Answer:
Where is the DNA template strand?
Explanation:
The correct answers are:
• The solar wind produces the northern and southern lights
• The solar wind disrupts communication systems.
• The solar wind distorts Earth's magnetic field.
Solar winds are geomagnetic stream of charged particles radiated from the outer atmosphere of the sun. All planets are protected from solar winds via magnetic fields but planets that are positioned closer to the sun can experience degeneration of the magnetic field.
Solar winds are capable to distort and even destroy the functioning of communication satellites in outer space, such as radio and tv communication and satellite based internet services.
The effects of solar winds that are visible to naked eye are the Aurora Borealis (the Northern lights) and the Aurora Australis (the Southern Lights).
Answer:
genes for flower color and edge shape are linked. They do not assort independently.
Explanation:
<u>Available data:</u>
- test cross between a purple-flowered pea plant having serrated leaves and a white-flowered pea plant having smooth edges.
- serrated leaves → dominant trait
- smooth edges → recessive trait
- purple color → dominant trait
- white color → recessive trait
- F1: 4 purple-serrated:1 purple-smooth:1 white-serrated:4 white-smooth.
There are two genes involved in the cross. The expected ratios are 1:1:1:1 because we assume genes assort independently. However, we see a different phenotypic distribution. When phenotypic ratios differ from the expected ones, it means that genes are linked.
To know if two genes are linked in the same chromosome, we must observe the progeny distribution. If individuals, whose genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. But if instead of this distribution, we observe a different one, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent
Answer:
C) 2.0 kb
Explanation:
It is given that out of the 4 nucleotides A, T, C & G each one has equal probability to occur at any position on the DNA molecule which simply means that the probability of occurrence of any nucleotide at a position is 1/4.
Also, it is given that probability of occurrence of either A or T at 3rd position is equal which means that the probability at that particular position will be 2/4 = 1/2.
Now, GA(A/T)TC is the DNA sequence where Restriction enzyme HinfI cleaves so the total probability of an average HinfI cleavage fragment will be = 1/4 x 1/4 x 1/2 x 1/4 x 1/4 = 0.00195 = 0.2 i.e. 2 kb.
