Answer:
<u>Part A</u> : 70 secondary oocytes will be formed.
<u>Part B</u> : 70 first polar bodies will be formed.
<u>Part C</u> : 70 ootids will be formed.
Explanation:
During oogenesis growth maturation of a single oogonium produces one primary oogonium.
the primary oogonium then undergoes meiosis -1 and produces one secondary oocyte and first polar body.
The secondary oocyte then undergoes meiosis - 2 and forms an ootid and second polar body.
The ootid then differentiates into the ovum.
As in the above scenario , 70 primary oocytes are present , they undergo meiosis-1 and produces 70 secondary oocytes and 70 first polar bodies. Hence answers of part A and B is 70.
As 70 secondary oocytes are formed , they undergo meiosis -2 and forms 70 ootids which then differentiate in 70 ovums.
Answer:
Zero (0)
Explanation:
According to the given information the genotype of the woman with blood type "AB" would by I^AI^B. The genotype of the man with blood type O would be "ii". Here, the alleles I^A and I^B are dominant over the allele "i".
A cross between parents with genotype I^AI^B and ii would give 50% of children with I^Ai genotype and 50% of children with I^Bi genotype. The children with "I^Ai genotype" would have blood type "A" and the children with I^Bi genotype would have blood type "B". This couple is never likely to have any child with blood type "O" since the mother does not carry allele "i".
Cross: I^AI^B x ii = 1/2 I^Ai : 1/2 I^Bi
Answer:
The answer is C
Explanation:
Because if its organelles are damaged it can not function properly and will soon eventually end end of course the cell dying
Fossil fuels are found
B. Below Ground
