BIOC33H3 Lecture Notes - Allergic Rhinitis, Allergic Contact Dermatitis, Type I Hypersensitivity

44 views4 pages

ceruleanllama424

25 Mar 2013

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Chapter 14: genetics, altered immune responses, and transplantation. The different types of genetic testing include direct testing, linkage testing, biochemical testing, and karyotyping. Gene therapy is an experimental technique used to replace or repair defective or missing genes with normal genes. Stem cells are cells in the body that have the ability to differentiate into other cells. Stem cells can be divided into two types: embryonic and adult. Immunity is a state of responsiveness to foreign substances such as microorganisms and tumor proteins. Immune responses serve three functions: defense, homeostasis, and surveillance. Immunity is classified as innate (natural) or acquired. Acquired immunity is the development of immunity, either active or passive. The immune response involves complex interactions of t cells, b cells, monocytes, and neutrophils. These interactions depend on cytokines (soluble factors secreted by wbcs and a variety of other cells in the body) that act as messengers between the cell types.

Related Questions

Answer the following Linked to sex genetics problems:

1.) In Drosophila the recessive mutation linked to the X chromosome, scalloped ( sd ), causes the edge of the wings to be irregular. Present the results (you must prepare the crossing scheme) phenotypic for F1 and F2 for a cross between:

A) female scallop and a normal male (regular edges)

B) A normal (homozygous) female and a scallop male .

C) Compare these results with those that would be obtained if scallop were an autosomal recessive mutation.

2.) In Drosophila the autosomal recessive ebony mutation (e) is located on chromosome 3 and produces a dark body in the fly that expresses it. What will be the phenotypic proportions for F1 and F2 that will be obtained from a cross between a scallop female and homozygous for a normal body, with an ebony male with normal wings? Solve by the bifurcated lines method.

3.) In Drosophila the recessive mutation linked to the vermilion X chromosome (v) produces bright red colored eyes in contrast to the wild phenotype that has brick red eyes. An autosomal recessive mutation, suppressor vermilion (su-v) causes homozygous or hemizygous males to females present vermilion red brick eyes. In the absence of the vermilion allele su-v has no effect on the phenotype of the eye. Determine the genotype and phenotype of F1 and F2 at a cross between a homozygous female for vermilion and su-v, with a vermilion male and wild homozygous for su-v (su-v + su-v + ) . Present the diagram of bifurcated lines for these crosses.

4.) At a crossroads with Drosophila involving recessive mutations white (linked to the X chromosome) and sepia (autosomal recessive) that produces dark eyes, predict the results of F1 and F2 when crossing pure parents with the following phenotypes.

A.) White females with sepia males.

B.) Sepia females with white males

5.) Two mothers have given birth to their children at the same time in an urban hospital with an excess of patients. The child of couple 1 has hemophilia, a condition caused by a recessive allele linked to the X chromosome. Neither parent has the condition. Couple 2 has a normal child even though the father is a hemophiliac. Several years later, couple 1 filed a lawsuit alleging that the babies were changed after birth and you are called to testify as a genetic counselor. What information would you give the jury in relation to the allegation of the couple 1?

Question 1 Select one answer.

Two heterozygous individuals are crossed. What percentage oftheir offspring will be heterozygous?

a)50%

b)75%

c)100%

d)25%

Question 2 Select one answer.

In peas purple is dominant to white for flower color. A purpleflower is crossed with a white flower. All the offspring (over 100plants) produce purple flowers. What is the genotype of the purpleflower?

a)pp

b)Either PP or Pp

c)PP

d)Pp

Question 3 Select one answer.

A heterozygous germ cell undergoes meiosis. The gametes will

a)only have the recessive allele

b)only have the dominant allele

c)either have the dominant or the recessive allele

d)have both the dominant or the recessive allele

Question 4 Select one answer.

Sex-linked disorders

a)are always dominant are

b)always recessive

c)can be either dominant or recessive

d)are neither dominant nor recessive

Question 5 Select one answer.

A nucleus is built around chromatids which have separated. Thisoccurs during which stage of meiosis?

a)Telophase II

b)Metaphase I

c)Metaphase II

d)Telophase I

Question 6 Select one answer.

The condition of having anincorrect number of chromosomes is called ____________.

a)nondisjunction

b)aneuploidy

c)epistatis

d)trisomy

Question 7 Select oneanswer.

In humans females are

a)XY

b)YY

c)none of these

d)XX

Question 8 Select oneanswer.

Autosomal means that the gene isfound on the

a)X chromosomes

b)the Y chromosome

c)either the X or the Ychromosome

d)neither on the X nor theY

Question 9 Select oneanswer.

What organism did Mendelstudy?

a)Roses

b)Ants

c)Peas

d)Yeast

Question 10 Select oneanswer.

Huntington's disease is anautosomal dominant disorder. If two individuals who areheterozygous at this locus breed, what are the chances that theiroffspring will develop Huntington's?

a)100%

b)75%

c)50%

d)25%

Question 11 Select oneanswer.

Plum gametes have 24 chromosomes.How many chromosomes would a zygote have?

a)12

b)24

c)36

d)48

Question 12 Select oneanswer.

Hydrangeas are blue in acidicsoil with available aluminum, and they are pink in alkaline soilwithout available aluminum, but they cannot be orange under anysoil condition. The color of hydrangeas is affect by

a)genetics only

b)the environment only

c)genetics and theenvironment

Question 13 Select oneanswer.

Somatic mosquito cells have 6chromosomes. How many chromosomes do gametes have?

a)3

b)6

c)12

d)24

Question 14 Select oneanswer.

Alleles are different versions ofa

a)Gene

b)Phenotype

c)Genotype

d)Chromosome

Question 15 Select oneanswer.

A female producesgametes:

a)that have either an X or a Ychromosome

b)That have 2 Xchromosomes

c)that have a Y chromosomeonly

d)that have an X chromosomeonly

Question 16:

use the following infromation to answer the next 4 questions

In dogs, colored fur (E) is dominant to un-colored fur (e).Blackfur color (B) is dominant to brown fur color (b). The E gene isepistatic to the B gene. A yellow male dog is mated to a blackfemale. the male's parents were both brown dogs. The female'smother was yellow and her father was brown

Q1 of 4: what is the genetype of the male dog in thisexample

a)eeBb

b)eeBB

c)ee

d)eebb

Q2 of 4: What is the genotype of the female dog in thisexample?

a)EeBb

b)EEBB

c)EeBB

d)EEBb

Q3 or 4: What percentage of their puppies will be yellow?

a)100%

b)75%

c)50%

d)25%

Q4 of 4: what percentahe of the puppies will be homozygous forthe B gene.

a)100%

b)75%

c)50%

d)25%

carmineprairie-dog596

Some people suffer from adenosine deaminase (ADA) deficiency.ADA is an enzyme necessary for the immune systems T-cells tofunction properly. Helper T-cells activate B cells which secreteantibodies which are proteins that help fight against infection.When ADA is lacking due to an autosomal recessive mutation, theimmune system is extremely impaired and afflicted individuals aresusceptible to chronic infections. There are several differenttechniques that have been utilized in an attempt to treat ADAdeficient individuals: Chelsea received a bone marrow transplantfrom her father at four months of age. His healthy cells populatedher bone marrow, replacing her defective cells with normal, healthycells that produce the needed enzyme. She is now doing quite welland has functioning T and B cells in circulation. Michael receivedADA manufactured using recombinant DNA technology. Unfortunately,the enzyme stays in the blood for only a few minutes before beingbroken down - not long enough to restore immunity. Michael hassince passed away from a fatal viral infection. Shelley wasselected for gene therapy when a bone marrow donor could not befound for her. Her T-cells were separated and treated with aretrovirus which delivered a functional ADA gene to the T-cells.These T-cells were then re-infused back in to her body. Her immunesystem gradually began to function, but the treatment had to berepeated every few weeks as the genetically altered T-cells beganto die off. Todd had gene therapy like Shelley, but it wasperformed on hematopoietic stem cells from his bone marrow ratherthan on mature, circulating T-cells. Todd soon had many normallyfunctioning T cells and antibodies circulating in his blood for thevery first time. He even grew tonsils, which are typically absentin folks with ADA deficiency. Todd's ADA activity rose to 25% ofnormal levels, enough to provide limited immunity. To date hehasn't required any additional gene therapy.

Shelley needed to repeat her treatment (gene therapy) whereTodd and Chelsea did not because:

		Shelley s treatment altered only the phenotypic deficiency notthe genotypic mutation.
		Shelley s treatment altered the genotypic deficiency not thephenotypic mutation.
		Shelley s treatment involved germ-line therapy, which is moretransient.
		Shelley s treatment didn't activate the proper genes in herT-cells.

silverworm125

BIOC33H3 Lecture Notes - Allergic Rhinitis, Allergic Contact Dermatitis, Type I Hypersensitivity

Document Summary

Get access

Related Documents

EDP 132 Chapter 3: Chapter 3: Lifespan Development ed. 6

NURS312 Lecture Notes - Lecture 15: Frameshift Mutation, Genetic Disorder, Autosome

ATS2458 Lecture Notes - Lecture 3: Mendelian Inheritance, Haplotype, Autosome

Related Questions