Answer:
It depends on the lens of the Anthony´s microscope, but It is likely that he saw artifacts
Explanation:
An optical microscope contains lenses of focal lenght that enable to see different structures. In this case, it is likely that Antony saw artifacts since living cells are not easy to focus with the optical microscope, it requires operator experience
I'm having trouble on this too :(
Chromosome duplications and deletions frequently result in abnormal phenotypes or inviable gametes. Gene dosage is modified is a major contributor to this phenomenon.
In the field of genetics, we can define gene dosage as the quantitative measure or copies of a particular gene that is present in an organism. Abnormalities in the gene dosage at a particular location can cause severe damage to the resulting phenotype.
Gene dosage can lead to chromosome duplications if the copy number or gene product is more and it can cause deletions if the copy number or gene product is less. Such complications will result in abnormal phenotypes or inviable gametes. For example, in Down's syndrome, the person has a modification of the 21st chromosome as there is one extra 21st chromosome present. This leads to a variety of diseases and defects in the person.
Although a part of your question is missing, you might be referring to this question:
Chromosome duplications and deletions frequently result in abnormal phenotypes or inviable gametes. Which factor is a major contributor to this phenomenon?
a. Recessive diseases are unmasked by additional copies.
b. The genes are found in a novel arrangement.
c. Gene dosage is modified.
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Answer:
The correct answer is 160.
Explanation:
Based on the given question, the total number of fruit flies is 1000, of which red eye flies are 640. Thus, the number of sepia eye flies will be 1000-640 = 360.
On the basis of the Hardy-Weinberg equation, p+q = 1, in which p is the rate of recurrence of one allele, that is, dominant and q is the rate of recurrence of the other allele, that is, recessive.
q^2 for the population is 360/1000 = 0.36, q = 0.6.
Therefore, p = 1-q = 1-0.6 = 0.4.
The frequency of homozygous dominant allele is denoted by p^2 = 0.16.
Hence, the number of homozygous dominant for red eye color would be 16% of 1000 which will be equal to 160.