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Helen [10]
3 years ago
9

If there is complete linkage with genes Ss (for smooth or wrinkled seeds) and Gg (for green or yellow seeds), what phenotypes wo

uld you expect in the offspring of a cross with one heterozygous parent (Ss, Gg), which would all produce all smooth and green seeds, and a homozygous recessive parent (ss, gg), which would produce all wrinkled yellow seeds?

Biology
1 answer:
KonstantinChe [14]3 years ago
5 0

Answer:

Smooth and green seeds

Smooth and yellow seeds

Wrinkled and green seeds

Wrinkled and yellow seeds

Explanation:

This is a typical dihybrid cross involving two linked genes; one gene coding for seed texture and the other for seed colour. The allele for Smooth seed (S) is dominant over the allele for Wrinkled seed (s) in the first gene while the allele for green seed (G) is dominant over the allele for yellow seed (g) in the second gene.

The cross is between a heterozygous smooth, green plant and a homozygous recessive wrinkled, yellow plant

According to the law of independent assortment, each parent will undergo meiosis to produce the following gamtes with genotypes:

SsGg: SG, Sg, sG, sg

ssgg: sg, sg, sg, sg

Using a punnet square to perform a cross between these gametes ( check attached image), a total of 16 offsprings with four different phenotypes will be expected. The phenotypes are as follows:

SsGg: Smooth and green offsprings (4)

Ssgg: Smooth and yellow offsprings (4)

ssGg: Wrinkled and green offsprings (4)

ssgg: wrinkled and yellow offsprings (4)

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If you count 52 purple rough corn kernels and 50 yellow smooth corn kernels how many people smooth corn kernels should you expec
Vitek1552 [10]

Answer:

17

Explanation:

In corn,

Smooth kernel (S) is dominant to rough kernel (s)

Purple kernel (P) is dominant to yellow kernel (p)

Since no information has been given about the cross here let us assume it to be a standard dihybrid cross.

The phenotypic ratio of dihybrid cross is 9:3:3:1. 9/16 will show dominant phenotype for both traits, 3/16 will show dominant phenotype for one trait, other 3/16 will show dominant phenotype for the second trait, 1/16 will have recessive phenotype for both the traits.

Here,

Purple rough (P_ss) = 52

Yellow smooth (ppS_) = 50

Both of them show dominant phenotype for one trait so they each form 3/16 of the progeny.

If the total progeny was z, 3/16*z = 50

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So approximately 17 corn kernels will be yellow and rough.

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