BISC 202 Lecture Notes - Lecture 7: Mns Antigen System, Glycoprotein, Zygosity

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whiterat187

8 Oct 2017

School

Simon Fraser University

Department

Biological Sciences

Course

BISC 202

Professor

Carol Wenzel

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Document Summary

A(cid:374) itali(cid:272) upper(cid:272)ase letter (cid:894)d(cid:895: recessive alleles are usually indicated by: A(cid:374) itali(cid:272) lo(cid:449)er(cid:272)ase letter (cid:894)d(cid:895: if (cid:374)o do(cid:373)i(cid:374)a(cid:374)(cid:272)e e(cid:454)ists, itali(cid:272) upper(cid:272)ase letters a(cid:374)d supers(cid:272)ripts are used to de(cid:374)ote alter(cid:374)ati(cid:448)e alleles (r1, r2, cw, cr) The (cid:449)ildt(cid:455)pe allele is ofte(cid:374) just (cid:449)ritte(cid:374) as (cid:858)+(cid:859) or as the ge(cid:374)e s(cid:455)(cid:373)(cid:271)ol (cid:449)ith a supers(cid:272)ript. I. e. , there is a (cid:373)utatio(cid:374) a(cid:374)d a phe(cid:374)ot(cid:455)pe (cid:272)aused (cid:271)(cid:455) the re(cid:272)essi(cid:448)e a allele a(cid:374)d the wildtype a+ allele is dominant: ge(cid:374)ot(cid:455)pes: Ca(cid:374) (cid:271)e (cid:449)ritte(cid:374): a/a or a+/a or +/a [they all tell the same information] Slashes are helpful to disti(cid:374)guish (cid:271)et(cid:449)ee(cid:374) the t(cid:449)o (cid:373)e(cid:373)(cid:271)ers of the ho(cid:373)ologous pair of chromosomes. Carrier predi(cid:272)tio(cid:374)s i(cid:374) a pedigree (cid:1006)/(cid:1007):(cid:1005)/(cid:1007: more (cid:272)o(cid:373)ple(cid:454) o(cid:374)es. But this is (cid:374)ot al(cid:449)a(cid:455)s the (cid:272)ase i(cid:374) (cid:374)ature When the expected phenotypes or ratios are not observed: i(cid:374)tera(cid:272)tio(cid:374)s (cid:271)et(cid:449)ee(cid:374) alleles of a si(cid:374)gle ge(cid:374)e- variations on dominance. Lethal (cid:373)utatio(cid:374)s (cid:894)re(cid:272)essi(cid:448)e lethal alleles; so(cid:373)e (cid:272)ause dominant visible phenotypes: i(cid:374)tera(cid:272)tio(cid:374)s (cid:271)et(cid:449)ee(cid:374) ge(cid:374)es/protei(cid:374)s.

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

BISC 202 Lecture Notes - Lecture 7: Mns Antigen System, Glycoprotein, Zygosity

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