Answer:
a) The genes present on separate chromosomes exhibit indepedent assortment.
b) The genotype of tall and purple flower parent= TTPp
The genotype of short and purple flower parent: ttPp
Explanation:
a) In a dihybrid cross, if the genes are present on the same chromosome, they will not exhibit independent assortment. The genes present on the same chromosome are called linked genes. Linked genes are inherited together. The absence of independent assortment of a linked gene gives more of parental types in F2 progeny. Therefore, if genes are not present on a separate chromosome, Mendel’s ratios deviate.
b) Alleles for tallness and purple flower are dominant. In the progeny, all the plants are tall. This means that the dominant tall plant is homozygous. With respect to flower color, the trait exhibits segregation of dominant and recessive alleles and both purple and white flower progeny are obtained. This makes both the plants heterozygous for the flower color gene.
The genotype of tall and purple flower parent= TTPp
The genotype of short and purple flower parent: ttPp
Answer:
The correct answer is -
1. totipotent
2 multipotent
3. pluripotent
Explanation:
The cells that are able to develop or differentiate a different kind of cells in the body of the individual are known as totipotent cells and this ability is known as totipotency. An example of such cells is found in the zygote.
Multipotent on the other hand are the cells that are able to develop into a specific kind of cell of tissues these cells are present and the example of this type of cell is adult stem cells. Whereas Embryonic stem cells are Pluripotent cells that are characterized as the cells that develop into any type of cell of the body of an organism.
The most logical answer that I can think from Darwin's Theory of Evolution is that that species had been isolated from other possible panthers to mate with. So after generations of that isolated group of panthers their offspring would eventually become increasingly similar.
How many chromosomes would be present in a cell after anaphase of Mitosis?
During anaphase, sister chromatids separate to opposite sides of the cell. Therefore the 4n after doubling returns to 2n at each end of the dividing cell after anaphase. But until cytokinesis (1 cell pinching into 2), it's still one cell therefore 4n = 28