determining the vertical sequence of rock layers
B.
measuring the radioactive decay of isotopes within rock layers
C.
determining the horizontal sequence of rock layers
D.
identifying index fossil in rock layers
B. Sexual reproduction
C. Regeneration
D. Vegetative propagation
Answer:
A. Asexual reproduction
Explanation:
Because it requires only one cell.
However sexual reproduction require two cell for reproduction.
Answer:
Inflammation occurs only in live tissues and multicellular organisms
Explanation:
Inflammation is a response to foreign antigen by the bodies immune system. Where chemicals from the white blood cells are released to the blood or affected tissue.
It cannot occur in dead bodies because the cells are dead too.
Amoeba is a single celled organism that cannot posses the complex components of the immune system.
Inflammation helps eliminate the initial cause and initiates tissue repair
Due to increase of blood flow to the affected region, it may cause redness and pain.
B. Two oxygen atoms
C. Sodium and bromine
D. Fluorine and chlorine
Answer:
Sodium and bromine
Explanation:
Ionic bond is a type of chemical bond and binds the chemical molecules together. Ionic bond is formed by the transfer of electrons from an electropositive element ( metal) to the electronegative element (non metal).
Sodium is an electropositive element and contains one positive charge. Bromine is an electronegative element with a single negative charge over it. Sodium gives the electron to bromine and results in the formation of ionic bond.
Thus, the correct answer is option (C).
75% of the offspring have normal wings, and 25% of the offspring have short wings.
50% of the offspring are heterozygous for normal wings. 50% of the offspring are homozygous recessive for short wings.
75% of the offspring are heterozygous for normal wings. 25% of the offspring are homozygous recessive for short wings.
Answer:
25% of the offspring are homozygous for normal wings. 50% of the offspring are heterozygous for normal wings. 25% of the offspring are homozygous recessive for short wings.
Explanation:
Let dominant allele be A and recessive allele be a
Parent 1: heterozygous normal wings = Aa
Parent 2: heterozygous normal wings = Aa
Parent 1 X Parent 2 :
A a
A AA Aa
a Aa aa
Genotype of progeny will be as follows:
1/4 or 25% = AA = homozygous dominant for normal wings
1/2 or 50% = Aa = heterozygous normal wings
1/4 or 25% = aa = homozygous recessive for short wings