X-inactivation.html: 15_11XChromInactivation_CL.jpg
X inactivation effects.
The tortoiseshell gene is on the X chromosome, and has 2 different alleles, one for
orange fur and one for black fur.
In a heterozygous "mosaic" female,
orange patches are formed by populations of cells
in which the X chromosome with the orange allele is active;
black patches have cells
in which the X chromosome with the black allele is active.
(“Calico”
cats
also have white areas, which are determined by yet another gene.)
XLinkedA.html: 15_10XLinkedInheritanceA.jpg
A father with the disorder will transmit the mutant allele to all daughters but to no sons.
If the mother is a dominant homozygote,
the daughters will have the normal phenotype but will be carriers of the mutation.
XLinkedB.html: 15_10XLinkedInheritanceB.jpg
If a female carrier mates with a normal male,
there is a 50% chance that each daughter will be a carrier like her mother,
and a 50% chance that each son will have the disorder.
XLinkedC.html: 15_10XLinkedInheritanceC.jpg
If a carrier mates with a male who has the disorder, there is a 50% chance that each
child will have the disorder, regardless of sex.
Daughters who do not have the disorder will be carrier,
whereas males without the disorder will be completely free of the recessive allele.
XY.html: 15_06aSexDetermination-L.jpg
Human somatic cells have 22 pairs of homologous autosomes
plus one pair of sex
chromosomes.
XX individuals are female, while XY are male.
barr_body.html: ../../../bio3400/Locked/media/ch07/07_09-Barr_body.jpg
chromosome.html: 15_01TaggedGene_UP.jpg
Chromosomes can be tagged
to reveal a specific gene.
chromosome5.html: 15_chromosome5.jpg
A deletion in the p arm of chromosome 5 is associated with cri du
chat
syndrome.
chromosome_F1.html: 15_02MendelAndChromosmesB.jpg
Overall, F1 plants produce equal numbers of all four kinds of gametes
because the alternative chromosome arrangements at metaphase I are equally
likely.
The result is a 9:3:3:1 phenotype ratio in the F2.
chromosome_F2.html: 15_02MendelAndChromosmesC.jpg
Overall, F1 plants produce equal numbers of all four kinds of gametes
because the alternative chromosome arrangements at metaphase I are equally likely,
resulting in a 9:3:3:1 phenotype ratio in the F2.
chromosome_P.html: 15_02MendelAndChromosmesA.jpg
The chromosomal basis of Mendel's laws.
The arrangement of chromosomes at metaphase I and their movement in anaphase I account for the
segregation and independent assortment
of the alleles for seed color and shape.
continue
chromosome_map.html: 15_chromosome_map.jpg
chromosome_mutations.html: 15_14ChromosomalMutations.jpg
Alterations of chromosome structure.
A deletion removes a chromosomal segment.
A duplication repeats a segment.
An inversion reverses a segment within a chromosome.
A translocation moves a segment from one chromosome to another, nonhomologous one.
A reciprocal translocation results from the exchange of DNA fragments between nonhomologous chromosomes .
cones-rods.html: ../ch49/49_23VertebrateRetina.jpg
Hereditary color blindness usually involves defects in the
medium (Green) or long (Red) wavelength sensitive
cones
in the retina.
cones.html: Cones_color_graph.gif
Spectral absorption curves of the short (Blue), medium (Green)
and long (Red) wavelength pigments in human cone and rod cells.
dihybrid.html: 15_09LinkedGenes_4-L.jpg
wild-type crossed with black, vestigial-winged flies
produce heterozygous
F1 dihybrids, all of which appear wild-type.
A testcross of F1 females with black, vestigial-winged males
produced 2,300 F2 offspring.
drosophila-conclusion.html: 15_04WildMutantDrosB.jpg
Since all F1 offspring had red eyes, the mutant
white-eye trait (w) must be recessive
to the wild-type red-eye trait (w+).
Since the recessive white-eye trait was expressed only in males in
the F2 generation, Morgan hypothesized that the eye-color gene is sex-linked
and
located on the X chromosome.
drosophila-cross.html: 15_04WildMutantDrosA.jpg
Morgan mated a wild-type
red-eyed
female with a mutant
white-eyed
male. The F1 offspring all had red eyes.
Morgan then bred the F1 flies to produce the F2 generation.
The F2 showed a typical Mendelian 3:1
phenotype ratio of red:white eyes.
However, all females had red eyes.
Half the males had white eyes, and half had red eyes.
conclusion
fish.html: 15_FISH_chromosome.jpg
Fluorescent In Situ Hybridization
Probes containing modified nucleotides hybridize to denatured target DNA.
Fluorescent
antibodies then bind to the antigens in the probes, and are observed with fluorescence
microscope.
fruitfly.html: 15_drosophila.jpg
_bed_/sketches/sounds/Arthropoda/fly-musca_domestica.wav
link_recoma.html: 15_10aRecombLinkedGenes-L.jpg
Recombinant phenotypes can be explained by crossing over of homologous chromosomes
In the female, crossing over can occur between b and vg in
Meiosis I, which then segregate in Meiosis II and yield recombinant gametes.
In the male, crossing over yields no new allele combinations.
continue
link_recomb.html: 15_10bRecombLinkedGenes-L.jpg
The recombinant chromosomes from the
female explains the small percentage of recombinant
phenotypes in the F2 offspring.
linkage.html: 15_07ConstructLinkageMap.jpg
The recombination frequencies between 3 gene pairs
(b–cn 9%, cn–vg 9.5%, and b–vg 17%)
best fit a linear order in which cn is positioned about halfway between
the other 2 genes.
Recombination frequencies can be used to construct a linkage map
of the chromosome, since the farther apart genes are, the more
likely they are to be separated during crossing over, and the higher
their frequency of recombination.
linkage_map.html: 15_08DrosLinkageMap_L.jpg
A partial linkage map of Drosophila chromosome II.
This map shows map units between some genes and the locus for aristae length.
One map unit, or centimorgan, represents a 1% recombination frequency.
Notice that more than one gene can affect a given phenotypic characteristic, such as eye
color.
linked.html: 15_UN278DrosophilaTestX_L.jpg
If these two genes were on different chromosomes, the alleles from the F1 dihybrid
would sort into gametes independently, and we would expect to see equal
numbers of the 4 types of offspring.
If these two genes were on the same chromosome, we would expect each allele combination,
B+ vg+ and b vg, to stay together as gametes formed,
and only offspring with parental phenotypes would be produced.
Since most offspring had a parental phenotype,
the genes for body color and wing size must be on the same chromosome.
The small number of offspring with nonparental phenotypes is due to some mechanism that occasionally breaks the
linkage between genes on the same chromosome.
nondisjunction.html: 15_12Nondisjunction_L.jpg
Nondisjunction in either meiosis I or meiosis II can produce
gametes with an extra or missing chromosome.
The resulting offspring has an abnormal number of the chromosome in a condition called
aneuploidy.
An example is Down
syndrome, usually the result of an extra chromosome 21, or trisomy 21.
sex_determination.html: 15_09SexDeterminationB.jpg
Some chromosomal systems of sex determination.
Numerals indicate the number of autosomes. In Drosophila, males are XY,
but sex depends on the ratio between the number of X chromosomes and the number of autosome sets,
not simply on the presence of a Y chromosome.
translocation_reciprocal.html: 15_16TranslocationCML.jpg
A reciprocal translocation produces a short chromosome 22,
(Philadelphia chromosome), and a long chromosome 9,
allowing in the cell to escape control of the cell cycle
and becoming cancerous.