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goblinko [34]
3 years ago
7

Which is not correct about cancer risks?

Biology
1 answer:
rusak2 [61]3 years ago
8 0
The lifetime risk of developing cancer is a chance a person has over his or her lifetime and can be diagnose of having a cancer. In the scenario about the cancer risks, the answer would be c. the more you drink, the lower your risk is. The more alcohol one's drink does not mean it will lower one's risk of developing cancer.
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Use the following sentence to answer the question.
vfiekz [6]

Answer:

Opossums eat at night and eat mostly plants.

Explanation:

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2 years ago
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Why is the sympathetic nervous system referred to as the fight or flight system?
Nat2105 [25]

As you can see from the diagram below, the human nervous system is divided into two main components, the peripheral nervous system and the central nervous system.

The peripheral nervous system is further divided into two components. The somatic nervous system involves the voluntary movement of skeletal muscle. When you decide to raise your cup of coffee to your mouth, that is the somatic nervous system at work activating the appropriate muscles. The second component of the PNS is the autonomic nervous system, which controls processes that are out of our conscious control.

The autonomic nervous system is further subdivided into the parasympathetic nervous system and the sympathetic nervous system (which you are interested in). In very generalized terms, they have antagonistic effects on each other. The sympathetic nervous system generally activates the body, whereas the parasympathetic nervous system primarily acts to relax the body. This is a simplified view of things, but it is useful to think of it in this way.

The sympathetic division of the autonomic nervous system increases heart rate, inhibits digestion, relaxes bronchi in the the lungs to allow more airflow, causes the release of glucose from the liver, and relaxes the bladder. All of this activates the body, prepares it for a "fight".

In contrast, the parasympathetic division reduces heart rate, constricts bronchi in the lungs, and increases blood flow to the digestive tract. It has been called the "rest and digest" system.

As you can see from the diagram, the nerves of the parasympathetic nervous system come directly from the brainstem and spinal cord. However, with the sympathetic nervous system, there are "ganglia" (groups of cell bodies) outside the spinal cord itself.

Interestingly, the parasympathetic nervous system only uses Acetylcholine (ACh) to exert its effects on organs, but the sympathetic nervous system uses both ACh and norepinephrine (NE) to activate the body.

Its important to remember that the sympathetic and parasympathetic nervous systems are autonomic, meaning we generally have little control over them.

6 0
2 years ago
How can you describe the habitat in which an organism lives?
solong [7]
A niche describes where the organism lives and how it interacts with its environment .
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5 0
3 years ago
Which statement is not one of the axioms of Euclidean geometry?. . In space, there are at least four points that are not in the
GalinKa [24]
The correct answer for this question is "In space, there are at least four points that are not in the same plane." The <span>statement that is not one of the axioms of Euclidean geometry is that i</span><span>n space, there are at least four points that are not in the same plane</span>
5 0
3 years ago
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In Drosophila melanogaster, vestigial wings (vg) is recessive to normal wings (vg+), black body (b) is recessive to gray body (b
yan [13]

Correct progeny phenotype:

  • 1779 vestigial wings, black body and purple eyes, vg b pr  
  • 1665 normal wings, a grey body and red eyes, vg+ b+ pr+
  • 252 normal wings, an black body and purple eyes, vg+ b pr
  • 241 vestigial wings, a gray body and red eyes, vg b+ pr+
  • 131 normal wings, an black body and red eyes, vg +b pr+
  • 118 vestigial wings, a gray body and purple eyes, vg b+ pr
  • 13 vestigial wings, an black body and red eyes, vg b pr+
  • 9 normal wings, a gray body and purple eyes, vg+ b+ pr

Answer:

  • The order of these genes is vg --- pr --- b
  • Map distances between the genes vg/pr = 12.2 MU
  • Map distance between the genes pr/b = 6.4 MU
  • Map distances between the genes vg/b = 18.6 MU

Explanation:

We know that

•Normal wings expressed by vg+ is dominant over vestigial wings, vg

•Gray body b+ is dominant over black body

•Red eyes, pr+, is dominant over purple ayes, pr

We have the number of descendants of each phenotype product of the tri-hybrid cross.

•1779 vestigial wings, black body and purple eyes vg b pr  

•1665 normal wings, a grey body and red eyes vg+ b+ pr+

•252 normal wings, an black body and purple eyes vg+ b pr

•241 vestigial wings, a gray body and red eyes vg b+ pr+

•131 normal wings, an black body and red eyes vg +b pr+

• 118 vestigial wings, a gray body and purple eyes vg b+ pr

•13 vestigial wings, an black body and red eyes vg b pr+

• 9 normal wings, a gray body and purple eyes vg+ b+ pr

The total number of individuals is 4208.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

<u>Parental)</u>

  • 1779 vestigial wings, black body and purple eyes vg b pr  
  • 1665 normal wings, a grey body and red eyes vg+ b+ pr+

<u>Double recombinant)</u>

  • 13 vestigial wings, an black body and red eyes vg b pr+
  • 9 normal wings, a gray body and purple eyes vg+ b+ pr

<u>Simple recombinant)</u>

  • 252 normal wings, an black body and purple eyes vg+ b pr
  • 241 vestigial wings, a gray body and red eyes vg b+ pr+
  • 131 normal wings, an black body and red eyes vg +b pr+
  • 118 vestigial wings, a gray body and purple eyes vg b+ pr

Comparing parental with the double recombinants we will realize that between  

  • vg b pr (parental)
  • vg b pr+ (double recombinant)

and  

  • vg+ b+ pr+ (Parental)
  • vg+ b+ pr (double recombinant)

They only change in the position of the alleles pr/pr+. This suggests that the position of the gene pr is in the middle of the other two genes, vg and b, because in a double recombinant only the central gene changes position in the chromatid.  

So, the order of the genes is:

---- vg ---- pr -----b ----

Now we will call Region I to the area between vg and pr and Region II to the area between pr and b.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between vg and pr genes, and P2 to the recombination frequency between pr and b.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.  

So:

Parental)

• 1779 vestigial wings, black body and purple eyes vg pr b  

• 1665 normal wings, a grey body and red eyes vg+ pr+ b+  

Double recombinant)

• 13  vg pr+ b  

• 9  vg+ pr b+  

Simple recombinant)

• 252  vg+ pr b  

• 241 vg pr+ b+  

• 131  vg+ pr+ b  

• 118  vg pr b+  

P1 = (R + DR) / N

P1 = (252+241+13+9)/4208

P1 = 515/4208

P1 = 0.122

P2= (R + DR) / N

P2 = (131+118+13+9)/4208

P2 = 271/4208

P2 = 0.064

Now, to calculate the recombination frequency between the two extreme genes, vg and b, we can just perform addition or a sum:

P1 + P2= Pt

0.122 + 0.064 = Pt

0.186=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).  

The map unit is the distance between the pair of genes. Every 100 meiotic products, one of them results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.122 x 100 = 12.2 MU

GD2= P2 x 100 = 0.064 x 100 = 6.4 MU

GD3=Pt x 100 = 0.186 x 100 = 18.6 MU

---- vg ---------------------- pr ---------------------b ----

                    R1                                 R2

-----vg----12.2MU---------pr—

                                   ----pr--------6.4 MU----b—

-----vg ----------------18.6 MU--------------------b----

                                   

3 0
3 years ago
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