BIOL10005 Lecture Notes - Lecture 8: Sister Chromatids, Mendelian Inheritance, Epistasis

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plumrabbit83

31 Oct 2018

School

University of Melbourne

Department

Biology

Course

BIOL10005

Professor

Dawn Gleeson

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Document Summary

Shows if a gene is assorting independently or autosomally linked. If the genes are assorting independently -> ratio of 1:1:1:1: gene interaction. When two or more genes influence the same phenotypic trait e. g. ) One gene coats for coat colour (b= black b= brown) Other gene coats for the intensity of the colour (d= dense pigment d= dilute) An allele at one gene locus masks the phenotypic expression of the alleles of another gene locus. Can be dominant or recessive: recessive epistasis. When homozygous ee is present, no matter what the other gene (b or bb) is present, the coat colour is always cream: dominant epistasis. Genes on the same chromosome does not assort independently. Cis linkage: when both dominant alleles are on the same chromosome. During meiosis, cross over takes place which leads to formation of ab ab in trans arrangement. Crossing over (chiasma) -> exchange of genetic materials between homologous pairs or sister chromatids.

Related Questions

Q.In the non-9:3:3:1 ratio dihybrid cross below, there are 3 classes of phenotypes: white, pink, red which of the following is NOT true?

9 A_B_ white

3 A_bb pink

3 aaB_ red

1 aabb red

a.) phenotype is dependent on the genotype of gene B

b.) This is an example of recessive episasis

c.) aa is epistatic to B_, and bb

d.) phenotype is dependent on the genotype of gene A

Q,When choosing a selectable marker, you...

a.) choose a enhancer for your gene of interest

b.) choose a rescue gene to save your gene of interest

c.) choose a resistant gene that will allow for colony screening

d.) choose a gene that encodes for DNA replication

Q.The following steps detail an experiment resulting in transgenic mice with a knockout gene for fur color. Recall that agouti (brown) mice are A/A;M/M and black fur coats are a/a;M/M . List the steps in their correct chronological order.

1.Inject transformed agouti embryotic cells into an a/a;M/M embryo to produce A/A;M/m and a/a;M/M altered embryo

2.Self-cross hemizygous mice to produce homozygous transgenic mice (A/-;M/m)

3.Create a a targeting vector containing tk+ for ganciclovir sensitivity, neoR for neomycin resistance, and the gene encoding fur color.

4.Observe mice with chimeric phenotype and cross them with black a/a;M/M mice to produce transgenic hemizygous progeny (A/a;M/m)

5.Isolate all transformed embryonic stem cells that show homologous recombination and inject them into a embryonic stem cells of an agouti mouse.

a.) 3,5,4,1,2

b.) 1,5,3,4,2

c.) 3,5,1,4,2

d.) 5,3,2,4,2

If dog 1 has is black with alleles (A_B_) , and dog 2 is brown with (aaB_), and dog 3 is yellow with (_ _bb) alleles, what kind of epistasis is this and what color would dog 4 be if he was (AAbb).

1. Recessive epistasis, Yellow

2. Recessive epistasis, Black

3. Dominant epistasis, Yellow

4. Dominant epistasis, Black

A gene interaction in which pair of recessive genes at one locus prevents expression of dominant allele at another locus called

a. incomplete dominance

b. epistatis

c. complete dominance

d. polygenic inheritance

Problem A. You cross two homozygous parents: RRYY (round yellow) and rryy (wrinkled green) to produce an F1 generation. Round seeds are dominant, yellow seeds are dominant. You cross two F1s to produce the F2 generation. In the F2 generation, you find an individual with round yellow seeds. You do a testcross to find out the genotype of this âmysteryâ dominant/dominant phenotype. In the offspring of the cross, you find 1?2 have round yellow seeds and 1?2 have wrinkled yellow seeds.

1. Based on these data what was the genotype of the mystery F2?

Problem B. You carry out a dihybrid cross between a red-flowered rose with long pollen grains (A1A1 LL) and a white-flowered rose with round pollen (A2A2 l l). The flower color gene (A) shows incomplete dominance, and long pollen is completely dominant to round pollen. The genes for these traits are on different chromosomes so assort independently

2. What would be the phenotypes of the F1 generation?

3. What fraction of the F2 generation is expected to have red flowers and long pollen? (As always when traits are independent, what rule can you apply to save time?)

Problem D. In the example of epistasis presented in class, the agouti allele in mice (A) codes for striped hairs and is dominant to the allele for all black hairs (a). A loss-of-function allele for the pigment gene (b) produces no pigment and is recessive to the normal, dominant allele (B). The recessive allele at b masks the effect of the A gene. A dihybrid cross of AABB and aabb is carried out.

5. Name all of the genotypes in the F2 generation that can produce an albino (no pigment) phenotype.

Problem E. A cross is made between homozygous fruit flies that differ for eye color (the st gene) and

body color (the e gene). The F1 double heterozygote is shown below using chromosome notation.

st+ e

--------------------------------------

st e+

st = the recessive allele for scarlet eyes; st+ = the dominant wildtype allele for red eyes e = the recessive allele for ebony body; e+ is the dominant wildtype allele for gray body

6. Are the chromosomes shown in the genotype above in cis or trans configuration?

7. Based on this notation, what were the genotypes and phenotypes of the two parents of the F1? (Think carefully here...)

8. With what phenotype would you cross this F1 to measure recombination frequency?

cyansnail823

GENETICS : You perform a dihybrid cross between two dogs that are each BbFf. Where not specified otherwise, assume black fur B is completely dominant to white fur b and furry F is completely dominant to hairless f

1. What genotypes at one locus will be epistatic to what genotypes at the other locus, if hair color and hair presence were the only measured phenotypes? (you may find it helpful to write out the genotypes and the phenotypes each would produce, and remember that epistasis is involved in this question).

What phenotypic ratios would you expect in the following scenarios (assume, unless otherwise noted, that the epistasis in part 1 still applies)?

2. Black B is completely dominant to albino b, furry F is completely dominant to hairless f

3. Black displays incomplete dominance to albino , and furry is codominant with hairless

4. Animals that are heterozygous at either locus are black and furry, overriding the other locus' geneotype

5. The black allele B displays epistasis when present at even one copy that masks the effects of the hairless f allele

6. It takes two black B alleles to mask the effects of the hairless f allele

7. Both the black alleles B and furry alleles F are lethal in their homozygous dominant states

8. Both the albino alleles b and hairless alleles f are lethal in their homozygous recessive states

9. Female dogs die when they have homozygous recessive or homozygous dominant states for either locus (dogs use XY inheritance for sex like we do). Note sex of doss in your phenotype ratios.

10. The black allele B although completely dominant, is the only 70% penetrant in heterozygotes and is 90% penetrant in homozygotes for the B allele

11. The hairless allele f is dominant in males and recessive in females

BIOL10005 Lecture Notes - Lecture 8: Sister Chromatids, Mendelian Inheritance, Epistasis

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