Answer:
it woud have tow different rezalts
Explanation:
The female should be gently wrapped with dry blankets due to exposure of temperature of about 25 degrees Fahrenheit.
Explanation:
Prolonged exposure to extreme heat or cold conditions can impair the body’s temperature control system.
In the above case the patient is exposed to extreme cold condition in a wood with a freezing temperature of 25 degrees Fahrenheit (-3 degrees Celsius) which leads to hypothermia and frostbite.
Extreme cold conditions can freeze the tissue layers of the parts of the skin exposed like those of the face, hands and feet especially. Due to this the skin becomes white (even grayish yellow or blue) and waxy, cold and numb, and the outer surface may become hard and blistered. There will also be extreme pain, stinging and tingling sensation in the exposed areas.
These changes are mainly due to impaired blood circulation and hypothermia. Hypothermia which is lowering of body temperature causes confusion and lethargy.
The first-aid given for frostbite primarily includes warming the body. The affected person’s body should be covered with warm and dry blankets gently and loosely and any wet clothes should be removed.
So when the body is warmed, the body temperature also will increase and correct the hypothermic symptoms like lethargy and confusion.
To calculate the frequency of the heterozygote genotype (Pq) for this gene we must use the Hardy-Weinberg equation ( p2 + 2pq + q2 = 1 ). This equation relies on the Hardy-Weinberg principle, a model in population genetics that states that the frequency of the alleles in a population is never changing, only the combinations (the genotypes) are changing.
If there are only two alleles (variations) of this gene in a population, then their frequencies should add up to 1 (100%). From this, we can calculate the frequency of the q allele.
p +q=1
0,3 +q=1
q= 1-0,3
q= 0,7
Now hat we have the frequency of the q allele we can use the HW equation to calculate the frequency of the heterozygotes.
0,09 + 2pq +0.49= 1
2pq +0,58= 1
2pq= 1-0.58
2pq=0,42
The freqency of the heterozygotes in this population is 0.42
I’m pretty sure it’s oxygen gas
Answer: Antenna
Explanation:
The morphogenesis is a process by which the egg get transformed into a matured adult insect. The egg transform into several transitional stages such as embryo, larva, pupa and finally to adult insect.
The fruit flies undergoes three development stages during morphogenesis these includes the egg, larvae and pupa. The fruit flies can develop into adult within 1-2 weeks at room temperature. The egg as well as larval stages takes a span of eight days whereas the pupal stage takes six days for development.
The embryo enclosed inside the egg exhibit three important parts the head, thorax and abdomen the antenna is the structure which develop in adult flies.
