BIS 101 Chapter Notes - Chapter 2: Heterogametic Sex, Mitosis, Chromosome

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ceruleanelk285

29 Sep 2016

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Document Summary

Properties: a characteristic feature of an organism, such as size, color, shape, or enzyme activity. Gene discovery: the process whereby geneticists find a set of genes affecting some biological process of interest by the single- gene inheritance patterns of their mutant alleles or by genomic analysis. The heritable variants observed in an organism that differs from the wild type are mutants: individual organisms having some abnormal form of a property. The alternative forms of the property are called phenotypes (the form taken by some character in a specific individual; the detectable outward manifestations of a specific genotype). Mutation: the process that produces a gene or a chromosome set differing from that of the wild type; the gene or chromosome set that results from such a process. Polymorphisms: the occurrence in a population of multiple forms of a trait or multiple alleles at a genetic locus. The general steps of functional analysis by gene discovery:

Related Questions

In what phases is the genetic material in the cell correctly referred to as chromatids?

	A.	metaphase and telophase
	B.	anaphase and metaphase
	C.	interphase and telophase
	D.	interphase and prophase
	E.	metaphase and prophase

Consider two traits for an organism, determined by two genes, each of which is governed by at least two alleles. In the case of a dihybrid individual, the gametes formed will be of either the parental type or the recombinant type. Recombinant type gametes are formed because of

	A.	the principle of dihybrids.
	B.	multiple alleles.
	C.	heterozygosity.
	D.	incomplete dominance.
	E.	independent assortment.

An allele is

	A.	an alternate form of a gene.
	B.	always recessive.
	C.	the main factor determining a trait.
	D.	always one of a pair.
	E.	the dominant form of a gene.

After the DNA is replicated, and it condenses in prophase, two identical rods of DNA are seen. These are

	A.	spindle fibers.
	B.	kinetochores.
	C.	chromatids.
	D.	chromatin.
	E.	centromeres.

Special cells found in the gonads that give rise to gametes upon division are called

	A.	egg cells.
	B.	somatic cells.
	C.	germ cells.
	D.	stem cells.
	E.	basal cells.

Homozygous recessive loss of function mutations in any one of 4 different genes in C. elegans worms (genes A, B, C, or D) results in female worms that don't form proper egg-laying structures, called vulvas, as shown in Lines 2-5 of the data table below. In vulvaless females, the fertilized eggs hatch and baby worms develop inside the mother, killing her in the process.

	genotype	phenotype
1	AA BB CC DD (wild type)	normal vulva
2	aa BB CC DD	No vulva
3	AA bb CC DD	no vulva
4	AA BB cc DD	no vulva
5	AA BB CC dd	no vulva

You would like to figure out the order in which these genes act during the creation of the vulva (you cannot assume it will be A--->B--->C--->D as the gene names are arbitrary). Your friend tells you to perform epistasis analysis by making double mutants between the different homozygous recessive mutants and analyzing the phenotype of the double mutant. For example, she asks you to examine the phenotype of aabbCCDD (double mutant of genes A and B) and compare this to the single mutants to figure out whether gene A acts earlier than gene B, or vice versa. You know this won't work. Why not? Pick the ONE BEST choice:

a. The phenotype of the single mutant is already pretty severe. The double mutant will most likely be dead, therefore epistasis analysis won't be possible.

b. Each of the single mutants (aaBBCCDD) and (AAbbCCDD) has the same mutant phenotype i.e., no vulva. Epistasis analysis between any 2 genes is only possible when the mutant phenotypes for each gene is different.

c. Epistasis analysis involves making triple mutants (such as aabbccDD) in order to learn something about how the genes are ordered.

d. Epistasis analysis can never be carried out with null loss of function mutations. The mutations being analyzed all have to be dominant gain of function alleles.

To address your concern, you first generate overactive alleles of either gene A (denoted as A*) or gene B (denoted as B*). As shown in lines 6-7 of the table below, C. elegans that have one of these overactive alleles produce multiple vulvas.

	Genotype	Phenotype
6	A*A BB CC DD	multiple vulvas
7	AA B*B CC DD	multiple vulvas

To find out the order in which these genes act, you combine the overactive alleles with different loss of function alleles and observe the phenotype in double mutants (see Lines 8-11).

	Genotype	Phenotype
8	A*A bb CC DD	multiple vulvas
9	A*A BB cc DD	multiple vulvas
10	A*A BB CC dd	no vulva
11	AA B*B cc DD	multiple vulvas

Based on the data in the table, what is the order in which these 4 genes normally act in wild-type C. elegans in order to produce a wild-type/normal vulva?

a. A----> B----->C----->D ---> Vulva

b. D----> B----->C----->A----> Vulva

c. B----> C----->A----->D----> Vulva

d. C----> B----->A----->D----> Vulva

e. C----> B----->D----->A----> Vulva

f. don't have enough data to make any conclusions

Based on the vulva formation phenotype of the double mutants, which of the following statement(s) accurately describes the genetic interactions between the A* allele and alleles of other genes affecting vulva formation? Pick ALL that apply:

a. The A* allele is epistatic to homozygous recessive loss of function mutations in gene B

b. A homozygous recessive loss of function mutation in gene B is epistatic to the A* allele

c. Homozygous recessive loss of function mutation in gene D is epistatic to the A* allele

d. The A* allele is epistatic to homozygous recessive loss of function mutation in gene D

e. The A* allele enhances the homozygous recessive loss of function mutation in gene D

QUESTION 1

Jim is blood type A, his brother John is type O and his sister Jane is type AB. What are the genotypes of their parents?