Answer:
Explanation:
The observed frequency of the double crossovers (DCOs) is 20/1000, or 0.02. If there is no interference then this will be the same as the expected, which is just the product of the single crossover frequencies, that is, SCO frequency (a-b) x SCO freq. (b-c) = 0.02.
(a) Any combination of map distances whose product (a-b) x (b-c) = 0.02 is possible. However, three are most likely:
a 20 mu b 10 mu c
a 10 mu b 20 mu c
a 14.14 mu b 14.14 mu c
(b) The distances would be exactly the same.
(c) The expected number of DCO progeny would be 17.
Answer: 11.5, but it would be D because it's the closet
Explanation:
The correct answer should be, <span>Absorbed glucose.</span>
Answer:
a. Heterozygous individuals may pass on their copy of the disease-causing allele to offspring.
Explanation:
Tay-Sachs, which is a recessive lethal disease ---- Let say the recessive lethal diseases is s
∴ it only results when an individual posses two copies of the diseases-causing allele i.e two copies of the disease will be ss.
Now, when two hetrozygous individuals crossed , it is obvious that each can pass on their copy of the disease-causing allele to the offspring.
Let show an illustration for the above statement.
Let the heterozygous individual be Ts, if Ts cross with another Ts;
we will have:
Ts × Ts
T s
T TT Ts
s Ts ss
the offspring are TT,Ts,Ts,ss
We can now see how the Heterozygous individuals pass on their copy of the disease-causing allele to the offspring (Ts).
Think b is the right answer
