Chromosomal Inheritance & Cell Division – Mastering Genetics. – Flashcards

Bacterial genomes may consist of a single, circular chromosome or multiple chromosomes that may be linear or circular.

nucleoid

The beads are the nucleosomes, and the string is the linker DNA

(least compact) naked DNA -> nucleosome -> selenoid -> loop domains -> chromatid -> metaphase chromosome (most compact) Chromatin compaction is required for the nucleus to accommodate genomes that are often more than 1,000 times longer than the nuclear diameter. As the 2nm wide DNA is organized into chromatin, each level of compaction has a chromatin structure with a signature diameter (2nm DNA, 11nm nucleosome, 30nm solenoid, 300nm loop domains, 700nm chromatid, and 1400nm metaphase chromosome).

H1 The histone protein H1 plays a key role in stabilizing the 30-nm solenoid structure. The long N-terminal and C-terminal ends of the H1 protein attach to adjacent nucleosome core particles pulling them into an orderly solenoid array.

Histones in general have a net positive charge that allow them to bind to DNA. Acetylation of histones, decreases their positive charge and weakens the histone-DNA interaction. Histones are basic proteins that interact with negatively charged DNA. The strength of this interaction is modulated by epigenetic modifications. One of these modifications is acetylation. Acetylation adds an acetyl group to the positively charged amino group present on the side chain of the amino acid lysine effectively changing the net charge of the protein by neutralizing the positive charge. When the positive charge is reduced, the histones loosen their grip on the negatively charged DNA.

H1

23

telocentric

euchromatin

acrocentric

no

euchromatic region

M

sister chromatids

prophase I

Meiosis I in the mother Meiosis II in the mother Meiosis I in the father Meiosis II in the father During meiosis I, homologous chromosomes separate. If nondisjunction occurs, both homologous chromosomes migrate into the same daughter cell instead of different daughter cells. If this happens in the mother, both X chromosomes end up in one cell, with no X chromosome in the other cell. If this happens in the father, both the X and Y chromosome end up in one cell, with no sex chromosome in the other cell. In both cases it is possible for one parent to not give a sex chromosome to the child. If the other parent gives an X chromosome, the child will be XO and have Turner syndrome. During meiosis II, sister chromatids separate. If nondisjunction occurs, both sister chromatids migrate into same daughter cell, instead of different daughter cells. If this happens in the mother, two identical copies of an X chromosome end up in one cell, with no X chromosome in the other cell. If this happens in the father, either two identical copies of the X chromosome end up in one cell, with no sex chromosome in the other cell - or - two identical copies of the Y chromosome end up in one cell, with no sex chromosome in the other cell. In all cases, it is possible for one parent to not give a sex chromosome to the child. If the other parent gives an X chromosome, the child will be XO and have Turner syndrome.

Meiosis II in the father The key to solving this problem is to realize that the child got two Y chromosomes. Both of those Y chromosomes had to have come from the father, as the mother has no Y chromosomes. In order to get two Y chromosomes, nondisjunction would have to have occurred in meiosis II - identical Y sister chromatids migrated together into one cell. During meiosis II, sister chromatids should separate. If nondisjunction occurs, both sister chromatids migrate into same daughter cell, instead of different daughter cells. If this happens in the father, either two identical copies of the X chromosome end up in one cell, with no sex chromosome in the other cell - or as in this case two identical copies of the Y chromosome end up in one cell, with no sex chromosome in the other cell. If the sperm containing two copies of the Y chromosome fertilizes an egg containing an X chromosome, the child will be XYY.

Meiosis I in the mother Meiosis II in the mother Meiosis I in the father To produce a child with XXY, either mom gave two X chromosomes and dad gave Y OR mom gave one X chromosome and dad gave both X and Y. There are two ways mom could give two X chromosomes - either she gave both of her X's (meiosis I nondisjunction), or she gave two copies of one of her X's (meiosis II nondisjunction). In both these cases if dad gives a Y chromosome, the child is XXY and has Klinefelter's. There is only one way dad could give X and Y, and that's if X and Y did not separate in meiosis I. Then if mom gives an X chromosome, the child is XXY and has Klinefelter's. Nondisjunction in meiosis II in dad would give both sister chromatids - either XX or YY, which would produce either a XXX girl or a XYY boy... not a XXY boy.

H1