Answer:
The probability of sampling a homozygous recessive individual from a population of 100 tree frogs indicates that, the frequency of latent allele is 0.06 for example q2 is 0.06. Thus, it has been implies that each 6 tree frogs have the passive allele out of 100. So, the likelihood of accepting the tree frogs with homozygous passive tree frogs will be essentially 6%.
B. A species that does not normally live in an area
Answer:
Rudolf Virchow
Explanation:
During his studies on inflammatory exudates, Rudolf Virchow observed that the increased number of cells in the exudates was due to the formation of new cells by pre-existing cells. He also studied that the formation of new cells from the preexisting cells is responsible for wound healing and for normal growth and development of living beings.
Answer:
7%.
Explanation:
The recombinant progeny might occur due to the exchange of genetic material between non sister chromatids known as crossing over during the meiosis.
The chromosome sequence is DABC. A and B distance is 4 map units, B and C is 2 map units, B and D is 5 map units. The distance between A and D is 1 map unit that can be calculated by subtracting the distance between BD and AB ( 5-4). The CD map distance can be calculated by adding the distance between AD, AB and BC ( 1 + 4+ 2). The recombination frequency is equal to distance between them is 7%.
Thus, the answer is 7%.
I’m sorry but I do not understand the question. Furthermore, if you explain I will gladly help my guy :)
