False
reason -
Gene testing is not full proof as there are possibilities of discrepancies in it based on the testing procedures used and the condition of sample. Since we human beings have the same genomic structure there are chances that our DNA may match with other people who do not share a heredity with our family. Basically statistical probability is used to determine probability of paternity, relationship of any kind etc. among two individuals . usually A 99% or higher percentage of probability is considered conclusive. and thus it gene testing can not be considered as full proof
Im pretty sure that the answer is C, a method producing genetically identical offspring.
Asexual reproduction is where only one parent is involved..
The eukaryotic organisms have the process of mitosis but differently than the process of the prokaryotic because the prokaryotic organisms dont have the dna enclosed in a nucleus. Mitosis needs to occur in eukaryotic organims because the cell could keep growing an it is going to be less efficient in moving material across the cell membrane. They reason why mitosis happens is because volume and surface are do not increase at the same rate.
Explanation:
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)
Answer:
I think true statement is A.