Lec 29

Question 1

RNAi

Answer 1

in eukaryotes shuts off gene expression using double stranded RNAs

Question 2

miRNA

Answer 2

is transcribed from a distinct gene and targets other genes for regulation

Question 3

siRNA

Answer 3

comes from mRNA, transposons or viral RNA and targets the genes that it comes from

Question 4

The Immune System

Answer 4

purpose is to defend the body from disease
-Antibody Mediated Immunity:
T cells recognize antigens
B Cells make antibodies ( 5 classes depending on type of constant region on heavy chain)

-Cell-Mediated Immunity
Killer T-cells detect and destroy virus infected cells

Phagocytes ingest and destroy viruses, bacteria etc.

Question 5

Humoral Immunity

Answer 5

Antibodies produced by B cells

Question 6

Cellular Immunity

Answer 6

Immunity due to action of T Cells

Question 7

T cells

Answer 7

lymphocytes originate from stem cells in the bone marrow

• T cells mature in the Thymus and enter circulation
• They attack by binding host cells and lysing them

Question 8

B cells

Answer 8

mature in the bone marrow

-when b cells encounter antigens they mature into plasma cells, which secrete antibodies and bind to the antigen

Question 9

General Structure of antibody

Answer 9

-2 chains: Heavy and Light
• Each has a constant and a variable region
• Chains connected by disulfide bonds
• Antigen binds to variable region
• 5 classes of immunoglobulins are defined by the type of heavy chain that they contain.

Question 10

How can we make 10^6 to 10^8 antibody types

Answer 10

• somatic recombination in b cells during B cell differentiation
• DNA processing: to choose constant region, joining region and variable region for both the heavy and light chains.

-Alternative sites for recombination
• Somatic mutations (up to 2% of sequence is altered in variable area of chains which is the part that binds antigen)

Question 11

Somatic Recombination in B Cells

Answer 11

Germ line DNA (chr. 22)
Processed to mature B cell DNA
Stem-loop structure brings areas to be joined into proximity so that somatic recombination can occur at various alternative sites (eg. Between V2 and J3 here)
Pre-mRNA is processed
Somatic mutations also common

LOOK AT DIAGRAM

Question 12

Bruton’s Disorder (agammaglobulemia)

Answer 12

• X-linked recessive
• B-cells and plasma cells missing or nonfunctional so no antibodies made
• T-cells ok so still protection against viruses
• Bone marrow transplant to give new stem cell population which gives rise to functional B cells.

Question 13

Severe Combined Immunodeficiency Syndrome (SCID)

Answer 13

-Both B and T cells missing or nonfunctional
-See autosomal and X-linked forms of disease (may be due to interaction of several genes)
-ADA (adenosine deaminase) missing in one form of disease
– Gene therapy procedures – transfer good ADA gene to white blood cells using retrovirus. Cells with active ADA gene are put into patient in hopes that ADA enzyme will stimulate T and B cell activity.

Question 14

Epigenetics

Answer 14

Heritable changes in gene expression that occur without
alterations in DNA sequence
-can be due to modifications of the DNA (primarily METHYLATION) and modification of the HISTONES (methylation, phsophorylation, and acetylation)
-Histone acetylation often turns on gene expression.
-Increased methylation generally turns off gene expression
-Hypermethylation of tumor suppressor genes observed in some cancer cells
– Methylated cytosine is more likely to mutate to thymine than normal cytosine – possibly leading to mutations as well

Question 15

Sequencing the Epigenome

Answer 15

Efforts are underway to determine methylation patterns on DNA as well as methylation, phosphorylation and acetylation patterns on histones in an effort to understand more about gene expression and epigenetic control.

Question 16

Imprinting

Answer 16

-considered to be a epigenetic modification
-methylation of cytosine residues
-causes gene to be inactive
-Can be maternally or paternally imprinted
– Imprinting is a way to mark which parent the DNA came from

Question 17

Development

Answer 17

– Regulated growth resulting from interaction of genome and cytoplasm and environment
– Programmed sequence of events
– Usually not reversible

Question 18

Differentiation

Answer 18

– Aspect of development

– Forming different types of cells, organs, etc. through specific regulation of gene expression.

Question 19

Oogenesis

Answer 19

-Oogenesis gives rise to unfertilized eggs via meiosis. (Oogenesis occurs prior to development)
• In animals, this is an asymmetrical process in which 3 polar bodies and one oocyte are produced
• Cytoplasm of the oocyte contains critical regulatory molecules that control the early stages of development after fertilization

Question 20

Overview of Animal Development

Answer 20

1)Begins with fertilized egg that contains
– Paternally derived set of chromosomes
– Maternally derived set of chromosomes and substances needed for growth and differentiation
2)After fertilization, cleavage divisions occur to partition ooplasm in daughter cells. This can be asymmetrical.
3)Because cells generally have the same genetic material, differentiation is due to the differential activation or suppression of genes
4) Rapid cleavage divisions (by mitosis) increase the number of cells creating a hollow ball of cells called a blastula
5)Determination is the process by which cells are assigned developmental fates
6)At the blastula stage, some cells have been determined to become specific tissues in the animal’s body
– Confirmed by transplantation studies.
7)Gastrulation occurs as cells migrate to form 3 layers that will each give rise to specific structures
8)Differentiation occurs as cells become different from each other due to the expression of different sets of genes.

Question 21

Ectoderm

Answer 21

outer layer of gastrulation

Question 22

Mesoderm

Answer 22

Middler layer of gastrulation

Question 23

Endoderm

Answer 23

inner layer of gastrulation

Question 24

Early Embryological Drosophila Development

Answer 24

1) Maternal Genes (Egg Polarity Genes)
2) Segmentation Genes
3) Homeotic Genes

Question 25

Egg Polarity Genes

Answer 25

• Establish anterior/posterior polarity
  and dorsal/ventral polarity
  -Transcribed during egg development (Maternal Genes)
  -Translated after fertilization Anterior-Posterior –Axis Dorsal-Ventral Axis
  Proteins are asymmetrically distributed
  – mRNA localized in specific region of cell
  – Protein may be distributed asymmetrically by transport system or by selective degradation in certain cell regions

Question 26

Dorsal GENE

Answer 26

The Dorsal gene is transcribed and translated in the ovary so egg contains the mRNA and protein.
Dorsal protein is distributed to the nuclei of cells that will become the ventral side and remains in the cytoplasm of cells that will be the dorsal side
-Inside the nucleus, the dorsal protein acts as a transcription factor to regulate other genes important in ventral structure development. Absence of this gene in the dorsal nuclei allows genes for dorsal structure development to be expressed.

Question 27

Anterior-Posterior Axis Determination by Maternal Genes

Answer 27

• Set up by maternal effect genes bicoid and nanos to regulate zygotic translation of hunchback and caudal
• Hunchback and caudal mRNAs are distributed uniformly throughout the oocyte
• Bicoid RNA accumulates at anterior end
• Nanos RNA accumulates at posterior end

Question 28

Bicoid

Answer 28

–improtant in Anterior-Posterior Axis Determination
Bicoid mRNA is produced in ovary and becomes anchored to the anterior end of the egg. It then is translated.
• High concentration of bicoid protein at anterior end since its mRNA is located there and since protein has short life.
• Bicoid acts as a transcription factor and stimulates production of many genes including hunchback
• Bicoid inhibits translation of caudal RNA by binding to the 3’ UTR region of the caudal RNA.
• Hunchback protein is produced from its RNA and then acts as a transcription factor to regulate genes for differentiation in the anterior of embryo

Question 29

Nanos

Answer 29

-improtant in Anterior-Posterior Axis Determination
Nanos mRNA is deposited in the posterior of egg. After fertilization, it is translated and also has gradient along anterior posterior axis.
• Inhibits Hunchback translation by causing degradation of its mRNA causing anterior-posterior hunchback protein gradient
• Caudal RNA is translated in the posterior end of the embryo
• Caudal protein acts as a transcription factor to regulate genes for differentiation of the posterior part of the embryo.

Question 30

Segmentation Genes

Answer 30

• Affect number and organization of segments
• Transcribed after fertilization so no maternal effect
• Bicoid/Nanos gradient regulates these genes
• Gap Genes divide embryo into broad segments
• Pair-Rule Genes affect same part of the pattern in every
other segment
• Segment Polarity Genes affect anterior/posterior polarity of each segment

Question 31

Homeotic GENES

Answer 31

• Give specific identity to each segment
• 2 major gene clusters
– Antennapedia complex: Head and anterior thorax (5 genes)
– Bithorax complex: posterior thorax and abdomen (3 genes) – All on same chromosome
– Genes in order from anterior to posterior effect
– Each gene is “on” only in specific segments based on concentration of earlier gene products
-Found in most animal phyla
• Have common DNA sequences called homeoboxes (180 bp long)
• The region of the protein formed from the homeobox DNA is called the homeodomain (60 amino acids long)
• Proteins containing the homeodomain are DNA binding proteins. The homeodomain binds to specific DNA sequences and is thought to regulate transcription.