3) A test-cross is used to determine the genotype of an individual with a functi

Question

3) A test-cross is used to determine the genotype of an individual with a functional phenotype. An organism with a functional phenotype (but unknown genotype) is mated to a homozygous individual with a non-functional phenotype and the progeny are examined. If a red-flower producing plant is test-crossed to a homozygous white-flower producing plant, what would the phenotypic ratios be if: A) The red flowers were RR B) The red flowers were Rr? Punnett square please and detail information Thank you

9) Hunchback serves as a silencer protein for the gene encoding Ultrabithorax. A) Which of these two genes is epistatic to the other? B) How do you know?

Please give detailed information Thank you

Explanation / Answer

3) A test-cross is used to determine the genotype of an individual with a functional phenotype. An organism with a functional phenotype (but unknown genotype) is mated to a homozygous individual with a non-functional phenotype and the progeny are examined. If a red-flower producing plant is test-crossed to a homozygous white-flower producing plant, what would the phenotypic ratios be if: A) The red flowers were RR B) The red flowers were Rr?

Answer- A test cross is a cross between an individual with dominant phenotype and an individual with a recessive phenotype , to see if the individual with dominant phenotype is homozygous or heterozygous, test cross is done.

Here in case- A the red flowers were RR , so when these RR that is homozygous dominant type crossed with rr the homozygous recessive than the result will be as below in the Punnett square between RR X rr

R

R

r

Rr

Rr

r

Rr

Rr

So the offspring will be 100% Red having the genotype Rr that is heterozygous dominant for red color.

Case-B, when the Rr genotype crossed with rr type = Rr X rr

The test cross would be as below, in the Punnett square

R

r

r

Rr

rr

r

Rr

rr

Here half of the offspring will be red having a genotype of Rr and half of the offspring will be white having genotype rr that is homozygous recessive.

so the phenotypic ratio will be 1:1.

Rr (50%):rr(50%).

Q9) Hunchback serves as a silencer protein for the gene encoding Ultrabithorax. A) Which of these two genes is epistatic to the other? B) How do you know?

Answer- Hunch beck has (hb) gene and Ultrabithorax has (Ubx) gene. Ubx gene regulates the decision regarding the number of wings and legs the adult flies will have.

Epistatic gene is a gene whose presence, suppresses the effect of another gene present at another locus. These are also called the inhibiting genes.

Here in this case hb (hunchback) suppresses the expression of Ubx (Ultrabithorax) gene, Ubx is a homeotic gene used in the regulation of patterning in morphogenesis of Drosophila.

Ubx is activated when there is lack of hb (Hunchback) protein. In Drosophila high concentration of hb is found in the anterior and posterior part of embryo, so Ubx expresses itself in the middle region. This shows the epistatic nature of hb.

So here hb is the epistatic gene, which suppresses the effect of Ubx.

R

R

r

Rr

Rr

r

Rr

Rr

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