BIOA01H3 Study Guide - Midterm Guide: Sister Chromatids, Gel Electrophoresis, Dna Ligase

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The lagging strand is the daughter strand that has its _____ end pointed toward the replication fork and is therefore synthesized _____. A) they would not differ, and progeny of both crosses would have an array of phenotypes that vary continuously across a range. B) the progeny from the cross of the plant with the single-gene trait would all be identical, whereas the progeny from the cross of the plant with the multiple- gene trait would vary continuously across a spectrum. D) the progeny from both crosses would only have three phenotypes possible: three genes (a, b, and c) are in the same chromosome. B is 1 map unit, between b and c is 4 map units and between a and c is 3 map units. The correct order of genes in the chromosome is: D) topoisomerase: gel electrophoresis separates dna fragments based on: D) hydrogen bonding: consider the following two statements about protein structure:

Related Questions

Assume that plant height and flower size in Four O'Clock plants display antogonistic phenotypes. Four O'Clock plants can be tall or short and the flowers can be large or small. Assume each trait is governed by alleles of a single gene and that each follows Mendelâs first and second laws. You start with two pure breeding strains, one that is tall with large flowers and a second that is short with small flowers. You cross these two strains to each other to generate F1 plants. You self-cross an F1 plant and grow 342 of the resulting seeds into plants. You find that 195 plants are short with large flowers, 62 plants are short with small flowers, 59 plants are tall with large flowers, and 26 plants are tall with small flowers.

You take each of the 26 plants that are tall with small flowers, self cross each of them and examine thousands of progeny plants. Among those thousands of plants you would expect to observe plants with which phenotypes? (Select 'Yes' or 'No' for each)

Yes No Tall with large flowers
Yes No Tall with small flowers
Yes No Short with small flowers
Yes No Medium height with medium size flowers
Yes No Short with large flowers
Yes No Can't tell from the information provided

bluehare134

1. Genomic DNA is digested with a restriction enzyme that has a 4 bp recognition (restriction) site. What would be the expected average size of the fragments that would be produced by digesting DNA of random sequence with this enzyme? A) 8 bp B) 16 bp C) 125 bp D) 256 bp E) 512 bp

2.insertion or removal of one or more nucleotide base pairs in DNA within a gene often results in a ____________ mutation. A) transition B) frameshift C) reversion D) transversion E) suppressor

3. Which ingredient is NOT required in Sanger DNA sequencing? A) Primers for DNA polymerase. B) Reverse transcriptase. C) Dideoxy nucleotides. D) Deoxy nucleotides. E) DNA template.

4. What is the purpose of Taq polymerase in a PCR reaction? a. DNA denaturation b. Primer annealing c. DNA synthesis d. Heating of the reaction e. All of the above are associated with Taq polymerase.

5.You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny: Blue shell, long antenna 92 Green shell, short antenna 88 Blue shell, short antenna 31 Green shell, long antenna 29 Total 240 Assuming that the genes are linked, what is the map distance between them in cM? A) 75.0 cM B) 25.0 cM C) 12.1 cM D) 12.9 cM E) The genes are actually assorting independently. Genes A and B are linked with 10 map units (cM) between them. A heterozygous individual AB/ab (note that the dominants are in coupling) would be expected to produce gametes in which of the following frequencies?

A) 45% AA; 10% Ab; 45% bb

B) 5% AB; 45% Ab; 45% aB; 5% ab

C) 10% AB; 40% Ab; 40% aB; 10% ab

D) 45% AB; 10% Ab; 10% aB; 45% ab

E) 45% AB; 5% Ab; 5% aB; 45% ab

Please answer all

Why does a template-dependent DNA polymerase require a primer to initiate DNA synthesis?

	DNA polymerase require 5' phosphate to add a new nucleotide
	DNA polymerase require 3' hydroxyl group to add a new nucleotide

DNA polymerase require energy by hydrolizing the primers

Which of the following enzymes is template independent?

	Reverse transcriptase
	Klenow fragment

	DNA polymerase
	Terminal deoxynucleotidyl transferase

RNA polymerase

The natural function of a DNA ligase enzyme is:

	Joining together DNA sequences with compatible cohesive ends from two different organisms to yield a recombinant DNA molecule
	Sealing single stranded nicks in double stranded DNA molecules

	Joining together two blunt ended fragments of DNA within a cell
	Joining together two single stranded DNA molecules

The natural function of 3' -> 5' exonuclease activity of a DNA polymerase is to:

	Remove the 5' end of the DNA strand that is being copied
	Remove damaged nucleotide from the template strand

	Remove nucleotides from the end of DNA to generate blunt ends
	Remove incorrect nucleotides from the newly synthesized DNA

Which technique is used to resolve the different sizes of DNA fragments?

	Gene cloning
	PCR

	DNA sequencing
	Gel electrophoresis

Which of the following vectors does not contain bacterial origin of replication (oriC)?

	plasmid
	cosmid

	bacterial artificial chromosome
	lambda vector

Which process of bacteriophage is not used for gene cloning?

	Concatemer formation
	Lytic cycle

	Lysogenic cycle
	viral packaging

What vector would be best suited for creating a contig of bovine (cattle) chromosome 10?

	lambda vector
	Cosmid vector

	P1 vector
	Plasmid

E. coli takes up plasmid DNA by which of the following methods?

	Transduction
	Transformation

	Conjugation
	Transfection

The following times and temperatures are an example of the steps for PCR. You can use the Figure to help you answer the following questions.

Why is the first step is carried out at 94Â°C?

	To degrade the template
	To denature the template

	To activate the polymerase
	To facilitate the primer binding

What happens in the reaction when the temperature shifts to 55Â°C during cycling?

	the DNA polymerase will carry out DNA synthesis by extending the annealed primers
	the DNA polymerase will finish DNA synthesis

	the primers anneal to the single-stranded regions of the DNA
	the primers anneal to the double-stranded regions of the DNA

During cycling, what occurs when the temperature is at 72Â°C?

	the primers anneal to the double-stranded regions of the DNA
	the DNA polymerase will carry out DNA synthesis by extending the annealed primers

	the primers anneal to the single-stranded regions of the DNA
	the DNA polymerase will finish DNA synthesis

You designed a set of primers to amplify 1 kb DNA with polymerase chain reaction. By which cycle of polymerase chain reaction, we would expect to see the first double stranded DNA with expected size?

	The end of cycle 1
	The end of cycle 2

	The end of cycle3
	The end of cycle 4

BIOA01H3 Study Guide - Midterm Guide: Sister Chromatids, Gel Electrophoresis, Dna Ligase

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