Answer:
what are the options?
may I get brainliest please?
Answer: Colon cancer
Explanation:
All cancers, but especially colon and rectal cancers, commonly referred to as colorectal cancer (CRC), have hereditary factors that potentially increase one’s risk.
Genetic testing is determine if there is an increased risk of colon cancer. Apolipoprotein E (apoE) plays a major role in the metabolism of bile acids, cholesterol and triglycerides, and has recently been proposed as being involved in the carcinogenic process. Given the potential role of bile acids in colorectal cancer etiology, it is reasonable that colorectal cancer risk might be modified by apoE genotype. The absence of an e3 apoE allele significantly increased the risk of colon cancer especially for individuals above 50years.
A great amount of Antarctic animals feed on them. Whales, penguins, fish, birds, etc. They eat phytoplankton.
In ecology, the law of conservation of mass is applied by the principle of taking energy from the lower trophic levels and passing it to the higher trophic level by any means.
<h3>What is the law of conservation of mass?</h3>
The law of conservation of mass states that the actual mass in an ecosystem is neither created nor destroyed. Overall remains constant in a system.
In ecology, it illustrates that when an organism is ingested by other organisms, its mass is conserved. Some of the organisms may also be dependent on the waste products like urine and feces to make proteins and other compounds.
Therefore, the energy or mass is neither lost in any form. It simply is taken by other organisms through any means.
To learn more about Ecological principles, refer to the link:
brainly.com/question/7413811
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<span>In Drosophila + indicates wild-type allele for any gene, m is mahogany and e is ebony.
Female parents are m+/m+ and males are +e/+e.
F1 are m+/+e, all wild type. F1 females are crossed with me/me males - the test cross.
Offspring will be : non recombinant m+/me, mahogany wild type or +e/me wild type ebony. OR
recombinant me/me mahogany ebony or ++/++ wild type.
As the two genes are 25 map units apart, the percentage of recombinants will be 25% and therefore percentage parental types will be 75%.
75% 1000 is 750. There are two parental types, so you would expect 375 of each. Therefore, you would expect 375 m+/me and 375 +e/me.
25% of 1000 is 250 split between two recombinants =125 of each. Therefore you would expect 135 me/me and 125 ++/++</span>
