C
by using your cells to make copies of themselves
Answer:
Mosquitoes West Nile Virus St. Louis Encephalitis Western
<h2><u>
Full Question:</u></h2>
In the family tree below, people with the recessive trait of attached earlobes are shaded gray.
What must be true about the person labeled "A"?
A. It is a male with at least one dominant allele.
B. It is a male with two dominant alleles.
C. It is a female with at least one dominant allele.
D. It is a female with two dominant alleles.
<h2><u>Answer:</u></h2>
Its a male with atleast one dominant allele.
Option A.
<h3><u>Explanation:</u></h3>
The gene for the attached earlobe is recessive while the gene for the free earlobes is dominant. In the phylogenetic tree, we can see that both the father and mother aren't having attached earlobes. So both of them are having atleast one dominant allele which makes them have free earlobe.
In the F1 offsprings, one of the female and a male is having free earlobes. So both of them have atleast one dominant allele. The 2nd female is having an attached earlobe. So both the recessive allele have come form one parent each. So both of them are heterozygous.
Thus, the male marked as A atleast have one dominant allele. He can be a homozygous dominant, but the probability is 25%.
Answer:
Cellulose
Explanation:
trunk is the main axis of a tree that supports the branches and is supported by roots. Cellulose makes it sturdy. Cellulose is the main substance in the walls of plant cells, helping them to remain stiff and upright.
Answer:
(a) number of strands (n) = time (t) ÷ proportionality constant (k)
(b) The time needed for the bacterial to double its initial size is 3.36 hours.
Explanation:
(a) Let the rate (time) be represented by t and the amount (number) of strands of bacteria be represented by n
t is proportional to n, therefore, t = kn (k is the proportionality constant)
Since t = kn, then, n = t/k
(b) Initial amount of strands = 300
Amount of strands after 2 hours = 300 + (300 × 20/100) = 300 + 60 = 360
k = t/n = 2/360 = 0.0056 hour/strand
Double of the initial size is 600 (300×2 = 600)
Time (t) needed for the bacterial to double its initial size = kn = 0.0056×600 = 3.36 hours