Answer:
Louisiana
Explanation:
I got the answer wrong and this was the correct answer
Answer:
E) Either anaphase I or II
Explanation:
Failure of segregation of homologous chromosomes during anaphase I or failure of segregation of sister chromatids during anaphase II leads to the presence of the abnormal number of chromosomes in resultant gametes. In the given example, the egg mother cell with 48 chromosomes (24 pairs) would enter meiosis I but the failure of one pair of homologous chromosomes to segregate from each other followed by normal meiosis II would result in the formation of two gametes with one extra chromosome and two gametes with one less chromosome.
On the other hand, if the nondisjunction occurs at anaphase II of meiosis II, two normal gametes, one gamete with one extra chromosome and one gamete with one less chromosome will be formed. Therefore, nondisjunction at anaphase I or anaphase II would have resulted in the production of eggs with one extra chromosome.
latic acids. cause by milk
chittrus acids. cause by lemon juice
Answer:
<u>the genotype is (Hh)</u> <u>phenotype hybrid which is a Guinea pig with hair</u>.( hybrid means its not a pureblood haired Guinea pig. pureblood means same. EX: (GG). Guinea pig with hair. its also known as heterozygous, and zygous.)
witch also means its heterozygous. there is no percentage chance for something to have a different genotype of phenotype.
Answer:
An experimental control is used in scientific experiments to minimize the effect of variables which are not the interest of the study. The control can be an object, population, or any other variable which a scientist would like to “control.”.
Explanation:
