The cell theory teaches that all living things are composed of one or more cells; the cell is the basic unit of life.
Answer:
a) The equation describes the relationship between p and q is p + q = 1 for allelic frequencies
b) The frequencies of alleles A1 in decimals is 0.7 and in percents is 70%.
The frequencies of alleles A2 in decimals is 0.3 and in percents is 30%.
Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (Homozygous dominant), 2pq (Heterozygous), q² (Homozygous recessive). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the same way, the sum of genotypic frequencies equals 1, this is
p2 + 2pq + q2 = 1
Being
- p the dominant allelic frequency,
- q the recessive allelic frequency,
- p²the homozygous dominant genotypic frequency
- q² the homozygous recessive genotypic frequency
- 2pq the heterozygous genotypic frequency
In the exposed example,
- A1 codes for purple leaves, has a frequency of 0.7 and is dominant.
- A2 codes for white leaves, has a frequency of 0.3 and is recessive.
According to this equation: p + q = 1.
0.7 + 0.3 = 1
A1A1= p²= 0.7² = 0.49
A2A2= q² = 0.3² = 0.09
A1A2 = 2pq = 2 x 0.7 x 0.3 = 0.42
According to this equation: p² + 2pq + q² = 1
0.49 + 0.42 + 0.09 = 1
- The frequencies of alleles A1 in decimals is 0.7 and in percents is 70%.
- The frequencies of alleles A2 in decimals is 0.3 and in percents is 30%.
There aren't <span>any cures for genetic disorders because most people are born with genetic disorders. The genetic defect is actually encoded in the cells DNA and it makes it nearly impossible to cure those defects after birth. I hope that this is the answer that has actually come to your desired help.</span>
Answer:
a. 1/2 b. The organism is an animal
Explanation:
a. What is the medullary index for an organism that has a medulla width of 2cm and hair shaft width 4cm?
Medullary index = diameter of medulla/diameter of hair.
Given that diameter of medulla = medulla width = 2cm and diameter of hair = hair shaft width = 4cm,
Medullary index = diameter of medulla/diameter of hair = 2 cm/4cm = 1/2
b. Is this organism human or animal?
Since the medullary index = 1/2 > 1/3 (if it is ≤ 1/3 it is human hair and ≥ 1/2 it is animal hair), it is animal hair. So, the organism is an animal.
Answer:
carbon dioxide,glucose,oxygen
Explanation:
im like 90% sure im right lmk if not!
