Answer:
1) In the first step, we need to predict the possible alleles for the cross. The dominant allele will be written with a capital letter. The recessive allele will be written with a small letter. Hence, the allele for brown hair colour will be B and the allele for red hair colour will be b.
2) In the second step, we need to determine the genotype of the parents. The genotype of the homozygous dominant parent will be BB. The genotype of the heterozygous brown hair colour will be Bb.
3) The punnet square for cross between these parents can be shown as follows:
B b
B BB Bb
B BB Bb
4) In the fourth step, lets determine the phenotype of the children. The phenotype of all the offsprings born will be brown hair colour.
5) The genotype from the punnet square shows that there is a 50% chance that the offsprings will be heterozygous dominant (Bb) for brown hair colour and their will be a 50% chance that the child born will be homozygous dominant (BB).
The earth's inner core is the innermost layer of the earth. Primarily, it is a solid ball composed of an iron-nickel allow and some other elements, which holds a radius of about 1,220 kilometers.
Answer: One parent is IAi and the other parent is IBi
Explanation: As this trait is codominant, the child can inherited IA, IB or i.
Tina has type O, which means she is ii and her sister is AB, so her genotype is IAIB. Now, to have a child with a recessive trait both parents has to carry at least one allele for the recessive, i. Rosa is type AB which means she had to have inherited one allele IA from one of her parents and the other IB from the other parent. Thus, one parent is IAi and the other is IBi.
Answer:
it involves a physical exchange of chromosome segments in the tetrad (3)
Explanation:
def of crossing over
Crossing over is the swapping of genetic material that occurs in the germline
<u>Answer</u>:
A total of 32 ATP molecules per glucose molecule is generated in cellular respiration and glycolysis. In substrate-level phosphorylation, ATP is generated from the coupled reactions whereas in oxidative phosphorylation ATP is formed by using the oxidized NADH and FADH cofactors. The energy for oxidative phosphorylation comes from the movement of proton to the matrix of mitochondria through ATP synthase.
<u>Explanation</u>:
Each molecule of glucose utilizes 2 ATP molecules and breaks down into 2 pyruvate molecules and generates 4 ATP from substrate-level phosphorylation and 2 molecules of NADH.
Pyruvate converts into acetyl Co-A and produces 2 molecules of NADH. In citric acid cycle, the 2 molecules of acetyl Co-A produces 2 molecules of GTP/ATP, 6 molecules of NADH, 2 molecules of FADH2.
In oxidative phosphorylation 1 molecule of NADH contributes to the production of 2.5ATP, FADH2 gives 1.5ATP. Hence the total production of ATP is 4 + 2 + 25 + 3 = 34 – 2 ATP (used early in glycolysis) = 32 ATP (net ATP production).
