Because it was not tested multiple times by other scientists
A Boundary Is A Line That Marks A Limit Of Space. So If You Live In A Neighborhood, Then There Would Be Fences, And That Would Be A Boundary line. It Marks Space And Divides Space. Hope That Helps.
Answer:
True.
Explanation:
White blood cells cannot function properly and their reduced production can lead to fever and frequent infections. This is because the function of white blood cells in fighting germs has been disrupted.
Other signs and symptoms:
- Weak and tired.
- Lack of appetite and weight loss.
- Swelling and bleeding gums.
- Headache.
- Abdominal swelling is caused by enlargement of the liver and lymph.
- Bone pain, can cause weakness.
Joint pain.
- The glands are swollen. If the glands in the neck and chest are swollen, it can cause blood flow to be blocked. This causes the face to swell, difficulty breathing and snoring.
#If i'm wrong, i'm sorry
Answer:
The main function of the circulatory system is a To transport material throughout the body.
Explanation:
Blood in an important component of circulatory system.Blood contain Red blood cell that in turn contain hemoglobin which helps in the transport of oxygen and carbon dioxide to and from various body tissue.
The circulatory system also helps in the transport of nutrients such as glucose to various parts of our body such as heart,muscle,brain etc.The glucose molecules are utilized by those organs to perform their physiological functions.
The circulatory system also transport various hormone from the region of secretion to the target cell .
Answer:
The correct answer is - 1/41,493
Explanation:
Let assume the frequency of the two possible same allele genotype (dominant and recessive) in an inbred population is p and q. Then the frequency of heterozygotes (H) is denoted as:
2pq + 2pqF. ( where F is the inbreeding coefficient).
The frequency of the two different hoozygotes in inbred population can be calculated as:
p2 + pqF and q2 + pqF. (Where p and q are the allele frequency of the dominant and recessive phenotype.
Given: Frequency of Alkaptonuria (q 2) = 1:500, 000
=> q = 1/707
p = 706/707 ( Approx values)
solution:
Inbreeding coefficient (F) = 1/64
Therefore,
Frequency of Alkaptonuria in second cousins= q 2 + pqF
= 1/500, 000 + (706/707 x 1/707) x (1/64)
= 1/500, 000 + 1/45, 248
= 1/41,493 (approx)
