Lecture 3 Flashcards

1
Q

Are homologous pairs of chromosome present in mitosis? Explain your reasoning.

A

This question addresses a common misconception among students who often assume that homologous pairs of chromosomes exist in meiosis but not in mitosis. Homologous pairs of chromosomes are present in mitosis. However, they don’t pair up in mitosis like they do in meiosis, and sister chromatids of each chromosome separate independently.

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2
Q

List the stages of mitosis and the major events that take place in each stage.

A

(1) Prophase: The chromosomes condense and become visible, the centrosomes move apart, and microtubule fibers form from the centrosomes.
(2) Prometaphase: The nucleoli disappear and the nuclear envelope begins to disintegrate, allowing for the cytoplasm and nucleoplasm to join. The sister chromatids of each chromosome are attached to microtubules from the opposite centrosomes.
(3) Metaphase: The spindle microtubules are clearly visible and the chromosomes arrange themselves on the equatorial plane of the cell.
(4) Anaphase: The sister chromatids separate at the centromeres after the breakdown of cohesin protein, and the newly formed daughter chromosomes move to the opposite poles of the cell.
(5) Telophase: The nuclear envelope reforms around each set of daughter chromosomes. Nucleoli reappear. Spindle microtubules disintegrate

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3
Q

Why are the two cells produced by the cell cycle genetically identical?

A

The two cells are genetically identical because during S phase an exact copy of each DNA molecule was created. These exact copies give rise to the two identical sister chromatids. Mitosis ensures that each new cell receives one of the two identical sister chromatids. Thus, the newly formed cells will contain identical daughter chromosomes.

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4
Q

What are the major results of meiosis?

A

Meiosis involves two cell divisions, thus producing four new cells (in many species). The chromosome number of a haploid cell produced by meiosis I (haploid) is half the chromosome number of the original diploid cell. Finally, the cells produced by meiosis are genetically different from the original cell and genetically different from each other.

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5
Q

How does anaphase I of meiosis differ from anaphase of mitosis?

A

In anaphase I of meiosis, homologous chromosomes separate whereas in anaphase of mitosis the sister chromatids separate.

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6
Q
A cell in G1 of interphase has 12 chromosomes. How many chromosomes and DNA molecules will be found per cell when this original cell progresses to the following stages?  
A. G2 of interphase
B. Metaphase I of meiosis
C. Prophase of mitosis
D. Anaphase of meiosis
E. Anaphase II of meiosis
F. prophase II of meiosis
G. After cytokinesis following mitosis
H. After cytokinesis following meiosis II
A

The number of chromosomes and DNA molecules depends on the stage of the cell cycle. Each chromosome contains only one centromere, but after the completion of S phase, and prior to anaphase of mitosis or anaphase II of meiosis, each chromosome will consist of two DNA molecules.

a. G2 of interphase
Solution:
G2 of interphase occurs after S phase, when the DNA molecules are replicated. Each chromosome now consists of two DNA molecules. So a cell in G2 will contain 12 chromosomes and 24 DNA molecules.
b. Metaphase I of meiosis
Solution:
Neither homologous chromosomes nor sister chromatids have separated by metaphase I of meiosis. Therefore, the chromosome number is 12, and the number of DNA molecules is 24.
c. Prophase of mitosis
Solution:
This cell will contain 12 chromosomes and 24 DNA molecules.
d. Anaphase I of meiosis
Solution:
During anaphase I of meiosis, homologous chromosomes separate and begin moving to opposite ends of the cell. However, sister chromatids will not separate until anaphase II of meiosis. The number of chromosomes is still 12, and the number of DNA molecules is 24.
e. Anaphase II of meiosis
Solution:
Homologous chromosomes were separated and migrated to different daughter cells at the completion of meiosis I. However, in anaphase II of meiosis, sister chromatids separate, resulting in a temporary doubling of the chromosome number in the now haploid daughter cell. The number of chromosomes and the number of DNA molecules present will both be 12.
f. Prophase II of meiosis
Solution:
The daughter cells in prophase II of meiosis are haploid. The haploid cells will contain six chromosomes and 12 DNA molecules.
g. After cytokinesis following mitosis
Solution:
After cytokinesis following mitosis the daughter cells will enter G1. Each cell will contain 12 chromosomes and 12 DNA molecules.
h. After cytokinesis following meiosis II
Solution:
After cytokinesis following meiosis II, the haploid daughter cells will contain six chromosomes and six DNA molecules.

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7
Q

What is the difference between sister chromatids and homologous chromosomes?

A

Sister chromatids are identical copies (unless crossing over takes place) of the same original chromosome. Homologous chromosomes are different chromosomes, containing information for the same traits (homologous information) but not the same genetic information. One homolog comes from one parent; the other homolog comes from the other parent.

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8
Q
A cell has eight chromosomes in G1 of interphase. Draw a picture of this cell with its chromosomes at the following stages. Indicate how many DNA molecules are present at each stage. 
A. Metaphase of mitosis
B. Anaphase of mitosis
C. Anaphase II of meiosis
D. Prophase I of meiosis
A

A. 8 C; 16 DNA
B. 16 C; 16 DNA
C. 8 C; 8 DNA
D. 8 C; 16 DNA

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9
Q

The fruit fly Drosophila melanogaster has four pairs of chromosomes, whereas the house fly Musca domestica has six pairs of chromosomes. In which species would you expect to see more genetic variation among the progeny of a cross? Explain your answer.

A

The progeny of an organism whose cells contain more homologous pairs of chromosomes should be expected to exhibit more variation. The number of different combinations of chromosomes that are possible in the gametes is 2^n, where n is equal to the number of homologous pairs of chromosomes. For the fruit fly with four pairs of chromosomes, the number of possible combinations is 2^4 = 16. For Musca domestica with six pairs of chromosomes, the number of possible combinations is 2^6 = 64.

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10
Q

A cell has two pairs of submetacentric chromosomes, which we will call chromosomes Ia, Ib, IIa, and IIb (chromosomes Ia and Ib are homologs, and chromosomes IIa and IIb are homologs). Allele M is located on the long arm of chromosome Ia, and allele m is located at the same position on chromosome Ib. Allele P is located on the short arm of chromosome Ia, and allele p is located at the same position on chromosome Ib. Allele R is located on chromosome IIa and allele r is located at the same position on chromosome IIb.

A. Draw these chromosomes, identifying genes M,m,P,p,R, and r, as they might appear in metaphase I of meiosis. Assume no crossing over

B. Taking into consideration the random separation of chromosomes in anaphase I, draw the chromosomes (with genes identified) present in all possible types of gametes that might result from this cell’s undergoing meiosis. Assume no crossing over.

A

No picture available, sorry

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11
Q

On average, what proportion of the genome in the following pairs of humans would be exactly the same if no crossing over took place? (For the purposes of this question only, we will ignore the special case of the X and Y sex chromosomes and assume that all genes are located on nonsex chromosomes.)

A. Father and child
B. Mother and child
C. Two full siblings (offspring with same two biological parents
D. Half siblings (offspring with only one biological parent in common)
E. Uncle and niece
F. Grandparent and grandchild

A

a. Father and child
Solution:
The father will donate one-half of his chromosomes to his child. Therefore, the father and child will have one-half of their genomes that are similar.

b. Mother and child
Solution:
The mother will donate one-half of her chromosomes to her child. Therefore, the mother and child will have one-half of their genomes that are similar.

c. Two full siblings (offspring that have the same two biological parents)
Solution:
The parents can contribute only one-half of their genome to each offspring. So it is likely that the siblings share one-fourth of their genes from one parent. Because each sibling would share one-fourth of their genes from each parent, their total relatedness is one-half (or ¼ + ¼).

d. Half siblings (offspring that have only one biological parent in common)
Solution:
Half siblings share only one-fourth of their genomes with each other because they have only one parent in common.

e. Uncle and niece
Solution:
An uncle would share one-half of his genomes with his sibling, who would share one-half of his or her genome with his or her child. So, an uncle and niece would share one-fourth of their genomes (½ × ½).

f. Grandparent and grandchild
Solution:
The grandparent and grandchild would share one-fourth of their genomes because the grandchild would share one-half of her genome with her parent and the parent would share one-half of her genome with the child’s grandparent.

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12
Q

List as many similarities and differences in mitosis and meiosis as you can. Which differences do you think are most important and why?

A

Similarities
• Both involve chromosome and cell division.
• Both are preceded by DNA replication.
• Both use spindle fibers to separate chromosomes.
• Both have a stage where sister chromatids separate.
Differences
• Meiosis normally involves two cell divisions; mitosis usually has only a single cell division.
• Chromosome reduction occurs in meiosis but not in mitosis.
• Resulting daughter cells are genetically different in meiosis but not in mitosis.
• Crossing over occurs in meiosis but does not normally take place in mitosis.
• Random assortment of chromosomes occurs in anaphase I of meiosis but does not occur in anaphase of mitosis.
• In metaphase I of meiosis, pairs of homologous chromosomes line up; in metaphase of mitosis (and metaphase II of meiosis) individual chromosomes line up.
• In anaphase I of meiosis, homologous chromosomes separate; in anaphase of mitosis (and anaphase II of meiosis) sister chromatids separate.

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13
Q

An RNA molecule has the following percentages of bases: A = 23%, U = 42%, C = 21%, G = 14%.

a. Is this RNA single stranded or double stranded? How can you tell?
b. What would be the percentages of bases in the template strand of the DNA that contains the gene for this RNA?

A

A.
The RNA molecule is likely to be single stranded. If the molecule was double stranded, we would expect nearly equal percentages of adenine and uracil, as well as equal percentages of guanine and cytosine. In this RNA molecule, the percentages of these potential base pairs are not equal, so the molecule is single stranded.

B.
Because the DNA template strand is complementary to the RNA molecule, we would expect equal percentages for bases in the DNA complementary to the RNA bases. Therefore, in the DNA we would expect A = 42%, T = 23%, C = 14%, and G = 21%.

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14
Q

The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit. The bottom strand is the template strand. Give the sequence found on the RNA molecule transcribed from this DNA and label the 5′ and 3′ ends of the RNA.
5′–ATAGGCGATGCCA–3′
3′–TATCCGCTACGGT–5′ Template strand

A

The RNA molecule would be complementary to the template strand, contain uracil, and be synthesized in an antiparallel fashion. The sequence would be: 5′–A U A G G C G A U G C C A–3′. The RNA strand contains the same sequence as the nontemplate DNA strand except that the RNA strand contains uracil in place of thymine.

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15
Q
14.	Write the consensus sequence for the following set of nucleotide sequences. 
A G G A G T T 
A G C T A T T 
T G C A A T A 
A C G A A A A 
T C C T A A T
 T G C A A T T
A

The consensus sequence is identified by determining which nucleotide is used most frequently at each position. For the two nucleotides that occur at an equal frequency at the first position, both are listed at that position in the sequence and identified by a slash mark: T/AGCAATT

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16
Q

A strain of bacteria possesses a temperature-sensitive mutation in the gene that encodes the sigma factor. The mutant bacteria produce a sigma factor that is unable to bind to RNA polymerase at elevated temperatures. What effect will this mutation have on the process of transcription when the bacteria are raised at elevated temperatures?

A

Binding of the sigma factor to the RNA polymerase core enzyme forms the RNA polymerase holoenzyme. Only the holoenzyme binds to the promoter. Without the sigma factor, RNA polymerase will be unable to bind the promoter and transcription initiation will not occur. Any RNA polymerase that has completed transcription initiation and has begun elongation will complete transcription because the sigma factor is not needed for elongation. However, no further initiation will be possible at the elevated temperature.

17
Q
  1. Draw a prokaryotic gene and its RNA product. Be sure to include the promoter, transcription start site, transcription termination site, untranslated regions, and labeled 5’ and 3’ ends.
A

no picture :(