Since it's been crossed with a homozygous wrinkled green, the offspring has a genotype for heterozygous round and yellow. As round and yellow are dominant traits, they're expressed in the phenotype. But when self pollinated in the f2 generation, the recessive ones will show as well
Hope it helps :')
Answer:
B: only amino acids join together with peptide bonds to form proteins.
Plant A would have one allele from each parent, so the leaf color will be the dominant one.
Plant B would have a more complex percentage for traits. And it'll probably "cancel-out" some alleles only showing the info given in one of them.
Hope it helped,
BioTeacher101
Cells produced at the end of telophase II have half as many replicated chromosomes as cells that started the process.<span>Telophase I is the stage double stranded chromosomes arrive at poles of each cell.</span>
Answer:
The answer is mother in meiosis II
Explanation:
nondisjunction can be defined as the failure of the segregation of the child chromosomes in meiosis I and meiosis II resulting in gametogenes. This results in abnormal gametes with some chromosomal imbalance being formed and subsequent fertilization of these gametes results in a generation of abnormal individuals.
According to exercise, red-green color blindness follows an X-linked recessive pattern and the phenotype is only expressed. the father cannot bring the possibility of color blindness to his children, and the nondisjunction, defined earlier, cannot occur in either meiosis I or meiosis II. As a conclusion we can say that it may have a place in meiosis I or meiosis II corresponding to the mother and although the mother's condition may be normal, the first child would be compromised with color blindness, therefore, the mother is the carrier.
