. 6

Homozygosity for extremely rare mutations in a human gene called SCN9A causes complete insensitivity to pain and a total lack of the sense of smell. The gene encodes a 1977-amino acid protein. Different point mutations in the SCN9A gene were identified in three families with the condition. IN family one, a G to A substation in Exon 15 yields a 898-amino acid protein. In family two, deletions of a single base results in a 766-amino acid protein. In family 3, a C to G substitution in exon 10 yields a 458 amino acid protein. What type of mutation produces truncated proteins in these three families? A. Nonsense mutation B. Missense mutation C. Frameshift mutation D. Both A and C E. Both B and C

D. Both A and C

Which enzyme(s) would you need to make a cDNA library? A. RNA polymerase B. Reverse Transcriptase C. DNA ligase D. Both B and C E. A, B, and C

D. Both B and C

Which of these mutations would arrest the cell cycle and prevent progression into the next phase? A. Receptor is active in absence of growth factor B. Cyclin is not degraded C. Cyclin is continuously made D. CDK cannot bind cyclin

D. CDK cannot bind cyclin

Hachimoji DNA is an artificial DNA made of eight building blocks. IN addition to the four nucleotides found in standard DNA (A, T, G, C), hajimoji DNA contains four additional nucleotides that base pair through hydrogen bonding: P pairs with Z, and B pairs with S. For a region of DNA that is 10 base pairs (bp) in length, which DNA has greater potential for diversity and by how much? A. Standard DNA has twice as many unique sentences as hachimoji DNA does B. Standard DNA has 2^10 times as many unique sentences as hachimoji DNA does. C. Hachimoji DNA has twice as many unique sentences as standard DNA does. D. Hachimoji DNA has 2^10 times as many unique sequences as standard DNA does. E. The two types of DNA have equal potential for diversity.

D. Hachimoji DNA has 2^10 times as many unique sequences as standard DNA does.

You want to test a hydroxylating agent that adds an -OH group for mutagenicity by doing the Ames test. Hydroxylamine adds -OH to cytosine, and hydroxylated C pairs with A instead of G. Given that the rate of forward mutation is much higher than the rate of reverse mutation, why does the Ames test use the reversion rate to measure mutation? A. Forward mutations occur in response to environmental change, but reverse mutations occur spontaneously and thus give a more accurate measure of mutation B. Forward mutations can generate many different mutant alleles, but reverse mutations occur rarely to restore the wild type allele. C. It is much easier to screen for his+ revertants by using the minimal medium supplemented with histidine D. It is much easier to select for his+ revertants by using the minimal medium. E. The Ames test detects either forward or reverse mutation

D. It is much easier to select for his+ revertants by using the minimal medium.

What is a feature unique to retroviral vectors and not shared with adeno-associated viral (AAV) vectors? A. Packaging the therapeutic DNA into virus-like particles requires "helper DNA" which expresses all the proteins required to make viral particles. B. Therapeutic gene is transcbried and translated in patient cells using the host transcription and translation machinery. C. Gene therapy needs to be repeated periodically to provide a constant supply of the therapy protein. D. Therapeutic gene integrates randomly into the patient's genome. E. Packaging cells do not make active viruses because helper DNA lacks packaging sequences and therapeutic DNA lacks essential viral genes.

D. Therapeutic gene integrates randomly into the patient's genome.

In cats, the dominant big O allele of the X linked orange gene is required to produce orange fur; the recessive little o allele of this gene yields black fur. Tortoiseshell cats have coats with patches of orange fur alternating with patches of black fur. Which of the following crosses could produce a XX female tortoiseshell cat? A. Orange female x black male B. Black female x orange male C. Orange female x orange male D. Two of the above could produce a female tortoiseshell cat E. All of the above could produce a female tortoiseshell cat.

D. Two of the above could produce a female tortoiseshell cat

Which gene therapy CANNOT be used to treat LOF diseases such as cystic fibrosis? A. Use retroviral vectors to express a wild-type copy. B. Use adeno-associated viral (AAV) vectors to express a wild-type copy. C. Use CRISPR to replace the mutant copy with a wild-type copy. D. Use RNAi to knock down the mutant gene expression.

D. Use RNAi to knock down the mutant gene expression.

Would you expect the DNA sequences of the five species in BLAST figure to have higher or lower degree of similarity than the protein sequence? Explain.

DNA sequences would have lower degree of similarity than the protein sequences because nucleotide sequences that do not affect the protein sequence (e.g. introns, intergenic regions, non-coding exons) and expression (e.g. non -regulatory elements) are less constrained by evolutionary selection pressure and therefore more prone to accumulate mutations over time. On the other hand, protein sequences would be more conserved (higher degree of similarity) for the function to be maintained in different organisms.

In radishes, color and shape are each controlled by a single locus with two incompletely dominant alleles. Color may be red (R1R1), purple (R1R2), or white (R2R2), and shape can be long (L1L1), oval (L1L2), or round (L2L2). What is the proportion of the purple and oval radishes among the offspring of a cross between two plants heterozygous at both loci? A. 9/16 B. 3/16 C. 1/16 D. 1/8 E. 1/4

E. 1/4

Why is it essential to sequence many clones from multiple cDNA libraries to annotate a genome? A. Different genes are transcribed in different tissues B. A gene may be spliced differently in different tissues C. Some genes are transcribed rarely in certain tissues D. Every cDNA library will represent an identical set of transcribed genes and multiple sequencing will support correctness of sequence data E. A, B, and C

E. A, B, and C

A modified uridine xo5U in the wobble position can base pair with A, G, U, or C. If the anticodon is mutated to 5'xoUGC3' after the tRNAA has been charged with the correct amino acid (that is, amino acid for anticodon 5'xoUAC3'), what could happen? A. Val in proteins could be replaced by Ala B. Ala in proteins could be replaced by Val C. Proteins could be misshaped D. A and C E. B and C

E. B and C

Suppose you did not have the cDNA close representing the above sequence. What piece of evidence is likely to indicate a protein coding gene from the genomic DNA sequence? A. The DNA sequence is at least 50% similar to its closest match in mouse, chicken, and zebrafish, and the putative protein sequence is at least 80% similar B. It contains an ORF, which is 60 bp in length C. There are many copies of the sequence found in the same genome, and they have similar function D. Both A and B E. Both A and C

E. Both A and C

What is the best evidence that shows redundancy of the genetic code? A. Single base substitutions change only one amino acid B. A gene's nucleotide sequence is collinear with the corresponding amino acid sequence C. Different point mutations that affect the same amino acid can recombine to give wild type protein D. Artificial messages with different base sequences give rise to different proteins E. Mutations in a codon do not always change the identity of an amino acid

E. Mutations in a codon do not always change the identity of an amino acid

Which of these transgenic techniques relies on homologous recombination? A. Knock out via CRISPR B. Knock in via CRISPR C. Knock out using targeting DNA construct containing an antibiotic resistance gene D. Pronuclear injection E. Two of the above

E. Two of the above

CYP gene encodes an enzyme that belongs to cytochrome P450 family and is involved in estradiol metabolism. Mutant CYP enzyme is defective in converting estradiol to 4-hydroxy estradiol. One wild type copy of CYP gene is sufficient to break down estradiol to 4-hydroxy estradiol. A different enzyme called aromatase produces estradiol from testosterone. Juvenile glaucoma is associated with increase in estradiol level. Assume normal phenotype for individuals unable to make estradiol. ------------(Aromatase) Testosterone ----------> Estradiol (CYP) -------> 4-OH- Estradiol What type of gene interaction is illustrated for CYP and aromatase?

Epistasis

How can gene therapy be used to treat GOF diseases such as Huntington disease? A. Use retroviral vectors to express a wild-type copy. B. Use adeno-associated viral (AAV) vectors to express a wild-type copy. C. Use CRISPR to replace the mutant copy with a wild-type copy. D. Use RNAi to knock down the mutant gene expression. E. Both A and B F. Both C and D

F. Both C and D

Two genes, A and B are antagonistic in function. The functional copy of one gene encodes a protein (protein 1 in the figure below) that produces a compound that protects from glaucoma. The functional copy of another gene encodes a protein (Protein 2 in the figure below) that inhibits the production of this protective compound. For each gene, the nonfunctional allele is recessive to the dominant, functional allele. Which gene, A or B, has a protective function against glaucoma? (note: either the dominant allele of gene A or the recessive allele of gene B can cause glaucoma) PROTEIN 1 Precursor ------->Glaucoma Protective -------------------|Compound PROTEIN 2 -- | PROTEIN 3

Gene B

People who have mutations in BRCA1 or BRCA2 genes have an increases lifetime risk of developing breast and ovarian cancers, as well as other cancer types. These mutations impair one of the two repair pathways for DNA double strand breaks (DSBs). The repair pathway mediated by the BRCA proteins is an accurate method of repairing DSBs. What pathway is this? To what type of radiation are the BRCA mutants most sensitive?

Homologous recombination (HR) X-Rays and other ionizing radiation

The enzyme lactase is expressed exclusively from the intestinal epithelial cells. How would the levels of transcription factors (activators and repressors) that regulate the LCT gene expression compare between the intestinal and non intestinal cells?

Intestinal cells would express activators but little to no repressors. Non-Intestinal cells would express repressors but little to no activators

How do you know CYP is autosomal or X linked?

It is autosomal because a father passes the trait to his sons and a normal father has an affected daughter

How do you know MYOC is autosomal or X linked?

It is autosomal because father does not pass the trait to his daughters

Poly(ADP-ribose) polymerase or PARP enzyme mediates the repair of single strand breaks (SSBs). Unrepaired SSBs can become DSBs during replication. Note that cells, including cancer cells, subject to extensive amount of DNA damage can undergo apoptosis. Would an anti-cancer drug promote or inhibit PARP activity?

It would inhibit PARP

Lactase is a membrane bound glycoprotein and glycosolation is essential for the function of lactase. What would be the phenotype of an individual with loss of function mutation in the glycosylating enzyme: LP or LNP? And is glycosylation an example of transcriptional or post-transcriptional regulation?

LNP post-transcriptional

In most mammals including humans, down regulation of the LCT gene occurs around the time of weaning and results in low lactase activity throughout adult life. If you knock down the expression of the DNA methyltransferase in adults via RNA interference, what would be the phenotype: LP or LNP? and is RNA interference an example of transcriptional or post-transcriptional regulation?

LP post-transcriptional

People who have mutations in BRCA1 or BRCA2 genes have an increases lifetime risk of developing breast and ovarian cancers, as well as other cancer types. These mutations impair one of the two repair pathways for DNA double strand breaks (DSBs). If this accurate repair of DSB is defective, a cell relies on a more error prone mechanism of DSB repair. What pathway is this? Why is it more error prone?

Non-homologous end joining (NHEJ) The repair does not rely on sequence homology so any two broken ends can be joined together. NHEJ can also cause deletion or insertion of one or a few nucleotides during repair.

A third gene C encodes a protein (Protein 3 in the figure below) that prevents protein 2 from functioning. What is the phenotype of an individual who is heterozygous for all three genes A, B and C? PROTEIN 1 Precursor ------->Glaucoma Protective Compound --------| PROTEIN 2 -- | PROTEIN 3

Normal

Cisplatin is a major anti-cancer drug that kills cancer cells by damaging their DNA and inducing apoptosis. In its reactive form, cisplatin covalently binds to DNA bases, particularly purines, forming DNA adducts to distort the helix. What repair mechanism would remove the bulky DNA adducts or lesions generated by cisplatin? What type of radiation can be tolerated by cells resistant to cisplatin?

Nucleotide Excision Repair (NER) UV radiation

The box containing a couple vertical bars illustrates the RNA-Seq expression data for 53 tissues averaged from 570 donors. What is the starting material for generating this data?

RNA or mRNA extracted from each tissue type

Right Side: ATTCTA / TAAGAT Left Side: TCGGCC / AGCCGG Which side, left or right, would require more energy to unwind based on the given sequences? Why?

Right side because it has a higher GC content

In humans, the ability to digest lactose, the sugar in milk, declines after weaning because of decreasing levels of the enzyme lactase encoded by the LCT gene; this condition is knows as the lactase non persistent (LNP) trait. However, some individuals maintain high levels of lactase and are able to digest lactase into adulthood; this condition is known as the lactase-persistent (LP) trait. Although the change in LCT gene expression is not completely understood, it involves both transcriptional and post transcriptional regulation. Many polymorphisms associated with the LP vs. LNP trait have been identified. The key variant associated with the LP trait is the 'T-allele" derived from the ancestral 'C-allele' associated with the LNP trait, which are located 13,910 base Paris upstream of the LCT gene transcription start site. What is the name of this polymorphism?

Single Nucleotide Polymorphism (SNP)

TT homozygotes have high lactase activity, CC homozygotes have a low lactase activity, and CT heterozygotes have intermediate lactase activity. Considering the transcriptional control of the LCT gene, which allele at -13910 is preferentially bound by the activator: C-allele or T-allele?

T-allele

Cisplatin is a major anti-cancer drug that kills cancer cells by damaging their DNA and inducing apoptosis. In its reactive form, cisplatin covalently binds to DNA bases, particularly purines, forming DNA adducts to distort the helix. The damaged strand is cleaved on both sides of the lesion by the endonuclease complex XPF-ERCC1 during repair. Testicular cancer cells have lower levels of the XPF-ERCC1 complex than ovarian cancer cells. Which cancer type would be more sensitive to the cisplatin chemotherapy (i.e. killed more efficiently)?

Testicular cancer cells

Why are there two possible recombinant DNA molecules when you use BamHI for cloning? If you use two different restriction enzymes to cut the insert and the vector (e.g. the insert has a BamHI site at one end and another restriction site, found once in the polylinker, at the other end), how many recombinant DNA type(s) is/are possible?

The insert can go in either orientation because of identical BamHI overhangs at both ends. one recombinant DNA

If the LCT gene is on the right arm of chromosome two, what is the direction of transcription: toward or away from the centromere? And where is the promoter located with respect to the upper part of the figure where exons and introns are depicted: to the left or to the right of the diagram?

Toward the centromere to the right of the diagram

The nucleotides surrounding the -13910 confer both temporal and spatial regulation of the LCT expression. What is the name of the cis-acting regulatory element containing the -13910 locus? And what is the name of the cis-acting regulatory element that limits the influence of this element from reaching wrong genes?

enhancer insulator

Expression of a protein called replication protein A (ROA) is increased in a number of tumors due to mutation. Is this loss of function or gain of function mutation? Does the mutation likely affect the coding sequence or the regulatory sequence of the RPA gene?

gain of function mutation regulatory sequence

Which enzyme would make the replication bubble grow in size overtime by unwinding the two strands of DNA? What type of chemical bond is broken during the process?

helicase hydrogen bond is broken

Cisplatin is a major anti-cancer drug that kills cancer cells by damaging their DNA and inducing apoptosis. In its reactive form, cisplatin covalently binds to DNA bases, particularly purines, forming DNA adducts to distort the helix. A compound that covalently modifies RPA enhances the effect of cisplatin in causing cancer cell death and therefore is a promising anti-cancer drug. Would the compound activate or inhibit RPA activity?

inhibit RPA activity

People who have mutations in BRCA1 or BRCA2 genes have an increases lifetime risk of developing breast and ovarian cancers, as well as other cancer types. These mutations impair one of the two repair pathways for DNA double strand breaks (DSBs). Various point mutations found in tumor cells produce BRCA proteins with reduced or no activity. Are these loss of function mutations or gain of function mutations? Do these mutations likely affect the coding sequence or regulatory sequence of the BRCA gene?

loss of function mutation coding sequence

People who have mutations in BRCA1 or BRCA2 genes have an increases lifetime risk of developing breast and ovarian cancers, as well as other cancer types. These mutations impair one of the two repair pathways for DNA double strand breaks (DSBs). In certain breast cancers, mutations within the 5' UTR of the BRCA gene result in reduced translation of the BRCA protein. Are these loss of function mutations or gain of function mutations? Do these mutations affect the coding sequence or regulatory sequence of the BRCA gene?

loss of function mutation non-coding sequence (regulatory)

Beta-glucosidase is similar to LPH in function in that it releases glucose by breaking covalent bonds, but it uses a different substrate (not lactose). Humans also have the beta-glucosidase enzyme. Would you categorize the human and plant beta-glucosidase genes as orthologs or paralogs? And which protein would have the greater degree of amino acid similarity to the human beta-glucosidase: human LPH or the plant beta-glucosidase?

orthologs plant beta-glucosidase

Suppose you identified a new human gene that shares a high percentage of sequence similarity with the LCT. Would you categorize these two human genes as orthologs or paralogs? And what is the evolutionary mechanism that can explain the origin of these two closely related species?

paralogs gene duplication and divergence through accumulation of mutations

What would be a 10 nt (nucleotide) primer you can use for sequencing the insert below? What is the template strands used for sequencing the primer, top or bottom? 5'AATTTCGGCCX1X2X3X4X5...3' 3'TTAAAGCCGGY1Y2Y3Y4Y5...5'

primer: 5'AATTTCGGCC3' template: bottom strand

In which cell type would you expect to find the shortest telomere fro the list: somatic cells of a toddler, germ cells of a toddler, somatic cells of a 100 year old woman, germ cells of a hundred year old woman?

somatic cells of a 100 year old woman

Which enzyme would relax the supercoils generated during replication? What type of chemical bond is broken during the process?

topoisomerase phospodiesteir bond is broken