All cells have DNA, cytoplasm, and ribosomes
Answer:
A. 21 rad/s
B. 200.5 rpm
C. 26.46 rad/s2
D. 0.299 Hz
Explanation:
Parameters given are:
Tangential velocity,v = 1.26m/s
Diameter (outer edge) = 0.120m
Outer edge radius, ro = 0.12/2
= 0.06m
A. Calculate Angular velocity
Angular velocity, w = v/r
= v/ro
= 1.26/0.06 = 21 rad/s
In rpm,
2pi/60 rad/s = 1 rpm
1 rad/s = 60/2pi rpm
Sp, 21 rad/s = 60 * 21/2pi rpm
= 200.5 rpm
B. Calculate the inner edge radius, ri given w = 500 rpm
Converting rpm to rad/s,
= 2pi/60 * 500 rad/s
= 52.34 rad/s
ri = v/w
= 1.26/ 52.36
= 0.024m
C. Calculating centripetal acceleration, a from the outer edge, ro
a = v2/r = w2r
= (1.26)2/0.06
= 26.46 rad/s
D. Calculate the frequency, f of the outer edge angular velocity, w = 21 rad/s
f = 2pi/w
= 2pi/21
= 0.299 Hz (or per second)
Answer:
C. 40
Explanation:
Pure-breeding means that the individuals are homozygous for the genes being analyzed.
From Mendel's Law of Dominance we know that the traits that appear in the F1 are the dominant ones.
I will call:
P_ = purple flowers
pp = red flowers
L_ = long pollen
ll = round pollen
Initial cross:
P Pl/Pl x pL/pL
F1 Pl/pL
<u>Test cross</u> (cross with a homozygous recessive individual):
Pl/pL x pl/pl
<u>Expected progeny:</u>
Pl/pl = Parental (purple flowers, round pollen)
pL/pl = Parental (red flowers, long pollen)
PL/pl = Recombinant (purple flowers, long pollen)
pl/pl = Recombinant (red flowers, round pollen)
20% of the offspring have purple flowers and long pollen (PL/pl).
Every time crossing over happens in the meiosis of the F1 individual, both a <em>PL</em> gamete and a <em>pl</em> gamete form. That means that 20% of the offspring will also be pl/pl, and the total proportion of the offspring that will be recombinants will be 40%.
A distance of 1 map unit corresponds to a recombinant frequency of 1%.
A recombinant frequency of 40% therefore means that 40 map units separate the glower color and pollen shape genes.
Answer is D
hope this helps:)sorry if it doesnt
brainliest plzzzzzzzz
