Apply-Extend 2 January 22-24
From Jan 22 lecture
*
In normal meiosis, the chromosomes
separate from each other as they go
into the gametes.
Members of the same homologous pair
separate at the end of Meiosis I. The
twin chromatids separate from each
other at the end of Meiosis II.
What results if there is non-disjunction
= failure of chromosomes (either a
homologous pair or twin chromatids) to
move apart during meiosis?
aneuploidy ”not good ploidy” =
abnormal chromosome number
AE 2. Relevant learning outcomes (from 2/22 lecture)
1-16. What is aneuploidy? What is trisomy 21?
1-17. What is the genetic (chromosomal) basis of sex
determination (male or female) in placental mammals?
1-18. Given information about parental phenotypes for an
X-linked recessive trait, draw a Punnett Square (including
X and Y chromosomes) and predict the percentage or
fraction of
(a) all their offspring and (b) their sons that would
express the recessive trait in their phenotype.
self-check: what is an example of epigenetic “marking”?
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Trisomy 21
Down
Syndrome
Review the normal chromosome movements
during meiosis –>
Match the non-disjunction event with the
resulting number of chromosomes in the
daughter cells resulting from meiosis.
Human example, n = 23.
From Jan 22 lecture
LO. 1-16. What is trisomy 21? (Down
Syndrome phenotype)
3 copies of chromosome
21, not a pair, in this
person’s genome; how
many chromosomes
total?
(a) 4 daughter cells contain. 22, 22, 24, 24
(b) 4 daughter cells contain 23, 23, 23, 23
(c) 4 daughter cells contain 23, 23, 22, 24
A person with Trisomy 21 results from the
fertilization of a gamete with the normal
number of chromosomes (23) with a
gamete having an extra chromosome
(24).
In most trisomies, individuals die very
early in embryonic development. Trisomy
21 individuals are the exception. The
second most common trisomy observed
in human fetuses is Trisomy 18, the fatal
Edwards Syndrome.
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1____no nondisjunction in meiosis
2.____one nondisjunction event
in the first meiotic division
3.____one nondisjunction event
in the second meiotic division
Explain how a child
with Down Syndrome
could develop?
[not just the gamete]
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Skill: Predict offspring genotypes & phenotypes
for X-linked traits using a Punnett square.
Punnett Squares for X-linked traits must show
X and Y as well as the X-linked alleles (H/h)
Trisomy is when there are 3 chromosomes instead of
2 in a homologous pair (there’s an extra
chromosome in the individual’s genome.)
e.g. a human trisomy individual has __
chromosomes. (23, 24, 46, 47?)
Polyploidy is when there are multiple (duplicate or
triplicate) copies of ALL the chromosomes in the
genome.
e.g., if there were such thing as a polyploid human
(there’s not!), that individual could have ____
chromosomes. 46, 47, 92
this is an extra slide from 1/22 lecture
more later in the semester about polyploidy
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Practice Punnett square for X-linked traits.
5. A man with red-green color blindness (a
recessive, X-linked trait) marries a woman with
normal vision whose father was color-blind.
From the Punnett Square tell what is the
probability that their first child will be
(a) a color-blind daughter?
(b) a color-blind son?
[each birth is independent]
A. 0, 1/4
B. 1/4, 1/4
C. 1/2, 1/4
D. 1/4, 1/2
E. 1/4, 0
4. A woman is heterozygous for
the recessive X-linked gene for
Duchenne Muscular Dystrophy (DMD).
Her husband is not affected by DMD.
What are the parent genotypes
(with X and Y chromosomes)?
Draw the Punnett Square for this mating.
(a) What proportion of their sons will have DMD?
(b) What proportion of their daughters will be
carriers for the DMD allele?
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Epigenetic inheritance.
Epigenetic inheritance. [slides will be in 1/22 lecture]
(p. 315) Inheritance of traits transmitted by mechanisms
NOT involving the nucleotide sequence are called epigenetic
inheritance.
Because genetically identical twins
can accumulate different epigenetic
markings over time, they can
express genes differently. Many
quite normal developmental
processes require epigenetic marking.
Epigenetic mechanisms
DNA methylation
histone acetylation
are normal methods
of control of gene expression
—remember from BIO 311C?
However, there are external, non-normal factors contributing
to epigenetic change in individuals: environmental chemicals,
diet, and lifestyle factors. We also know epigenetics plays a
role in aging and cancer.
Whereas mutations in DNA are permanent, modifications to
the chromatin can be reversed.
For example, DNA methylation patterns in the parent
chromosomes are largely erased during gamete formation,
then re-established.
Important: epigenetic ”marking’ affects gene expression but
does not change the gene itself.
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A large-scale example of normal epigenetic marking happens
in X-chromosome inactivation, initially from some
short, non-coding RNAs, but then by long-term by
DNA methylation all over one of the X chromosomes.
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Example (Instapoll in 1/22 lecture)
2.5 In embryos that are exposed to alcohol during
development, some genes are turned off
epigenetically.
Which would be an example?
A. alcohol causes a mutation in the base sequence
B. alcohol triggers DNA methylation of a base
sequence
C. alcohol inactivates important enzymes in the
cytoplasm
D. two of these would be epigenetic changes
Do you understand epigenetics?
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[slide will be in 1/22 lecture]
X chromosome inactivation occurs
in female mammals at a time when
the embryo is composed of a few
cells.
Inactivation is random, in the sense
that either the maternal X or the
paternal X can be inactivated initially
in any given cell.
All daughter cells have the same
inactivated X.
Thus, the body of an XX female
develops as a genetic mosaic in
which some tissues express the
maternal X chromosome and other
Gene EXPRESSION
tissues express the paternal X.
difference, not inheritance
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Black and
Orange alleles
are
co-dominant at
the same locus
on the X
chromosome.
X inactivation happens in any
mammal that has more than one X
chromosome
A cell producing
black fur
contains:
A. only genes
for black fur
B. active gene
for black fur;
inactive gene
for orange
fur;
C. only genes
for orange fur
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The inactive X shows up under a
microscope as a Barr body.
stained nucleus
with one Barr
body
Gene EXPRESSION
difference, not inheritance
END OF WED 1/22 lecture
ONLY IF TIME--What could
be the evolutionary
advantage of X inactivation
in mammals???
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AE2 #1. A woman with blood type A gave birth to a
child with blood type AB. It is uncertain which of
three men is the father of the child. Person X has
blood type A, person Y has blood type O, and
person Z, has blood type B.
Can one or more of these men (X, Y, Z) be
eliminated as the father of the child based on ABO
blood group data?
A. only Person X can be eliminated
B. only Person Y can be eliminated
C. only Person Z can be eliminated
D. Persons X and Y can be eliminated
E. Persons X and Z can be eliminated
F. Persons Y and Z can be eliminated
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Modified from text p. 238, #6
AE2 #2. Hemochromatosis is an inherited
disease caused by a recessive allele (not
sex-linked). If a woman and her husband,
who are both heterozygous carriers, have
three children, what is the probability that all
three children have hemochromatosis?
A. 0
B. 1/64
C. 1/16
D. 1/8
E. 1/4
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AE3 #3 A woman is heterozygous for the recessive
X-linked gene for hemophilia. Her husband is not
affected by hemophilia. Determine the the parent
genotypes (with X and Y chromosomes), and draw
the Punnett Square for this mating.
AE2 #4. If a baby girl was born color-blind,
which of the following can be assumed to be
true about her parents’ phenotype?
What proportion of their sons will have hemophilia?
A. 0 (none)
B. 1/4
A. Both of her parents must be color-blind
B. Her father must be color-blind
C. Her mother must be color-blind
D. None of these above about her parents’
phenotype can be assumed
C. 1/2
D. 3/4
E. 1 (all)
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How many Barr bodies in:
XX individual?
XY individual?
XXX individual?
XXY individual?
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AE2 Q5. Very few protein-coding genes are on
the Y chromosomes, and there is virtually no
crossing-over between X and Y.
Y-linked genes include the sry gene that
expresses the testis-differentiation protein and
determines male development. A male with this
gene will
A. usually pass it to his sons, but about 10% of
the time also pass it to a daughter
B. only pass the gene to his sons
C. only pass the gene to his daughters
D. only pass the gene to his grandsons, not sons
E. pass the gene to all of his children if the
mother is a carrier.
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AE2. Q6.
Why do X-linked recessive traits appear in the
phenotype of males more than in females?
A. Males always carry more recessive traits
B. Males inherit more recessive traits from their
mothers than females do.
C. Males inherit more recessive traits from their
fathers than females do.
D. Females have two X chromosomes so if one has
a recessive gene a normal allele on the other X
chromosome can be expressed.
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