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kow [346]
3 years ago
6

In sweet peas blue flowers are the result of a dominant factor (R) and red flowers of its recessive allele (r). Long pollen is d

ue to the dominant factor (L) and round pollen to its recessive allele (l). A plant homozygous for blue flowers and round pollen is crossed with one homozygous for red flowers and long pollen. The F1 individuals were crossed to red flowered, round pollen plants (test cross).Diagram this cross as though these genes were independently assorting and give the expected ratios.The actual results were: 23 blue, long; 153 blue, round; 155 red, long; 21 red, roundBased on the observed results are these genes assorting independently?If they are linked compute the cross-over frequency (map distance).
Biology
1 answer:
adelina 88 [10]3 years ago
6 0

Answer:

a) If the genes are independent, the expected phenotypic ratios are:

1/4 blue, long; 1/4 blue, round; 1/4 red, long; 1/4 red, round

b) The genes are linked and separated by 12.5 map units.

Explanation:

Blue flowers (R) is dominant to red flowers (r)

Long pollen (L) is dominant to round pollen (l).

A plant homozygous for blue flowers and round pollen (Rl/Rl) is crossed with one homozygous for red flowers and long pollen (rL/rL).

The F1 individuals (Rl/rL) were test crossed (rl/rl).

<u>If the genes assort independently, the F1 individuals can produce 4 types of gametes, all with the same proportions:</u>

  • 1/4 RL
  • 1/4 Rl
  • 1/4 rL
  • 1/4 rl

The homozygous recessive individual can only produce <em>rl </em>gametes.

<u>The expected phenotypic ratios would be: </u>

  • 1/4 blue, long
  • 1/4 blue, round
  • 1/4 red, long
  • 1/4 red, round

<u>However, the actual results were: </u>

  • 23 blue, long (RL/rl)
  • 153 blue, round (Rl/rl)
  • 155 red, long (rL/rl)
  • 21 red, round (rl/rl)

Total: 352

As you can see, the individuals that inherited the parental gametes (Rl and rL) are the most abundant, and the other two are the recombinant gametes. The proportions of the four types of gametes are very different, so these two genes are linked.

Map distance = 100 × Number of recombinants/Total

Map distance = 100 × (23 + 21) / 352

Map distance = 12.5 map units.

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