1. TT1 = 6.5
TT 2 = 6
TT 3 = 1.5
TT4 = 8
TT5 = 3
2. Answers by order: 4,1,2,5,3,3
3.
TT 4 (5mL pepsin & 5mL sodium bicarbonate)
TT 3(10mL of hydrochloric acid (HCl))
TT 2(10 mL of pepsin)
TT 5(5mL Pepsin & 5mL (HCI))
TT 1(10mL of water)
4. Tube 5 is the best representation for the stomach.
Answer:
The correct answer is - XDXd and XD Y
Explanation:
Sex-linked disorders are the disorders that are expressed only when there is two copy of the mutated allele or have not masked by the dominant allele of the gene. In DMD, the dominant allele is XD and the mutated allele is Xd. To express it there are the following genotype possible-
in women XdXd, and in men XdY, so these two genotypes are not possible for the parents in this case. If a mutated allele is present with the dominant allele it will be considered as carrier women but diseased will not be found in them.
So the correct genotype would be
mother: XDXd ( a carrier woman)
father : XDY (normal man)
While mitosis produces 2 daughter cells from each parent cell, meiosis results in 4 sex cells, or gametes in males and 1 in females. Unlike the cells created by mitosis, gametes are not identical to the parent cells. In males, meiosis is referred to as spermatogenesis because sperm cells are produced.
The conversion of DNA information into the messenger RNA is transcription. In the transcription process the genetic code sequence is transcribed into the mRNA by the enzyme RNA polymerase. Hence the process conversion DNA sequence into mRNA is called as transcription.
