Answer:
See the explanation
Explanation:
Answer 1.
As given that in F1 all are short and white then it can be said according to Mendel's law that short and white are dominant over tall and purple.
Let S for short s for tall and W for white and w for purple allele.
So the genotype of short purple will be Ssww or SSww. So In first case Ssww self crossed then resultant offsprings will be,
................ Sw .................... sw
Sw ........ SSww ............. Ssww
sw ........ Ssww .............. ssww (tall and purple)
So from this 1/4 will be tall and purple while 3/4 will be short and purple.
In second case SSww only short purple progeny will appear.
Answer 2.
2. a) The female progeny will not show any trait because there are two X chromosome in females , so female offspring can be carrier but not show any trait in case of X- linked trait.
2. b) Half of male offsprings show trait because X is inherited from mother. So the chance of having X-linked recessive allele is 1/2.
2. c) The chance of having X linked affect allele in daughter is 1/2. So the chance of inheriting that X to son will be 1/2 so in total there is chance of 1/4 that son will be affected.
2. d) the chance of first child show this trait will be 1/4 in case of male offspring while 0 in case of female offspring.
Hope this helps!
Answer:
yes because fajamarutation procedure
Explanation:
C.2
.........................
Answer: 1: allergic disorders
2. Immunodeficiency disorders
3. Autoimmune disorders
4. Cancers
The uninoculated tube is considered to be a negative control
because a negative control is an effect of having to show no response. Am uninoculated
control is where organisms are not being
inoculated such as having a culture bacteria of not being inoculated with an
organism.
