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ale4655 [162]
3 years ago
14

Which forms of viral hepatitis are transmitted through the fecal-oral route?

Biology
1 answer:
e-lub [12.9K]3 years ago
5 0

Answer:

Hepatitis A

Explanation:

  • Amongst the several types of viral hepatitis, it is hepatitis A that can spread through the fecal-oral route.
  • Usually, a person can acquire hepatitis A if he/she ingests contaminated food or water, further it can also spread through direct contact with the infected person.
  • due to its transmission through the fecal-oral route, lack of good sanitation practices and hygiene are the major causes for its spread.
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Approximately how much of Earth’s surface is water? 100% 75% 50% 25%
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3 years ago
Assume that the mother's genotype is AZ/az, and the father's genotype is Az/aZ, and the recombination rate is 10%. What are the
attashe74 [19]

Answer:

0.45 from mom and 0.05 from dad.

Explanation:

The recombinant progeny might occur due to the crossing over at the time of meiosis in which the exchange of chromosomes occur in the homologous chromosomes of non sister chromatids.

The progeny receives half of their chromosome from the mother and half from the father. The mother 50% AZ and 50% az chromosomes respectively. The 10% recombination has occured due to which the mother chromosomes is reduce upto 45%. The recombinant  10% will be aZ and Az and has 5% frequency. Mom has the ability of az chromosomes is 0.45 %. The father has the genotype 50% Az and 50% aZ. The 10% recombination has occured due to which the father chromosomes is reduce upto 45%. The recombinant will be az and AZ with 5% frequency. So, father has az with 0.05 % probability.

Thus, the answer is 0.45 from mom and 0.05 from dad.

8 0
3 years ago
A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 70% of the gametes produced in the population con
IceJOKER [234]

Answer:

Option C

Explanation:

Given ,

A1 allele is carried by 70 % people

Let us assume A1 s dominant genotype

This means p= \frac{70}{100} = 0.7

Thus, frequency of allele in the given population is 0.7

It is also given that the population is in Hardy-Weinberg equilibrium thus

p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3

Frequency of fruit fly with genotype A2A2 will be

q^2\\= (0.3)^2\\= 0.09

As per Hardy-Weinberg's second equation of equilibrium

p^{2} + q^{2} + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42

Hence, option C is correct

7 0
3 years ago
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