Answer:
Random segregation of homologous chromosomes makes the two siblings differ from each other for 0-23 chromosomes.
Explanation:
Sexual reproduction adds genetic variations in the progeny by crossing over, independent segregation of homologous chromosomes and random fusion of gametes. Without crossing over, independent segregation of homologous chromosomes towards opposite poles during anaphase I of meiosis may result in two siblings to vary from each other for 0-23 chromosomes.
There is an equal probability of each of the two siblings to get a chromosome from mother or father. Hence, irrespective of the variations provided by crossing over, random segregation of homologous chromosomes makes the two siblings differ from each other for 0-23 chromosomes.
Answer:
The correct answer is option E. "DNA in both daughter cells would be radioactive".
Explanation:
Radioactive thymidine incorporation has irremediable effects in thymine and cytosine in bacteria such as E. coli. Although there are some reports of recovery of E. coli from UV-irradiation DNA damage, the presence of radioactive bases such as thymine is often irremediable. Therefore, after a dividing culture of E. coli exposed to radioactive thymine divides, DNA in both daughter cells would be radioactive.
Answer:
cant really answer much since you need the class results for these. but i can simplify question 1. so, does the class results prove that the traits made my dominant alleles are the most common? or in other words, are traits made by dominant alleles most common in the class results?
number 4 is yes because there are more possibilities that the dominant trait will occur than the recessive trait. therefore, it is more common and the conception is correct
Since each codon consist of 3 nucleotides, 4 cordons can be formed by 12 nucleotides. (but please make sure that you check this answer. I am not completely sure)
