What is the answer with explaining
1 answer:
Answer:
All EeWw (100%)
Explanation:
This question involves two genes; one coding for earlobes and the other for hairline. According to the question, one parent is homzygous recessive for earlobe but homzygous dominant for hairline i.e. (eeWW) while the other parent is homzygous dominant for earlobe but homzygous recessive for hairline i.e. (EEww).
The cross between the two parents is as follows: eeWW × EEww. The following gamete combination will be produced by each parent:
eeWW - eW, eW, eW, eW
EEww - Ew, Ew, Ew, Ew
Using this gametes in a punnet square, the following proportion of offsprings will likely be produced:
All EeWw
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Answer: The bacteria transformed with this particular plasmid will form white colonies on the plates containing ampicillin and Xgal.
Explanation: The lacZ gene produces an enzyme called β-galactosidase which is responsible for the breakdown of lactose into glucose and galactose. The lacZ gene is one of the three genes (the other two being lacA and lacY) of the lac operon which is responsible for the transport and mechanism of lactose in E. coli and many other bacteria.
In recombinant DNA technology, when a plasmid is to be used to transform a host cell, such markers are used to help screen the transformed cells from the ones that have not taken up the plasmid. Xgal present in the plates is an artificial substrate which is hydrolyzed by
β-galactosidase into 5-bromo-4-chloro-indoxyl which will dimerize and oxidise into 5,5'-dibromo-4,4'dichloro-indigo. This is a blue pigment which will give blue color to the bacterial cells. Introducing a DNA fragment in this lacZ gene will make it non-functional so it will not be able to produce the enzyme.
Therefore, when a bacterial cell is transformed with a plasmid containing ampicillin resistance gene and a DNA fragment introduced in the lacZ gene and then grown on plates containing ampicillin and Xgal, white colored colonies will appear. The white colonies will show the bacterial cells that have successfully taken up the plasmid with the DNA fragment incorporated in the lacZ gene as this will render the gene non-functional and will not produce β-galactosidase which will breakdown Xgal to give blue colonies. Since the plates contain ampicillin, only the bacterial cells that have been successfully transformed with the plasmid ( the ones that have the DNA fragment and the ones without it) will grow as the ampicillin resistance will give them resistance against ampicillin in the plates. The bacterial cells that have not taken up the plasmid will not be resistant to ampicillin and will not form colonies on the plate.
This is called blue-white screening which is used to identify successfully transformed host cells. A picture of this is given in the attachment, taken from the following website:
https://www.mun.ca/biology/scarr/Blue_&_White_Colonies.html
