Download Answer Key BioSci 93 Discussion Week 6 Online

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BioSci 93 Discussion Week 6
Online Worksheet
1) Why does a cell go through meiosis? Explain the main concept in one sentence.
A cell undergoes meiosis in order to halve the genetic material so it can carry out sexual
reproduction.
In other words, we are diploid organisms, and we receive one set of chromosomes from
our mother and another set from our father. If human gametes (sperm and egg) fused
without going reduction in chromosome numbers, then the resulting zygote would be a
tetraploid (46 + 46 = 92 chromosomes [or 23 x 4n]). Each subsequence generation would
also have an increase in chromosome number. Therefore, germ cells undergo meiosis and
reduce the chromosome number to half (23). When a sperm (23 chromosomes) and egg
(23 chromosomes) fuse, the resulting organism will have 46 chromosomes (23 x 2n).
In addition, the process of meiosis is also for the creation of genetic diversity which
occurs through crossing over and independent alignment of chromosomes.
2) Here is a gamete with three chromosomes
a. Draw what this cell looks like in the following phases:
Metaphase I (2n)
Anaphase I (2n)
Metaphase II (n)
Anaphase II (n)
In Metaphase I, homologous chromosomes align the metaphase plate. Anaphase I is the
separation of homologous chromosomes! Afterwards, Metaphase II/Anaphase II is the
same but with pairs of sister chromatids. Please note! Meiosis I produces TWO daughter
cells, and each of those daughter cells enter Meiosis II, producing another two daughter
cells! This results in a total of four daughter cells!
b. In the cells above, write an “n” next to the cells that are haploid and “2n” for diploid.
c. Does the first or second division cause haploidy? First
1 of 3 d. Consider your drawings above. The end of the first meiotic division (end of Meiosis
I) is typically referred to as a “reduction division.” Why?
Know that Meiosis I involves the separation of homologous chromosomes, which
halves the number of chromosomes in the cell. A cell after S-phase has 4 genome (2 x
2n since we’re working with diploid cells) copies. Therefore, this first Meiotic
division halves the number of genome copies as well (to 2 x 1n, in which this will
already be considered a haploid).
3) Nondisjunction is an error that can occur during mitosis or meiosis where members of a pair
of homologous chromosomes or pair of sister chromatids fail to separate.
a. At what stage would nondisjunction occur?
Anaphase
b. What would fail to separate during each of the following:
i.
Mitosis
Sister Chromatids
ii.
Meiosis I
Homologous Chromosomes
iii.
Meiosis II
Sister Chromatids
4) Kleinfelter syndrome is a human genetic disorder that results when the individual has two X
chromosomes and one Y chromosome (XXY genotype) instead of a XY genotype or XX
genotype. In addition, the rest of the chromosomes are normal. This trisomy (extra chromosome
in addition to the original pair) is most likely a result of:
a. Nondisjunction during anaphase I or II in the germ cells
b. Polyspermy
c. Failure of cytokinesis I or II (the splitting of the original cell into two cells)
d. Double duplication of the X chromosome during the second S phase
e. Incorrect separation of chromosomes during prophase I or II
5) You are provided with an actively dividing culture of E. coli bacteria to which radioactive
adenine has been added. What would happen if a cell replicates once in the presence of this
radioactive base?
a. One of the daughter cells, but not the other, would have radioactive DNA
b. Neither of the two daughter cells would be radioactive
c. All four bases of the DNA would be radioactive
d. DNA in both daughter cells would be radioactive
e. Radioactive adenine would pair with nonradioactive guanine
Remember that DNA is replicated during the S phase of cell cycle. We know that
DNA replication is semi-conservative; each strand serves as a template strand for a
new DNA duplex. With this, each new double stranded molecule is composed of one
“old” and one “new” strand. The radioactive adenine will be added through
complementary base pairing so it will be present in both DNA (where the
chromosome itself will consist of a pair of sister chromatids). When these chromatids
are separated during anaphase, each daughter cell will receive one of the chromatids,
meaning that (d) is the correct choice.
6) Mitosis, in contrast to Meiosis:
a. Produces gametes (Meiosis)
2 of 3 b. Requires a previous S phase (Both Meiosis and Mitosis goes through DNA
replication)
c. Is seen in gut epithelial cells
d. Has crossing over before Anaphase (Meiosis)
e. Results in haploid cells (Meiosis)
8) In eukaryotic cells, an mRNA transcript that exits the nucleus:
a. Contains introns and exons
b. Is double stranded
c. Is made up of nucleotides adenine, guanine, thymine and cytosine
d. Has a 5’-cap and poly-A tail
Answer
a
True/False
F
b
F
Explanation
When transcribing mRNA from DNA, the introns are spliced/removed
out of the sequence
RNA is a single strand
c
F
In RNA, uracil replaces thymine
d
T
The mature RNA that exits the nucleus will have a 5’-end cap and a 3’end poly(A) tail. These modifications will help the mRNA to be
exported from the nucleus and to help it from degradation.
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