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).
Among the listed options, we can conclude that option C which states that "<em>It results in the </em><em>diversity </em><em>of the </em><em>species</em><em> that enhances the probability of </em><em>survival</em><em>.</em>" is correct.
Sexual reproduction offers organisms an advantage over those who undergo Asexual reproduction. Though there are benefits and drawbacks to each method, Sexual reproduction is a method that is <u>better adapted to a changing environment and promotes the faster </u><u>evolution </u><u>of a </u><u>species</u><u>.</u>
The main benefit of sexual reproduction is that it allows for a<u> </u><u>greater diversity </u><u>of the </u><u>species</u><u> which enhances the probability of </u><u>survival</u><u>.</u> This is due to the fact that sexual reproduction leads to varying combinations of alleles for specific traits, which coupled with selective reproduction (<em>the ability of a member of the species to choose a spouse with desirable traits</em>) <u>leads to </u><u>offspring </u><u>who over time are increasingly better </u><u>adapted </u><u>to life in the given </u><u>environment</u><u>, with specific traits tailored for </u><u>survival</u><u>.</u>
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Shear causes horizontal movement along a fault plane in a strike-slip fault.
Answer:
228,585 gram
Explanation:
M h2so4 = 2 + 32 + 16*4 = 98 g/mole
C% o2 in h2so4 = (64 / 98) * 100 = 65,31%
m o2 = 350 * 65,31% = 228,585 gram
Done :))