麻省理工大学：遗传学（Genetics）讲稿_lecture32

Lecture 32: Numerical Chromosomal Abnormalities and Nondisjunction Meiosis l Meiosis‖l Centromere-linked markers

Numerical Chromosomal Abnormalities and Nondisjunction Lecture 32: • Meiosis I • Meiosis II • Centromere-linked markers

Female XX Male 46.XY Human chromosomal abnormalities may be numerical or structural Numerical Total chromosomes/ cell Trisomy 3 copies of a single chromosome Monosomy 1 copy of a single chromosome 45 Triploidy 3N 69 Tetraploidy= 4N 92 Structural Deletion Duplication Translocation (involves 2 chromosomes

Female Male 46,XX 46,XY Human chromosomal abnormalities may be numerical or structural. Numerical Trisomy = 3 copies of a single chromosome 47 Monosomy = 1 copy of a single chromosome 45 Triploidy = 3N 69 Tetraploidy = 4N 92 Structural Deletion Duplication Translocation (involves 2 chromosomes) Total # chromosomes / cell

Chromosomal abnormalities manifest themselves in two ways 1)Spontaneous abortions 50% of human pregnancies--> spontaneous abortion or miscarriage nearly all during first trimester of pregnancy, with many during the first month, when pregnancy is recognized only y hormonal assays 50% of spontaneously aborted embryos and fetuses have chromosomal abnormalities Therefore 25% of all human embryos have chromosomal abnormalities

Chromosomal abnormalities manifest themselves in two ways: 1) Spontaneous abortions of human pregnancies --> spontaneous abortion or miscarriage of spontaneously aborted embryos and fetuses have chromosomal abnormalities nearly all during first trimester of pregnancy, with many during the first month, when pregnancy is recognized only by hormonal assays 25% of all human embryos have chromosomal abnormalities. 50% 50% Therefore

Breakdown of chromosomal abnormalities in spontaneous abortions Trisomy 16 15% 13, 18,21 9% XXX. XXY XYY 1% All others 27% Monosomy X (45, X or XO) 18% Triploidy 7% Tetraploidy 6% Other Total 100%

Trisomy XXX, XXY, XYY All others Monosomy X (45,X or XO) Triploidy Tetraploidy Other Total 15% 9% 1% 27% 18% 17% 6% 7% Breakdown of chromosomal abnormalities in spontaneous abortions: 100% 16 13, 18, 21

Chromosomal abnormalities manifest themselves in two ways 2 )Defects in newborns: 0.5% aggregate frequency among the most common XXY 1/1.000 males XYY /1.100maes 0双 1/7,500 females /1.200 females Trisomy 13 1/1500 Trisomy 18 1/1100 Trisomy 21 1/900 Structural anomalies 1/400

2) Defects in newborns: 0.5% aggregate frequency Among the most common: XXY XYY XO XXX Trisomy 13 Trisomy 18 Trisomy 21 Structural anomalies 1 / 1,000 males 1 / 1,100 males 1 / 7,500 females 1 / 1,200 females 1 / 15,000 1 / 11,000 1 / 900 1 / 400 Chromosomal abnormalities manifest themselves in two ways:

Fruitflies Female xx Male XY Is sex determined by presence/absence of Y? Is sex determined by number of xs?

Fruitflies 1916 Female Xx XXy o Male XY XO Sex is determined by number of xs

Sex chromosomes SEX DETERMINING SIGNAL MALE FEMALE Fruitfly XY XX of X: s Mammals XY XX Y(+or) Nematodes x0ⅩX of X: s hermaphrodite Birds ZZ ZW Some reptiles temperature

Fruitfly XY XX # of X’;s Sex chromosomes SEX DETERMINING SIGNAL MALE FEMALE Mammals XY XX Y (+ or -) Nematodes XO XX # of X’;s hermaphrodite Birds ZZ ZW ? Some Reptiles temperature

Trisomy 21 Down Syndrome Numerical chromosomal disorders are the result of nondisjunction= failure of chromosomes to separate normally during cell division Nondisjunction can occur during meiosis(before fertilization) or mitosis(after fertilization) How could you figure out whether nondisjunction for chromosome 21 had occurred during meiosis or mitosis?

Numerical chromosomal disorders are the result of Trisomy 21 Down Syndrome nondisjunction = failure of chromosomes to separate normally during cell division Nondisjunction can occur during How could you figure out whether nondisjunction for chromosome 21 had occurred during meiosis or mitosis? or meiosis (before fertilization) mitosis (after fertilization)

SSR 21.1 ABCD SSR212/8 D What can you conclude? At least two things 1. The presence in the affected child of two different maternal alleles for ssr 21.1 indicates that nondisjunction occurred before fertilization (in meiosis )in the mother 2. There has been recombination between the two chromosome 21's in the mother prior to nondisjunction

A B C D SSR 21.1 A’ B’ C’ D’ SSR 21.2 What can you conclude? At least two things: 1. The presence in the affected child of two different maternal alleles for SSR 21.1 indicates that recombination between the two chromosome 21’s in the mother prior to nondisjunction. nondisjunction occurred before fertilization (in meiosis) in the mother. 2. There has been