It can change multiple times depending on the studying being shown
A marmoset is a member of the family of primates and are known to be one of the smallest monkey. They are measured to have a body length of 12 - 15 inches and has a mass of approximately 100 - 125 g. Comparing to monkeys, marmosets have distinct characteristics. Instead of having nails, they have claws, they tactile hair or a sensitive hair found on their wrists. They do not have wisdom teeth and the layout of their brain is primitive.
Answer:
Frequency is inversely proportional to wavelength.
Explanation:
frequency is the number of oscillations in a unit of time.
Wavelength is the length between one oscillation and the next oscillation.
take two springs and 4 pins. (springs should have loops or hooks at terminals to attach to pins.)
Now get two pins and pin them at a certain distance apart.
take the first spring and attach its two terminals to the pins.
Repeat the above procedure, but double (increase the distance x2) the distance between two pins.
Now you have to measure how many coils are there within 5cm of the two springs.
you'll observe that the second spring has half the coils of the first spring within 5cm.
In here no. of coils is the frequency
5cm length is time.
distance between coil is the wavelength.
Answer:
<em>The correct answer is C) The organs mature</em>
Explanation:
During the 3rd trimester, the baby continues with its development. Structures like the bones have completely formed but are soft. The baby is capable of opening and closing its eyes. During the 3rd trimester, the organs of the baby are fully mature and can carry out their own functions. By the end of the third- trimester, the baby turns in head- down position. Hence, the correct option is C.
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)
