In this question, one parent is homozygous recessive (bbee) and other parent is dihybrid (BbEe)
B - black fur dominant trait
b - white fur recessive
E- black eyes dominant
e - red eyes recessive
a punnett square shows all the possible combinations of genotypes with the corresponding phenotypes that the offspring could inherit.
the genotypes and their phenotypes for the offspring are given below ;
BBEE / BbEe / BbEE / BBEe - black fur and black eyes
Bbee / BBee - black fur and red eyes
bbee - white fur and red eyes
bbEe / bbEE - white fur and black eyes
after analysing the punnett square the following results are obtained;
black fur and black eyes - BbEe- 4/16
black fur and red eyes -Bbee - 4/16
white fur and black eyes - bbEe - 4/16
white fur and red eyes -bbee - 4/16
B I’m pretty sure ! Hope this helped
Answer:
b: 8;8
Explanation:
Mitotic or meiotic cell division constitute the m phase of the cell cycle. At the end of the m phase, the new cells enter the interphase stage of the cell cycle. The interphase is further sub-divided into;
- <em>the phase,</em>
- <em>the phase,</em>
- <em>the S phase; and</em>
- <em>the phase</em>.
The phase is essentially a resting phase. Cells that do not need to divide except when necessary move into this phase after exiting the m phase.
Actively dividing cells enter the phase after exiting the m phase. Cell development and growth takes place. From there, the cells enter the S phase where DNA replication/synthesis takes place. The cells then enter the phase where proteins are synthesized in preparation for division or m phase.
At the S phase, the amount of DNA a cell carries is doubled but the chromosome number remains the same. For example, if a cell enters the S phase with 2 g of DNA containing 10 chromosomes, at the end of S phase, the amount of DNA would have come 4 g while the number of chromosomes will remain 10.
Hence, if the average amount of DNA in the assayed cells immediately after mitosis is 4 picograms, the amount would be 8 picograms at the end of S phase and will still remains 8 picograms at the end of phase.
The correct option is b.
Answer: see explanation
Explanation:
A. substrate
B. Active site
C. Enzyme binds with substrate
D. Active site of enzyme
E. Products leaving active site
Simplified enzymatic reaction. The substrate reversibly binds to the active site of the enzyme, forming the enzyme-substrate (ES) complex. The bound substrate is converted to product by catalytic groups in the active site, forming the enzyme-product complex (EP). The bound products are released, returning the enzyme to its unbound form, ready to catalyze another round of converting substrate to product.
Answer: nitrogen and hydrogen
N2 +3 H2 ---> 2NH3
Explanation: