Your answer is correct! The answer is clear sunny!
Answer:
Cellulose is the main substance found in plant cell walls and helps the plant to remain stiff and strong. Humans cannot digest cellulose, but it is important in the diet as a source of fibre.
Explanation:
Answer:
Label A. is probably Mitochondria, and Label B. is probably the nucleaus, but Im not sure about C. The ribisome or the endopplasmic reticulam, maybe??
Explanation:
Answer:
The correct answer is - option a.
Explanation:
Pubis or the pubic bone is one component bone of the right and left coxal bone articulate together. This bone makes the pelvis, in the female uretheral sponge covers the pubic bone from front.
The pubic bone is also makes the inferior and anterior obturator foramen. The pelvic bones are joint together to pubic symphyasis.
Thus, the correct answer is - option a.
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%.
