Download The diagrams below show two different scenarios for a pair of

PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
1.
The diagrams below show two different scenarios for a pair of homologous chromosomes, known
as a tetrad, undergoing a change where segments of DNA switch on parts of the chromosomes.
In each scenario, the cell containing the tetrad is about to undergo a meiotic division. The four
chromosomes are labeled 1 through 4 in the diagrams.
Which of these best predicts the outcome of the changes illustrated in the diagrams?
(A) In scenario A, DNA is switching in two different chromosomes, which may result in a missing
sex cell or sex cells with duplicate genes, whereas in scenario B, DNA is switching on
homologous chromosomes, which will likely result in no change since each chromosome still
contains the same set of genes.
Distractor Rationale:
This answer suggests the student may understand that DNA transfer between
chromosomes can result in missing or extra DNA on the chromosomes, but does not
understand that this would not occur if equal segments of DNA are exchanged on
homologous chromosomes because the student does not realize that homologous
chromosomes contain the same set of genes with potentially different versions (alleles) of
the genes, whereas sister chromatids are identical.
(B) There will be no change in the chromosomes in either scenario because, in scenario A, the
DNA is switching in homologous chromosomes, which have the same set of genes, and in
scenario B, DNA is switching on sister chromatids, which are identical chromosomes.
Page 1 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
Distractor Rationale:
This answer suggests the student may understand that homologous chromosomes contain
the same sets of genes and that sister chromatids are identical, but does not understand
that in scenario A there would be some genetic recombination because this exchange
occurs on homologous chromosomes, which contain the same set of genes but with
potentially different versions (alleles) of the genes.
(C) In scenario A, DNA is switching in homologous chromosomes, which will create new
combinations of alleles in the gametes, whereas in scenario B, DNA is switching on sister
chromatids, which will result in no change since the chromosomes are identical.
Rationale:
This answer suggests the student understands that crossing over in homologous
chromosomes increases variation in the chromosomes by combining alleles from the two
homologues, but a crossing over among sister chromatids would produce no change
because these chromosomes are genetically identical.
(D) There will be some genetic change in the gametes produced in scenario A because the DNA
is switching in two different chromosomes; however, there will be significant genetic
changes in the gametes produced in scenario B because genetic recombination is occurring
in two sets of chromosomes.
Distractor Rationale:
This answer suggests the student may understand that DNA transfer between
chromosomes usually results in increased genetic variation, but does not understand that
this would not occur if equal segments of DNA are exchanged on sister chromatids because
a crossing over between sister chromatids would exchange identical DNA segments,
resulting in no genetic change.
Aligned to: LO 3.14 CA 3.14: Apply Math to Genetics
Page 2 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
2.
The graphs below show changes in chromosome number for two different cells undergoing
meiosis.
Which description best explains the differences observed in the graphs?
(A) Cell 1 is undergoing normal meiosis, whereas cell 2 is not undergoing normal meiosis
because either homologous chromosomes or sister chromatids fail to separate to produce
haploid cells.
Rationale:
This answer suggests the student understands that normal meiosis produces daughter
cells with half as many chromosomes as the parent cell (2n
n) and identifies that the
graph for cell 1 displays this pattern. The student also understands that a failure for
homologous chromosomes or sister chromatids to separate during meiosis would result in
daughter cells with the same number of chromosomes as the parent cell, as displayed in the
graph for cell 2.
(B) Cell 2 is undergoing normal meiosis, whereas cell 1 is not undergoing normal meiosis
because it has too few chromosomes after replicating its chromosomes and completing the
second division.
Distractor Rationale:
This answer suggests the student may understand that DNA replication occurs before
meiosis, but does not understand that the replication of chromosomes occurs only before
meiosis I because sister chromatids are not separated during meiosis I and need to be
separated during meiosis II. The student may not recognize that the graph for cell 2 does
not show the pattern for chromosome number that would represent meiosis, because the
student does not fully understand the normal change in ploidy that occurs during meiosis
(diploid to haploid, 2n
n).
Page 3 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
(C) Cell 1 is undergoing normal meiosis, whereas cell 2 is completing meiosis but with too many
chromosomes because a pair of homologous chromosomes failed to separate and resulted
in a trisomy.
Distractor Rationale:
This answer suggests the student may understand that meiosis results in production of
haploid cells from a diploid cell and that nondisjunction may result in extra chromosomes in
gametes, but does not understand that nondisjunction of a pair of homologous
chromosomes would not result in a diploid daughter cell because there would be only one
additional chromosome, not an entire set. The student may be confusing the random event
of nondisjunction, which affects a single chromosome and causes a trisomy in that particular
chromosome, with the normal changes in ploidy that occur during meiosis, which affect all of
the chromosomes in the entire set.
(D) Cell 2 is undergoing normal meiosis, whereas cell 1 is undergoing mitosis followed by
meiosis because the cell first divided normally and then divided again to reduce the
chromosome number by half in the daughter cells.
Distractor Rationale:
This answer suggests the student understands that the graph of cell 2 could illustrate
mitosis, but does not understand that the graph of cell 1 illustrates meiosis, because there
is no replication of DNA between meiosis I and meiosis II.
Aligned to: LO 3.14 CA 3.14: Apply Math to Genetics
Page 4 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
3.
In fruit flies (Drosophila melanogaster), the allele for miniature wings (m) is recessive to the allele
for normal wings (M), and the gene for vermilion eyes (v) is recessive to the allele for normal
eyes (V).
A female who is heterozygous for both traits is crossed with a vermilion-eyed, miniature-winged
male. The phenotypes of their offspring are described in the table below.
Which statement best explains the results observed in the Drosophila cross?
(A) The data support Mendel’s Rule of Independent Assortment because the outcome is close
to the expected 50% for both dominant traits and 50% for both recessive traits for a cross
involving two independent traits.
Distractor Rationale:
This answer suggests the student may understand that independent assortment allows for
the segregation of different alleles in many dihybrid crosses, but does not understand that
the data do not support the model of independent assortment because almost all of the
offspring show the two sets of alleles from the parents remaining together instead of
assorting independently. The student may not understand that if recombination had
occurred, the data would have shown more offspring with phenotypes that differed from the
parents’ phenotypes (for example, offspring with a dominant allele from the wing-size gene
combined with a recessive allele from the eye-color gene).
(B) The data suggest that the two genes are linked since the dominant traits (normal eyes and
wings) normally appear together and the recessive alleles (miniature wings and vermilion
eyes) normally appear together in the offspring.
Page 5 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
Distractor Rationale:
This answer suggests the student may understand that the data show that the two genes
are likely linked, but does not understand that recombination has also occurred in a small
percent of the offspring because these offspring show a dominant allele from the wing-size
gene combined with a recessive allele from the eye-color gene, or vice versa. The student
may not understand that if recombination had not occurred at all due to the genes being
linked, all the offspring would have phenotypes that exactly match the parents’ phenotypes.
(C) The data indicate that the allele is most likely sex-linked because, in the F1 generation,
female offspring express different phenotypes than male offspring, which occurs when
genes are found on the X chromosome.
Distractor Rationale:
This answer suggests the student may understand that phenotype differences observed in
males and females may be due to sex linkage, but does not understand that the data do
not support the presence of sex linkage because significant gender differences are not
apparent in the offspring data. The student may not understand that if recombination had
occurred, the data would have shown more male and female offspring with phenotypes that
differed from the parents’ phenotypes.
(D) The data show that the two genes are linked and that they segregate only when crossing
over occurs at a location between them on the chromosome, resulting in a small number of
offspring showing one dominant and one recessive phenotype.
Rationale:
This answer suggests the student understands that the data support the claim that the
wing-size gene and the eye-color gene are linked and that crossing over and recombination
have occurred in some offspring because small percentages of the offspring show a
dominant allele from the wing-size gene combined with a recessive allele from the eye-color
gene and vice versa.
Aligned to: LO 3.14 CA 3.14: Apply Math to Genetics
Page 6 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
4.
Sordaria fimicola is a fungus that exists as a haploid organism through most of its life cycle and
becomes diploid when the fusion of two haploid nuclei forms a diploid nucleus. The diploid
nucleus must then undergo meiosis to restore its haploid state. Meiosis, followed by mitosis, in
Sordaria results in the formation of eight haploid ascospores contained within a sac called an
ascus (plural, asci). Each spore can develop into a new haploid fungus.
The diagram below shows meiosis in a nucleus created by the fusion of a wild-type Sordaria with
black ascospores (+) and a mutant strain with tan spores (tn). The ascus that develops contains
four black ascospores and four tan ascospores. The normal arrangement of the eight ascospores
in the ascus after meiosis and mitosis is shown below.
Crossing over often occurs in Sordaria during meiosis. A set of asci is shown below.
Page 7 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
Which statement best explains the meiotic process that produced these asci?
(A) The evidence shows that crossing over has not occurred in the asci since all the ascospores
are either black or tan, but not both at the same time.
Distractor Rationale:
This answer suggests the student may understand that crossing of two pure strains can
sometimes produce the expression of two different alleles in offspring, but does not
understand that two different alleles could not be expressed on one ascospore because the
ascospores are haploid and have only one allele each.
(B) The evidence shows that crossing over has occurred in two asci since there is an alternating
pattern of two black and two tan ascospores in the ascus, indicating that crossing over
occurred during meiosis I.
Distractor Rationale:
This answer suggests the student may understand that crossing over will reshuffle the
phenotypes of ascospores in each ascus, but does not understand that there are other
possible combinations that could be produced because crossing over could also produce
other patterns, such as 2 tan–4 black–2 tan or 2 black–4 tan–2 black.
(C) The evidence shows that crossing over has occurred in six asci since those asci have black
and tan phenotypes that are not in the expected location given a normal meiotic division.
Rationale:
This answer suggests the student understands that crossing over will reshuffle the
phenotypes of ascospores in each ascus and that any ascus that does not show four of one
color together and then four of the other color (for example, 4 black, then 4 tan) indicates
that crossing over must have occurred in meiosis I.
(D) The evidence shows that crossing over has occurred in fourteen of the asci since those asci
do not have four black ascospores on the top and four tan ascospores on the bottom of the
ascus.
Page 8 of 9
PROCTOR VERSION
3.14 B: Meiosis and Mendelian Models Quiz
Distractor Rationale:
This answer suggests the student may understand that crossing over has not occurred if
there is a pattern of four black ascospores on the top of the ascus and four tan ascospores
on the bottom, as seen in 14 of the resulting 16 asci, but does not understand that a lack
of meiosis could also produce four tan ascospores on the top and four black ascospores on
the bottom because the initial position of the homologous chromosomes is random and top
or bottom position is not relevant.
Aligned to: LO 3.14 CA 3.14: Apply Math to Genetics
Page 9 of 9