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.
The answer is "All of the above".
Answer:
Explanation:
A woman with type A blood (whose father was type O) meaning her genotype is AO mates with
Man that has type O blood (OO genotype)
Both are heterozygous for MN blood group and both also heterozygous for the FUT1 gene controlling the synthesis of the H substance (Hh)- which determines the expression of the A and B antigen.
Cross
A O M N H h
O AO OO M MM MN H HH Hh
O AO OO N MN NN h Hh hh
Type A- 1/2 O-1/2 type M- 1/4 MN-1/2 N- 1/4, type H- 3/4 h-1/4
Type A with M antigen:
1/2*1/4*3/4 = 3/32
Type A with M and N antigens:
1/2*1/2*3/4 = 3/16
Type A with N antigen:
1/2*1/4*3/4 = 3/32
Type O with M antigen:
1/2*1/4*3/4= 3/32
Type O with M and N antigens:
1/2*1/2*3/4 = 3/16
Type O with N antigen:
1/2*1/4*3/4 = 3/32.
The 3/4 value comes from the expression of Hh-3/4 (this determines if the A and B Angie will be expressed).
Denaturing. This step occurs at around 95C. This causes the double strand Helix to spectate
Answer:
C. They break down food materials into forms the cell can use.
