BIOL 360 – Which of the following can be a disadvantage

11 Jun BIOL 360 – Which of the following can be a disadvantage

Question
Question 33

Which of the following can be a disadvantage when analyzing human pedigrees?

a) There are many conditions coded by a single locus.

b) It can be difficult to know if a condition traces many generations back in a pedigree.

c) Humans are diploid.

d) We know the human genome sequence.

e) Many human conditions are not lethal.

Question 37

Assume that A and B are two linked genes on an autosome in Drosophila. A testcross is made where AB/ab flies are crossed to ab/ab flies and the progeny are counted and shown below. However, it is known that the Aa/bb genotype is lethal before the flies hatch and does not appear among the testcross progeny counted. What is the most precise map distance that can be calculated from these data?

Aa Bb = 235

aa bb = 225

aa Bb = 20

a) 4.0 cM

b) 8.0 cM

c) 4.2 cM

d) 50 cM

e) 16.4 cM

Question 42

Which of the following is a more likely outcome of a dihybrid cross between incompletely linked loci with full dominance ?

a) 100% of the progeny is identical because the loci are linked

b) 6:4:4:2 ratio.

c) a 1:2:1 ratio.

d) a 9:3:3:1 ratio.

Question 43

Linked genes can occasionally segregate from each other, due to the phenomenon known as_____

a) mutation.

b) complete linkage.

c) independent chromosomal segregation.

d) genetic repulsion.

e) genetic recombination.

Question 44

Which of the following best describes linked genes?

a) they fail to assort independently.

b) they are physically continuous in DNA.

c) they are distributed into gametes independently from one another.

d) they are actually the same gene

Question 45

Two loci in separate chromosomes are not linked because________

a) homologous chromosomes segregate independently in meiosis I.

b) there cannot be crossing over between non-homologous chromosomes.

c) all loci in one chromosome are linked to each other, but not between separate chromosomes.

d) they always segregate together into gametes.

Question 47

You identify an RFLP in mice by digesting genomic DNA with HindIII enzyme and radiolabeling a piece of probe DNA. Southern blot analysis shows that the probe detects a 2kb fragment in one strain of mice and a 4-kb fragment in another strain of mice. The two mice strains are crossed to produce an F1 generation. Two F1 siblings are mated to produce a dozen F2 progeny. You isolate genomic DNA from several F1 and F2 individuals, digest with HindIII, and perform a Southern blot with the same probe. How many bands would be seen in the Southern blot of an F1 individual?

a) 2

b) 3

c) 1 or 2

d) only 1

e) 4

Question 48

Which of the following provides an INACCURATE description of one of the basic components required for a bacterial cloning vector?

a) A bacterial origin of replication ensures that the plasmid is replicated while present within the bacterial cell.

b) Selectable markers provide a means for selecting cells that have been transformed with a recombinant plasmid.

c) Unique restriction enzyme sites provide a means for inserting the foreign DNA into the cloning vector at a specific, known sequence site.

d) Selectable markers provide a means for preferentially allowing growth of only those bacterial cells that have been transformed with the cloning vector.

e) Unique restriction enzyme sites allow for larger pieces of foreign DNA to be inserted into the bacterial cloning vector – smaller fragments insert spontaneously.

Question 49

Which of the following best describes knockout mice?

a) They have lower expression levels of a gene of interest.

b) They have higher expression levels of a gene of interest.

c) They have a point mutation in the gene of interest.

d) They have a gene of interest that has been fully disabled.

e) They have a gene removed which results in lowered fertility.

Question 50

What is the function of dideoxynucleotides in many DNA sequencing techniques?

a) They act as primers for reverse transcriptase.

b) They act as primers for DNA polymerase.

c) They allow only the specific sequencing of the RNAs of a genome.

d) They stop synthesis at a specific site, so the base at that site can be determined.

e) They cut the sequenced DNA at specific sites.