Answer:
0.949367 ≈ 0.95.
Explanation:
The chi square analysis can be done by the formula as follows:
Chi square = (O-E)²/E
Here, O is observed value and E is expected value.
The tall trait follows the Mendel's law, this means the cross between F1 results in the formation of progeny with ratio 3( tall) : 1 (short).
The observed values for set I is 63 tall, 16 short.
The expected value for tall plant = 
= 59.25 .
The expected value for short plant = 
= 19.75.
Put these value in the formula for chi square analysis:
For tall plants = ( 63 - 59.25)²/ 59.25= 0.237.
For short plants = (16 - 19.75)²/ 19.75 = 0.712
Total value for set I = 0.237 + 0.712 = 0.95.
Thus, the answer is 0.95.
The correct answer would be the endocrine system. Hope this helps!
Answer:
If this is a T or F question, it's probably true
Complete question:
Suppose "A" is a dominant gene for the ability to taste phenylthiocarbamide and "a" is a recessive gene for the inability to taste it. Which couples could possibly have both a child who tastes it and a child who does not?
a. father AA, mother aa
b. father Aa, mother AA
c. father Aa, mother Aa
d. father AA, mother AA
Answer:
c. father Aa, mother Aa
Explanation:
According to the given information, the ability to taste phenylthiocarbamide is a dominant trait and is imparted by the allele "A". This phenotype would be expressed in both homozygous and heterozygous conditions. The non-taster phenotype would be expressed in the homozygous recessive genotypes only.
To have both taster and non-taster children, both the parents should have at least one copy of the recessive allele. Among the given options, the father with genotype Aa and the mother with genotype Aa have the possibility to have both taster and non-taster children.
Aa x Aa= 3/4 taster (1/4 AA and 1/2 Aa): 1/4 non-taster (1/4 aa)
