1 kilometer = 6.68 × 
(<<<AU is in scientific notation here)
To convert 300 billion kilometer (3,000,000,000 km) into AU you must multiply the kilometer by the equivalent of 1 km in AU (6.68 × )
3,000,000,000 × (6.68 × ) = 20.04 AU
Hope this helped!
Answer:
<u>Red blood cell in humans -</u> it has no nucleus. Hence it offers the cell to carry more haemoglobin.
- they are disc shaped allowing them to pass through narrow capillaries.
<u>Root hair cells in plants-</u> they have a large cytoplasm which enables them to take water from the soil.
<u>White Blood Cells in humans -</u> they have lobed nucleus and so can change their shape to pass through narrow capillaries.
- <em><u>HOPE IT HELPS...</u></em>
Ligaments connect one bone to another bone. Therefore, they allow a joint to form, because joints are where two or more bones connect.
Take your knee joint for instance (***see attached pic***). The knee joint is formed by the connection of the femur (your thigh bone), the tibia (your shin bone), and the fibula (the other long bone in your lower leg). In order for all of these bones to connect there are many ligaments in the knee joint that keep the bones connected and in place. A well known example of one of these ligaments in the knee joint is the ACL (anterior crucate ligament), which is commonly torn in sports, namely football. Most people have heard of this ligament because it receives a lot of media attention since tearing it can greatly alter or even end professional athletes' careers.
Answer:
D)
Explanation:
Hoping it helps sorry if its wrong
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)
