Don't eat them? Or before cooking rinse them off before cooking ♀️
The correct answer is "megaloblastic anemia".
In the event of deficiency of folic acid and/or vitamin B12, there will be inhibition of DNA synthesis in red blood cell precursors as well as other cells in the myeloid lineage. This results to continuous cell growth without cell division leading to large, immature red blood cells (hence the term megaloblastic) with low hemoglobin levels. Vitamin B12 deficiency particularly causes neuropathic symptoms such as neuropathic pain and numbness.
If megaloblastic anemia is caused by the resection of the portion of stomach that produces the intrinsic factor (important in the absorption of vitamin B12), then the condition is called pernicious anemia.
Answer:
Nerve cells release chemical signals into synapses between them (short distance). They also transverse their lengths with an electrical signal that can result in signal travel along a series of cells (long distance).
Explanation:
Nerve cells release neurotransmitters in the synaptic cleft which are capable of affecting nearby cells such as other nerve cells and muscle cells. Neurotransmitter molecules include, among others, serotonin, acetylcholine, dopamine, norepinephrine and histamine. Moreover, the synaptic cleft is the space that separates a neuron cell and its target cell. On the other hand, neurons transmit signals through electrical impulses. Electrical impulses travel long distances in the body carried by axons of the nerves. Thus, nerve impulses connect the brain and spinal cord and they carry signals to different parts of the body.
Answer:
The genotype of the F1 was wy+/w+y.
Explanation:
One of the given options has a typo: the red eye-brown body offspring count should be 56 instead of 561.
<u>We have two genes with two alleles each:</u>
Red eyes (w+) is dominant over white eyes (w).
Brown body (y+) is dominant over yellow body (y).
The phenotypes of the F2 tesulting from a test cross (F1 x wy/wy) are:
- wy+/ey (white-eye, brown body): 670
- w+y/wy (red-eye, yellow body): 650
- wy/wy (white-eye, yellow body): 38
- w+y+/wy (red-eye, brown body 56
If the genes w and y are linked, two phenotypes in the F2 will be much more abundant than the other two. Recombination during meiosis is a rare event, so the most abundant phenotypes are the parentals (the ones present in the F1 parent).
Every individual in the offpsring has a <em>wy</em> chromosome, as this was the gamete inherited from the test cross individual.
In this case, the most abundant gametes are wy+ and w+y, so the genotype of the F1 was wy+/w+y.
Notice how when recombination occurs in the F1 parent, the recombinant gametes appear: wy and w+y+, which are the less abundant in the F2 progeny.
