Yes that answer does look right
Answer:
If the DNA wasn't duplicated, one of the daughter cells wouldn't have DNA and die.
Explanation:
The cell would split into a cell that had DNA and would survive and a cell that wouldn't survive. The one that doesn't have DNA wouldn't be able to go through cellular functions. Why? The DNA needed to direct functions wouldn't be present.
Answer:
Explanation:As previously stated, DNA is a macromolecule that's made up of individual subunits called nucleotides. Each nucleotide has three parts:
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A deoxyribose sugar.
A phosphate group.
A nitrogenous base.
DNA nucleotides can contain one of four nitrogenous bases. These bases are adenine (A), thymine (T), guanine (G) and cytosine (C).
These nucleotides come together to form long chains known as DNA strands. Two complementary DNA strands bond to each other in what looks like a ladder before winding into the double helix form.
The two strands are held together through hydrogen bonds that form between the nitrogenous bases. Adenine (A) forms bonds with thymine (T) while cytosine (C) forms bonds with guanine (G); A only ever pairs with T, and C only ever pairs with G.
Complementary Definition (Biology)
In biology, specifically in terms of genetics and DNA, complementary means that the polynucleotide strand paired with the second polynucleotide strand has a nitrogenous base sequence that is the reverse complement, or the pair, of the other strand.
Breastfeeding can reduce the mother's risk of breast and ovarian cancer, type 2 diabetes, and high blood pressure.
Answer:
a) 9%.
b) 16.8%.
Explanation:
a).
We are provided with the information that Two linked genes, A and B, are separated by 18 cM (centiMorgan). i.e the recombinant frequency is 18%
Also , the man's genotype is AB/ab... This only result to one explanation, that The man will definitely produce 18% of recombinant gametes which entails
9% Ab & 9% aB
i.e 0.09 Ab & 0.09 aB
On the other-hand, The mother ab/ab have tendency to produce just one single type of gamete which is ab
∴
The probability that their first child will be Ab/ab will be
Pr ( Ab/ab) = (0.09) x (1)
= 0.09
= 9%.
b).
If the father produces 18% of recombinant gametes which entails
9% Ab & 9% aB , this typically implies that the number of the non-recombinant gametes will be;
100%-18% = 82% ( non-recombinant gametes)
i.e genotype AB/ab = 82%
AB =41%; ab = 41%
AB = 0.41 ; ab = 0.41
Now, the probability that their first two children will both be ab/ab:
Using Multiplication Rule to calculate the probability that their first two children (ab/ab); we have:
(0.41)(1) ×(0.41)(1)
= 0.1681
= 16.8%.
