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ycow [4]
3 years ago
11

xxy Consider a pea plant- it has a gene controlling seed color- yellow Vs green; another gene controlling seed texture-smooth Vs

wrinkled. A true breeding smooth green plant is crossed to a true breeding wrinkled yellow. All the F1 progeny were smooth yellow. The F1 generations were self-fertilized. Assume the two genes (seed color and seed texture) are VERY TIGHTLY LINKED. In the F2 what is the expected ratio of the phenotypes- Smooth & yellow: smooth & green: wrinkled & yellow: wrinkled & green
Biology
1 answer:
emmasim [6.3K]3 years ago
3 0

Answer:

2 Smooth yellow : 1 smooth green : 1 wrinkled yellow : 0 wrinkled green

Explanation:

The cross between a true breeding smooth green and a true breeding wrinkled yellow produced a homogeneous smooth yellow F1 . According to Mendel's Law of Dominance, the traits that appeared in the progeny are dominant.

The parents were true breeding, so they are homozygous for all genes.

If the genes are linked, their genotypes can be written as:

     smooth green  X  wrinkled yellow

<h3>       Sy/Sy           X      sY/sY</h3><h3>F1:                    Sy/sY</h3><h3></h3>

The F1 is self-fertilized, so a cross Sy/sY x Sy/sY is done. If the genes are very tightly linked, almost no recombination happens between them and all gametes will be parental.

The F1 individuals can produce 2 types of gametes then: <em>Sy </em>and <em>sY</em>.

The expected offspring would be:

  • 1 Sy/Sy: Smooth green
  • 2 Sy/sY: Smooth yellow
  • 1 sY/sY: wrinkled yellow
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