Multiple choice 1. Which of the following types of plants would usually be a see
ID: 208738 • Letter: M
Question
Multiple choice
1.
Which of the following types of plants would usually be a seedless variety?
Tetraploid
Triploid
Diploid
Aneuploid
2.
Consider a normal chromosome with the following gene sequence along its length: FGHIJKL. Which of the following sequences represents an inversion?
FGHKL
FGHIJKJKL
JKLFGHI
FGJIHKL
3.
The house mouse Mus musculus has a diploid chromosome number of 40. Suppose that the first meiotic division of a germ cell is normal, but a single chromosome in one of the two daughter cells undergoes non-disjunction in meiosis II. How many chromosomes would be present in each of the four gametes that result from that meiosis?
20, 20, 21, 19
12, 12, 8, 8
10, 10, 12, 8
22, 20, 18, 16
21, 21, 19, 19
4.
What is the final stage of the Holliday model?
formation of a Holliday junction
formation of a double-stranded break
gene conversion
resolution
anticipation
5.
Short sequences of DNA synthesized on the lagging strand
lagging fragments
replication fragments
Okazaki fragments
restriction fragments
Tetraploid
Triploid
Diploid
Aneuploid
Explanation / Answer
1. Option D is correct.
Aneuploids do not produce seeds as they do not contain the exact set of chromosomes (they either contain n+ or n- chromosomes which can not properly segregate between daughter cells during meiosis).
2. Option D is correct.
Original sequence: FGHIJKL
Inverted sequence: FGJIHKL
here, the JIH sequence is inverted.
3. Option A is correct.
Meiosis produces haploid cells.
2n = 40
n = 20
But a single non-disjunction event produces defective cells with an abnormal number of chromosomes.
4. Option D is correct.
Resolution is the last step in holiday model
5. Option C is correct.
Short DNA fragments synthesized on the lagging strand are knwon as lagging strand.
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