BIOLOGY 305 Chapter Notes - Chapter 2: Wild Type, Zygosity, Gamete

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silvermoose544

22 Aug 2016

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Mutants: individual organisms having some altered form of a normal property. Wild type: normal form of any property of an organism, which is found in the wild, or in nature. The genetic approach to understanding a biology property is to discover the genes that control it. One approach to gene discovery is to isolate mutants and check each one for single-gene inheritance patterns (specific ratios of normal and mutant expression of the property in descendants) Mendel"s law of equal segregation: a hereditary factor called a gene is necessary for producing pea color, each plant has a pair of this type of gene, the gene comes in two forms called alleles. If the gene is phonetically called a wye gene, the two alleles can be represented by y (standing for yellow phenotype) and y (standing for green phenotype: a plant can be either y/y, y/y, or y/y. The slash shows that the alleles are a pair.

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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

In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u.. long. The X-linked gene for body color-with two alleles, y ⁺for gray body and y for yellow body-resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w ⁺for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f ⁺ for normal bristles and f for forked bristles, is located at map position 56.7. Each gene resides on the Xchromosome, and at each locus the wild-type allele is dominant over the mutant allele.

Predict the frequency of gray with white eyes and forked bristles progeny type produced by this mating.

Enter your answer as a percentage to three decimal places (example 0.235 or 0.230).

Frequency=_________

Phenotype	Number
Pale	648
Pale, oval	64
Pale, short	10
Pale, oval, short	102
Oval	6
Oval, short	618
Short	84
Wild type	98
	1630

A wild-type trihybrid soybean plant is crossed to a pure-breeding soybean plant with the recessive phenotypes pale leaf l, oval seed (r), and short height (t). The results of the three-point test cross are shown in the table. Traits not listed are wild type.

Calculate the interference value for these data.

Enter your answer to three decimal places (example 0.201).

scarletbadger940

If an organismâs diploid chromosome number is 18, how many different possible combinations of homologous chromosomes lining up during meiosis exist for the eggs or sperm produced by that organism?

	A.	512
	B.	9
	C.	18
	D.	128
	E.	36

At the end of metaphase I, _______________ separate.

	A.	sister chromatids
	B.	germ cells
	C.	homologous chromosomes
	D.	haploid chromatids
	E.	centrioles

Mendel observed that dominant traits

	A.	are seen in all of the F1 hybrid pea plants in his experiments.
	B.	are expressed in all plants.
	C.	were absent in the F1 generation of pea plants that he used in his experiments.
	D.	were the only traits seen in the F2 generation of pea plants in his experiments.
	E.	are only expressed in hybrids.

Cytokinesis in plant cells differs from cytokinesis in animal cells because

	A.	there is no difference.
	B.	in plant cells, the cell plate must also divide into two parts.
	C.	the contractile protein, actin, is important only in plant cells.
	D.	plant cells have a rigid cell wall.
	E.	a contractile ring forms only in plant cells.

Sickle cell anemia is an example of what type of inheritance?

	A.	complete dominance
	B.	incomplete dominance
	C.	codominance
	D.	multiple alleles
	E.	recessive dominance

Which of the following statements is true:

	A.	The dominant allele is masked in homozygous dominant individuals.
	B.	With recessive genetic disorders, if both parents are carriers, the offspring will all be affected.
	C.	In carriers, the recessive allele causes an intermediate phenotype.
	D.	In recessive genetic disorders, the mother and/or father of an affected individual must also be affected.
	E.	With dominant genetic disorders, the mother and/or father of an affected individual must also be affected.

Skin cancers typically develop in the

	A.	upper layers of the epidermis.
	B.	lower layers of the dermis.
	C.	subcutaneous layer.
	D.	lower layers of the epidermis.
	E.	upper layers of the dermis.

Duchenne muscular dystrophy (MD) is inherited from an X-linked recessive allele. What is the probability that a son with Duchenne MD inherited this disease from his biological father?

	A.	1/2
	B.	0
	C.	1/16
	D.	1/4
	E.	1/8

The genetic makeup of a particular trait in an individual is its

	A.	genotype.
	B.	heterozygosity.
	C.	phenotype.
	D.	filial.
	E.	dominance.

What structure holds the sister chromatids to the spindle fibers?

	A.	chromatin
	B.	kinetochore
	C.	MPF
	D.	centromere
	E.	cyclin

If you view a cell in which the genetic material is beginning to be visible as separate bodies, and the nucleus has disappeared from view, you may surmise that the cells is in

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

Gregor Mendel was successful in his analysis of the genetics of pea plants because

	A.	he examined and analyzed both the F1 and F2 generations.
	B.	he studied the parental plants to determine their differences.
	C.	he decided to only look at his results in an objective manner.
	D.	he studied a trait that had a strange inheritance pattern.
	E.	pea plants have genetics different from other organisms.

Tall corn plants (T) are dominant to dwarf plants (tt). Solid green leaves (G) are dominant to leaves with a white tip (gg). A cross between two corn plants yielded the following phenotypes: 51 tall plants with a white tip on their leaves; 43 dwarf plants with solid leaves; 48 dwarf plants with white tips on their leaves; 45 tall plants with solid leaves. What are the genotypes of the parents that produced these plants?

	A.	None of the above
	B.	TtGg x Ttgg
	C.	ttGG x TTgg
	D.	TtGg x TtGg
	E.	TtGg x ttgg

In humans, a gene that has been identified as causing a type of skin cancer is the

	A.	superwoman echidna.
	B.	mutant superman.
	C.	sonic hedgehog.
	D.	mutant mole rat.
	E.	superhero aardvark.

The segregation principle states that in sexually reproducing diploid organisms the two copies of each gene

	A.	segregate from each other during meiosis.
	B.	must always be the same allele.
	C.	separate from each other during mitosis.
	D.	will both wind up in either the sperm or egg.
	E.	move together as a unit during meiosis.

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.

BIOLOGY 305 Chapter Notes - Chapter 2: Wild Type, Zygosity, Gamete

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