If the cross from #1 is continued, what would be the expected outcome in the f2 generation?

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A test cross involves mating an unknown genotypic individual with a known homozygous recessive

• This is because recessive alleles will always be masked by the presence of dominant alleles
• Hence the phenotype of any offspring will reflect the genotype of the unknown parent

Testing an Unknown Dominant Phenotype

Test crosses can be used to determine whether a dominant phenotype is homozygous or heterozygous

• If the unknown parent is homozygous dominant, all offspring will express the dominant phenotype
• If the unknown parent is heterozygous, half the offspring should be dominant and half recessive

Testing for Gene Linkage

Test crosses can also be used to determine if two genes are linked or unlinked by mating with a known heterozygote

• If there is an equal ratio of the four potential phenotypes, the two genes are likely unlinked (independent assortment)
• If there are two phenotypes in high amounts and two phenotypes in low amounts (recombinants), the two genes are likely linked
• A chi-squared test for association can be used to determine the statistical likelihood of each scenario

Test crosses require large numbers of offspring to produce reliable data for meaningful conclusions

• With the advent of genetic screening and genome mapping, test crosses have become less commonly used

Tutorial to help answer the question

Tutorial

Predicting the genotype of offspring

Determine all possible combinations of alleles in the gametes for each parent. Half of the gametes get a dominant S and a dominant Y allele; the other half of the gametes get a recessive s and a recessive y allele. Both parents produce 25% each of SY, Sy, sY, and sy. (Review the tutorials for problems #1 and problem #4 if necessary).

Since each Parent produces 4 different combinations of alleles in the gametes, draw a 4 square by 4 square punnett square.

List the gametes for Parent 1 along one edge of the punnett square.

List the gametes for Parent 2 along one edge of the punnett square.

Fill out the squares with the alleles of Parent 1.

Fill out the squares with the alleles from Parent 2. The result is the prediction of all possible combinations of genotypes for the offspring of the dihybrid cross, SsYy x SsYy.

Predicting the phenotype of offspring

There are 9 genotypes for spherical, yellow seeded plants. They are: SSYY (1/16) SSYy (2/16) SsYY (2/16) SsYy (4/16)

Two recessive alleles result in green seeded plants. There are 2 genotypes for spherical, green seeded plants. They are: SSyy (1/16) Ssyy (2/16)

Two recessive s alleles result in dented seeded plants. There are 2 genotypes for dented, yellow seeded plants. They are: ssYY (1/16) ssYy (2/16)

A ssyy plant would be recessive for both traits. There is only 1 genotypes for dented, green seeded plants. It is: ssyy (1/16)

A phenotypic ratio of 9:3:3:1 is predicted for the offspring of a SsYy x SsYy dihybrid cross.

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