Meiosis and Common Chromosomal abnormalities

Question 1

Sequence of Meiosis I and II

Answer 1

1. Prophase I
   
   1. Laptotene and Zygotene: pairing of replicated homologous chromosmes establish connections (synapsis)
   2. Pachytene: contraction of chromosomes
   3. Diplotene: logitudinal cleft makes sister chromatids
2. Metaphase I: alignment of homologues in midline of cell
3. Anaphase I: separation of homologues
4. Interkinesis I: formation of nuclei around 2 daughter cells
5. Cytokinesis: formation of 1 daughter cell and polar body
6. Prophase II
7. Metaphase II
8. Anaphase II: separation of sister chromatids
9. Cytokinesis: fomration of 1 daughter cell and polar body

Question 2

Meiosis in Males vs. Females

Answer 2

In males:

• Meiosis I is followed by meiosis II starting at puberty

In females:

• Before birth –> stops at prophase I
• Ovulation –> meosis I completed with extrusion of first polar body
• Fertilization –> meiosis II with extrusion of 2nd polar body

Question 3

Nondisjunction in meiosis I

Answer 3

1. Homologous chromosomes do not separate
2. Results in
   
   1. 2 gametes that are disomic for that chromosome
      
      1. Have 1 maternal and 1 paternal copy
   2. 2 gametes that are nullisomic for that chromosome

Question 4

Nondisjunction in Meiosis II

Answer 4

1. Sister chromatids do not separate
2. Results in
   
   1. 2 gametes that are normal for the chromosome
   2. 1 gamete that is disomic for the chromosome
      
      1. Has either both maternal or both paternal copy
   3. 1 gamete that is nullisomic for the chromosome

Question 5

Kleinfelter’s Syndrome

Answer 5

1. XXY
2. 47 chromosomes
3. People are tall, hypogonadism, gynecomastia
   
   1. More X inc mental retardation

Question 6

Edward’s syndrome

Answer 6

Trisomy 18

Question 7

Patau’s syndrome

Answer 7

Trisomy 13

Question 8

Turner syndrome

Answer 8

1. X
2. 45 chromosomes
3. Cystic hygroma, gonadal dysgenesis, short stature, particular bone shape

Question 9

Determining if nondisjunction event occured in Meiosis I or Meiosis II

Answer 9

1. How?
   
   1. Can look at polymorphic markers (CA repeats) at the beginning of q or p arm to see if nondisjunction occured in meiosis I or II
2. Nondisjuntion in meiosis I
   
   1. Polymorphic markers will be different from each other
   2. If nondisjuntion occured in meiosis I, there is one maternal and one paternal copy in the disomic gamete that contributed to trisomy.
3. Nondisjunction in meiosis II
   
   1. Polymorphic markers will be identical to each other
   2. If nondisjunction occured in meiosis II, the sister chromatids did not separate so there is either two copies of maternal chromosomes or two copies of paternal chormosome.
   3. Can determine which parent’s gamete the nondisjuntion came from by comparing the identical polymorphic markers with those of the parent.

Question 10

Recombination in inversion loop during Meiosis I

Answer 10

1. Parent with inversion around centromere or a pericentric inversion is phenotypically normal
2. However, during meiosis I, it may result in a acentric (no centromere) and dicentric (two centromeres) chromatids with duplications/deficiencies of genetic information
3. When these align, they are incompatible for development

Question 11

Process of X-inactivation

Answer 11

1. X-inactivation occurs at the X-inactivation center called Xist.
   
   1. Tsix makes an antisense transcript for Xist in one X-chromosome to repress Xist. This chromosome remains active.
   2. In the other chromosome within the same cell, the Xist remains activated and creates a transcript that covers the entire X-chromosome inactivating it.
      
      1. Inactive X condenses in nucleus of cell
      2. In smear, you can see it as a barr body in females
      3. If male is XXY, he will show one barr body
      4. If female is XXX, she will have two barr bodies.

Question 12

Uniparental disomy

Answer 12

1. Occurs when a person receives two copies of a chromosome or a part of a chromosome from one parent and no copies from the other parent.
2. Can be the result of:
   
   1. Nondijunction in meiosis II
   2. Duplication of a single chromosome from one parent
   3. Trisomy rescue
      
      1. Fertilization with hyperhaploid gamete with subsequent loss of the normally inherited chromosome from the other gamete during mitosis

Question 13

Prader Willi

Answer 13

1. Paternal chromosome 15 normally active
2. Disease can result from:
   
   1. Mutation in paternal chromosome 15
   2. Maternal uniparental disomy for chromosome 15
   3. Defect in imprinted maternal gene that will prevent expression of paternal chromosome 15

Question 14

Angelman Syndrome

Answer 14

1. Maternal chromosome 15 is active
2. Disease can result from
   
   1. Deletion in maternal chromosome
   2. Paternal uniparental disomy
   3. Defect in paternally imprinted chromosome 15 which turns off expression of maternal chromosome