Answer: If all grasshoppers are removed from the food chain, what will happen to the bluebirds? ... The bluebirds will begin eating more plants.
Explanation:
These antimitotics (i.e. taxanes) target microtubule proliferation. The stage of mitosis where it acts is at the M-phase, particularly in the metaphase. The phases of mitosis are prophase, metaphase, anaphase, and telophase. Microtubles first appear at the metaphase.
Answer:
0.095
Explanation:
Phenylkentonuria is a disease caused by a recessive allele.
The frequency of the recessive allele + the frequency of the dominant allele equals 1.
The frequency of the recessive allele is q = 0.05
The frequency of the dominant allele then is p = 1 - q = 0.95
If people mate randomly, the frequency of the homozygous dominant genotype will be p², the frequency of the heterozygous genotype will be 2pq and the frequency of the homozygous recessive genotype will be q² .
2pq=2× 0.05 × 0.95
2pq=0.095
The heterozygote frequency in the population is 0.095
Answer:
The correct answer is option B. false.
Explanation:
The seminal vesicles are the site of production of the seminal fluid which makes 60 % of the total seminal vesicles. The seminal vesicles do not store the sperm.
The primary storage site of the sperm in seminiferous tubules and tubular storage sites for the sperm is epididymis.
Thus, the correct answer is option B. false.
Answer:
16/64 = 1/4
Explanation:
This is a typical trihybrid cross involving three genes T, P and Y. A plant with genotype TTPpYy is crossed with a plant recessive for all traits (ttppyy).
According to Mendel's law of independent assortment, each allele for each gene will get sorted into the following 8 gametes with only 4 different: TPY, TPy, TpY, Tpy, TPY, TPy, TpY and Tpy.
The recessive parent, ttppyy will produce tpy, tpy, tpy, tpy, tpy, tpy, tpy and tpy.
Hence, using a punnet square, 64 offsprings will be produced with only 16 of them heterozygous dominant for the three traits with genotype (TtPpYy). Hence, proportion is 16/64 equivalent to 1/4.
