Meiotic problems stemming

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Meiotic problems stemming from segregation of
trivalents, Including aneuploid gametes and
zygotes
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Most cases of Down syndrome, like the one shown
here, are due to meiotic nondisjunctions, but
there is a familial form of the syndrome as
well.
Down Syndrome, Trisomy 21, 47 XX or XY 21
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Familial Down Syndrome A Robertsonian
translocation involving chromosomes 14 and 21
14-21 Robertsonian translocation
Make sure you understand the relationship between
the translocation and chromosome dosage.
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Related species frequently have differences in
chromosome number but not in arm number.
Robertsonian translocations are therefore
implicated in the origin of these differences.
A familiar example Horse (Equus caballus)
2N 64 Donkey (Equus
asinus) 2N 62 Grevys zebra
(Equus grevyi) 2N 34 How many Robertsonian
translocations might separate each pair of
species? Note The differences among species
usually involve homozygous or fixed
Robertsonian translocations. But if these truly
lower the fertility of their
heterozygous carriers, how do we get from one
fixed chromosome number to the next?

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The mule is the classical example of a sterile
interspecific hybrid. What is a mule? What
would be its diploid chromosome number? Why is
it sterile? What is a hinny and how does it
differ from a mule?
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What is a mule? What would be its diploid
chromosome number? Why is it sterile? Ans A
mule is a hybrid obtained by crossing a jack
donkey with a mare. It has 63 chromosomes in
its somatic cells. It is a Robertsonian
translocation heterozygote and its sterility
stems (in part) from the formation of a single
trivalent. Other chromosome rearrangements
(inversions, etc.) are also involved. Even
though they are sterile, mules come in both
sexes. What is a hinny and how does it differ
from a mule? Ans A hinny is the reciprocal
hybrid, stallion X donkey mare. It has the same
meiotic problems as a mule but is said to
be somewhat easier to handle.
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A Robertsonian translocation was very likely
involved in the ancestry of our own species.
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Exam question (from koofer 1A)
Solution In a diploid cell, every fixed
Robertsonian fusion subtracts 4
telocentrics and adds 2 metacentrics (keeping the
same number of arms) so the diploid number
drops by 2 but the number of metacentrics
increases by 2. The species here are
separated by two rounds of fixed
fusions 38 - 8 4 metacentrics 30
telocentrics and 4 metacentrics
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M1 synapsis in a reciprocal translocation
heterozygote
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This multivalent can segregate at M1 in three
different ways. Try to identify these and
determine how many might yield viable gametes or
zygotes, without duplications or deficiencies.
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A
B
B
A
Two daughter cells of M1 showing pattern of
alternate segregation. Red arrows show the
two cases of adjacent segregation
B
A
This illustration is color-coded.
Follow red and green
Work through these segregation patterns for
yourself and be sure you understand them.
viable
often inviable
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The Philadelphia chromosome
Follow the red dots
A somatic translocation associated with a
particular form of cancer. Many forms of cancer
are accompanied by characteristic chromosome
abnormalities. This may mean that some forms
of cancer are due to position effects.
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An X1X2 Y sex chromosome system in a fish.

male 2N 35
male 2N 35
Note large Y chromosomes
male 2N 35
female 2N 36
M1 pachytene (testis) note trivalents
Translocations between autosomes and sex
chromosomes result in so - called multiple sex
chromosome systems. In some groups, these are
quite common.
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Break time..
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YOU CAN MOST EASILY
UNDERSTAND RECOMBINATION BY THINKING ABOUT THE
GENETIC COMPOSITION OF GAMETES Parents
AABB X aabb F1 AaBb Gametes
AB aB
Parental Recombinant
ab Ab
INDEPENDENT ASSORTMENT
PARENTAL RECOMBINANT In the
pool of gametes 1/4 AB 1/4ab 1/4 Ab 1/4 aB
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Effect of linkage Suppose genes A and B (and a
and b) are closely linked on the same chromosome.
The there would be NO recombinant
gametes Gametes from AaBb parent under very
tight linkage AB and
ab only. (parental) Effect of
crossing over Now suppose that a chiasma
occurred that involved one chromatid from the AB
chromosome and one from the ab homolog. Breakage
and reunion of the homologous chromatids occurred
between genes A and B (and between a and b).
When that cell completed meiosis, there would be
4 types of gametes AB (no chiasma)
(parental)
Ab due to chiasma or crossing
over (recombinant)
aB aB
(no chiasma) (parental)
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Chiasmata are random events. They occur
between two linked genes in some meioses but not
others Gamete pool Games With No
Chiasmata AB ab (parental)
and Gametes With Chiasmata Between
A and B AB ab
(parental) aB Ab (recombinant)
Net sum AB ab gt aB Ab
PARENTALS gt RECOMBINANTS
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INDEPENDENT ASSORTMENT PARENTALS
RECOMBINANTS
LINKAGE
PARENTALS gt RECOMBINANTS
This is a fundamental relationship and it is
very important that you work through it until
you understand it well.