The effect nervous system has on the heart rate is
Stimulation by parasympathetic nerves causes the heart rate to slow down.
The two branches of the autonomic (involuntary) nervous system regulate heart rate. The parasympathetic nervous system and the sympathetic nervous system (SNS and PNS) (PNS). To increase heart rate, the sympathetic nervous system (SNS) produces the catecholamines epinephrine and norepinephrine. The hormone acetylcholine is released by the parasympathetic nervous system (PNS) to reduce the heart rate. Your heart rate may briefly increase due to stress, coffee, and excitement, whereas it may temporarily decrease due to meditation or deep, steady breathing. Any amount of exercise will raise your heart rate, which will stay up as long as you keep exercising.
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Answer:
On the basis of physical traits.
Explanation:
Newer classification systems classify organisms on the basis of physical traits of an organism. All modern classification systems gets their concept and help from Linnaean classification system. In these new classification systems, those organisms that have similar characteristics are placed in one group because they evolve from the same ancestor or having common ancestry. These changes occurs in the organisms with the passage of time that helps organisms to survive in that environment.
Answer:
Consumers supply oxygen for producers to undergo aerobic respiration. Consumers produce carbon dioxide for producers to use during photosynthesis.
Consumers provide chemical energy needed by producers for cellular respiration.
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Answer:
<h2>
Allele frequencies for B and b.
</h2><h2>
"b"(q) allele frequency = 0.60
</h2><h2>
</h2><h2>
"B"(p) allele frequency = 0.40
</h2><h2>
</h2><h2>
Genotype frequencies;
</h2><h2>
BB = 0.16
</h2><h2>
Bb = 0.48
</h2><h2>
Bb = 0.36
</h2>
Explanation:
Given
Non-baldness (B) is dominant on baldness(b), so B is dominant over b.
Homozygous pattern baldness male (bb) = 360,
Heterozygous non- baldness (Bb)= 480,
Homozygous non-baldness (BB)= 160.
So, we can also denote then by genotypes only,
BB= 160;
Bb= 480;
bb= 360;
Total= 1000
Allele frequency q² (bb) = 360/1000=0.36
allele fequency for q( b)= √o.36=0.60.
Allele frequency for p²(BB) = 160/1000=0.16
allele frequency p(B)= √0.16 = 0.4
Expected genotype frequencies;
BB = 160/1000 = 0.16
Bb = 480/1000= 0.48
Bb = 360/1000= 0.36
