BIOL 205 Chapter Notes - Chapter 2: Chromosome, Petal, S Phase

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rosehornet886

23 Oct 2017

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Starts with the hunt for mutants in which the function under investigation is altered or destroyed: each mutant must be tested to see if it"s caused by single-gene mutation. Mate mutants w wild-type, then analyse f1 and f2. ** use of mutants in this way is referred to as genetic dissection --> property in question is picked apart to reveal its underlying genetic program. Forward genetics: strategy to understanding biological function starting w random single-gene mutants and ending w their dna sequence and biochemical function. The genetic approach to understanding a biological 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) Self pollination: carried out by allowing pollen from flower to fall onto its own stigma.

Related Questions

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

Canaries: Green (wild-type) Pied phenotype Yellow type

Canaries are popular pets and easy to breed in captivity. Additionally, they display great phenotypic diversity, which allows you to do some interesting genetic investigations with them.

When you take a canary from a true-breeding green strain and cross it with a canary from a true-breeding yellow strain, all offspring display the pied phenotype. Based on this you suspect that this color variation is inherited by a gene with a co-dominant inheritance pattern (showing both phenotypic traits in hybrid phenotypes).

You decide to test this hypothesis. To do so, you take pairs of your hybrid F1 individuals (of pied phenotype), and cross them to produce an F2 generation.

After many dozens of crosses and hundreds of offspring, you arrive at the peculiar ratio of:

3/16 green : 6/16 pied : 7/16 yellow

Which of the following hypotheses are consistent with these data? (Note that not all of these hypotheses are mutually exclusive. Assume that the two genes are independently assorting)

Hint: This problem is very hard to work from the given ratios to what is happening genetically. It is easier to "work backwards" and start with each choice below, and see if you can obtain all or important aspects of the results of your cross when making the assumption that the answer choice is correct.

Select all that apply:

Select one or more:

a. At least one of the genes involved is sex linked.

b. One gene controls production of darker pigment (melanin which appears green when mixed with yellow). A second gene acts as a patterning gene with one allele causing expression and the other non-expression of melanin and when in the heterozygous condition each allele controls melanin expression on different parts of the body so long as the bird's genotype of the first gene allows for the production of melanin.

c. this is a single-gene trait with co-dominance

d. one of the two genes involved displays complete dominance, while the other displays co-dominance.

e. the genes involved are epistatically interacting

f. more than one gene is involved

The Case: While Elizabeth is reading the morning newspaper, she notices an ad for a free genetic screening for breast cancer at the clinic next week.The ad specifically invites women of Ashkenazi Jewish ancestry to participate.According to the newspaper ad, subjects will be tested to see whether they have mutations in the BRCA1 gene which would predispose them to breast cancer. Elizabeth, age 27, had heard about the discovery of the gene and about the mutation linked to Jewish women. Her paternal grandmother had been diagnosed with breast cancer at age 51 and died two years later, and Elizabeth worried that she may have inherited a variant that would put her at increased risk. She also worried about her mother, age 52 and apparently cancer-free so far, and her 7- year old daughter. Her daughter is not allowed to participate in the screening, but Elizabeth convinces her mother to go with her to get tested. For the screening, a small amount of blood is drawn. DNA is isolated from the blood, and part of the BRCA1 gene is amplified by PCR. Using a Southern Blot method, the digested DNA is run on a gel, transferred to a neutral membrane. Once on the membrane, the samples are hybridized with specific probes corresponding to deleterous mutations known to be linked to increased breast cancer susceptibility, plus probes for wild type regions of BRCA1. The probe will only bind to the DNA if that mutation is present. Control DNA samples, known to have a specific mutation, are included in the run to verify probe . Probe binding is visualized by labels attached to the probe, and will depend on the lab running the analysis. Probes: 185delAG (AG deletion in exon 2) Normal 185 (no mutation at this site) 4184delTCAA (TCAA deletion in exon 11) Normal 4184 5382insC (C insertion in exon 13) Normal 5382 DNA samples: Elizabeth Mother Unrelated woman Control 185delAG Control 4184delTCAA Control 5382insC Control wild type BRCA1 (no mutations) If either Elizabeth or her mother test positive, is a prophylactic double mastectomy appropriate? (2 marks) What other issues does this type of testing raise, and how should these issues be addressed? (2 marks) What does it mean if a result came back as a Variant of Uncertain Clinical Significance? Provide a peer-reviewed journal article as a reference for this question. (1 mark answer, 1 mark reference) Does the lack of any of these mutations mean that the women will not get breast cancer? (2 marks) Who should have access to the test results? (2 marks) Does the daughter have the right to know the results? She is only seven now, but what about when she is 16? (3 marks) Should this type of screening be mandatory? (3 marks)

BIOL 205 Chapter Notes - Chapter 2: Chromosome, Petal, S Phase

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