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
Answer:
bb would be the phenotype white fur
Explanation:
B is dominant over b. It takes two recessive alleles to produce the phenotype of white fur.
Answer:
C. Four Haploid Sperm Cells
Explanation:
Meiosis is a cell division measure that creates four haploid cells (gametes) from a solitary diploid germline cell. In males, meiosis is essential for spermatogenesis, the cycle to produce sperm. Sperm are the haploid cells that are produced in meiosis.
Meiosis is a profoundly moderated cell division measure in eukaryotes that produces 4 cells (gametes) from a solitary cell (mother). Nonetheless, Nondisjunction in the male of Drosophila causes much of the time sperm with an addition or loss of chromosomes.
Answer: B) are excited by light that falls in the center of their receptive field.
Explanation:
There are two types of bipolar cells present in the retina which are differentiated on the way the centres of their receptive fields respond to light. The names of bipolar cells are ON-centre and OFF-centre.
If a light fall at the centre of the receptive field of a bipolar cell, it will experience an exciting effect on that cell, resulting in it being depolarized, it is an ON-center cell.
Hence, the correct option is B.