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The Chromosomal Basis of Inheritance 25 October, 2002 Text Chapter 15

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The behavior of chromosomes in meiosis and fertilization explains Mendel's rules ... a) The mother is homozygote for color blindness. ... – PowerPoint PPT presentation

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Title: The Chromosomal Basis of Inheritance 25 October, 2002 Text Chapter 15


1
The Chromosomal Basis of Inheritance25 October,
2002Text Chapter 15
2
Genes and Chromosomes
The behavior of chromosomes in meiosis and
fertilization explains Mendels rules of
inheritance.
Genes on different chromosomes assort
independently.
3
Morgans Mutant
4
Morgans experiments showed that some genes are
inherited along with the X chromosome.
Genes that are on the same chromosome are linked.
Those on the X chromosome are X-linked.
5
Crossing over allows linked genes to appear to
assort independently.
Remember, the purpose of crossing over is to
generate diversity in offspring arising from
sexual reproduction.
6
Crossing over results in recombinant gametes.
7
Recombinant Offspring
8
Recombination Distance
Genes that are far apart are more likely to have
a crossing-over event occur between them than
genes that are closer together on the chromosome.
9
Mapping
Knowing the recombination frequencies between
several linked genes allows the researcher to
construct a genetic map. Note that the maximum
genetic distance discernable in a cross is 50
(indistinguishable from independent assortment.)
Larger genetic distances are determined by adding.
10
Sex Determination
11
Sex-linked genes
(carried on the X
chromosome) show unique patterns of inheritance.
Since the Y chromosome has few genes, recessive
sex-linked alleles are not masked in males.
A carrier mother will pass the allele to 1/2 of
her offspring. Sons will be affected, daughters
will be carriers.
A father with an x-linked trait will transmit the
allele to all daughters, but no sons.
Carrier female x affected male produces affected
females
12
X Inactivation
13
Nondisjunction
14
Meiotic nondisjunction occurs when homologues or
sister chromatids fail to separate in anaphase.
This error leads to gametes that have too many or
too few chromosomes.
If these gametes are involved in fertilization,
the resulting offspring will have an abnormal
number of chromosomes (aneuploidy). They may
have one (monosomy) or three (triploidy) copies
of a chromosome.
This changes the gene dosage of the genes on the
affected chromosome
15
Alterations of Chromosome Structure
Gene dosage can also be changed by structural
alterations within a chromosome, including
deletion, duplication, translocation, and
inversion.
16
Down Syndrome
17
Genomic Imprinting
18
In a plant, leaf color and leaf shape are
controlled by two linked genes. Leaves of the
wild-type plant are red. A recessive mutation in
this gene causes white leaves. Wild-type leaves
are pointed, and a recessive mutation in this
gene causes them to be smooth. The following
crosses were performed pure breeding white,
smooth X pure breeding wild type gives F1 all
red, pointed Now, the next cross red, pointed
X pure breeding white, smooth gives F2 40
white, smooth 36 red, pointed 10 white, pointed
14 red, smooth What is the recombination
frequency between the gene for color and the gene
for leaf shape?
19
Parental types 76 Recombinants 24 Total is
100 Recombinants / Total is 100 / 24 0.24
20
If a man and a non-color blind woman jointly have
three sons, two of whom are color blind, which is
the most likely explanation? a) The mother is
homozygote for color blindness. b) The mother is
heterozygote for color blindness.
c) The father is color
blind.
d) None of the above.
A recessive sex-linked lethal mutation is
generated in Drosophila. If a female
heterozygous for the lethal allele is crossed to
a wild-type male, the ratio of males to females
expected among the living progeny would be (a)
01 (b) 11 (c) 21 (d) 12
21
Hemophilia in humans is due to an X-chromosome
mutation. What will be the results of mating
between a normal (non-carrier) female and a
hemophiliac male? A. half of daughters are
normal and half of sons are hemophilic. B. all
sons are normal and all daughters are carriers.
C. half of sons are normal and half are
hemophilic all daughters are carriers. D. all
daughters are normal and all sons are carriers.
E. half of daughters are hemophilic and half of
daughters are carriers all sons are normal.
22
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