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.)

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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.

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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.

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Human somatic cells have 22 pairs of homologous autosomes plus one pair of sex chromosomes. XX individuals are female, while XY are male.

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Chromosomes can be tagged to reveal a specific gene.

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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

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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 .

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Hereditary color blindness usually involves defects in the medium (Green) or long (Red) wavelength sensitive cones in the retina.

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Spectral absorption curves of the short (Blue), medium (Green) and long (Red) wavelength pigments in human cone and rod cells.

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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.

1. If the 2 genes were on different chromosomes, independent assortment should yield equal numbers of the 4 types of F₂.
2. If the 2 genes were on the same chromosome, each allele combination (B⁺ vg⁺ and b vg) should stay together and only yield parental phenotypes in the F₂.
3. The high percentage of parental phenotypes in the F₂, with some nonparental (recombinant) phenotypes imply the 2 genes are physically close to each other on the same chromosome.

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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

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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.

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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

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The recombinant chromosomes from the female explains the small percentage of recombinant phenotypes in the F₂ offspring.

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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.

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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.

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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.

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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.

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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.