A Mendelian gene generally has two types of alleles: dominant and recessive. An individual always contains two sets of chromosomes (homologous chromosomes) that contain the gene. If both chromosomes contain a dominant allele, then the dominant trait is expressed. If both contain the recessive allele, then the recessive trait is expressed.

If an individual carries both alleles (a heterozygous individual), then only the dominant trait is observed. This occurs because the recessive allele is silenced and is not expressed in the presence of a dominant allele. If the recessive allele is expressed in conjunction with the dominant allele, then it is called incomplete dominance. In normal heterozygous genes, however, the recessive allele is not expressed.

Each chromosome carries only one copy of each gene, and cannot accommodate two alleles for a single trait.

The passage states that the F1 generation only had round, yellow seeds. Test crossing an F1 offspring lead to an equal ratio of four different phenotypes. When you test cross, you are crossing the F1 offspring with a homozygous recessive individual; therefore, the test cross individual had recessive seed shape and seed color.

???? x aabb

If round and yellow were recessive, then the F2 offspring would all be round and yellow (because you would cross round/yellow with round/yellow).

aabb x aabb (all offspring round and yellow)

If only round was recessive then you wouldn’t get any wrinkled seeds in F2 generation; similarly, if only yellow was recessive then you wouldn’t get any green seeds.

Aabb x aabb (all offspring have one recessive trait, bb)

If round and yellow are both dominant, then means that wrinkled and green are recessive. If the F1 offspring is heterozygous for both traits, then we can see the observed ratios from the test cross.

AaBb x aabb

1 AaBb (round/yellow), 1 aaBb (wrinkled/yellow), 1 Aabb (round/green), 1 aabb (wrinkled/green)

We can conclude that green and wrinkled must be recessive to yellow and round.

The scenario described is a dihybrid cross. Draw a Punnett Square for the scenario, and identify the frequency of obtaining the genotype (wwrr), which is the only genotype that will display both recessive traits.

You should be familiar with the ratios represented in a dihybrid cross: 9:3:3:1. There will be nine inidividuals with both dominant traits (pointed leaves and white flowers), three individuals dominant for one trait (round leaves and pink flowers), three individuals dominant for the other trait (pointed leaves and white flowers), and only one individual recessive for both traits (pointed leaves and pink flowers).

The genotype to obtain pointed leaves is rr, while the genotypes to obtain white flowers are WW or Ww. Determine the frequence of obtaining WWrr or Wwrr using a Punnett Square.

You should be familiar with the ratios represented in a dihybrid cross: 9:3:3:1. There will be nine inidividuals with both dominant traits (pointed leaves and white flowers), three individuals dominant for one trait (round leaves and pink flowers), three individuals dominant for the other trait (pointed leaves and white flowers), and only one individual recessive for both traits (pointed leaves and pink flowers).

If a trait is expressed by complete dominance, the dominant allele will be expressed over the recessive allele. The only time that the recessive trait is expressed is when the organism has two recessive alleles. The genotype is the genetic makeup of an organism, while the phenotype is the physical expression of a trait.

Since there are only two alleles for the trait, there are two distinct phenotypes that the organism can express. A Punnett square reveals that two heterozygous organisms will yield a progeny with 75% displaying the dominant phenotype. 50% of the progeny will be heterozygous, and the other half will be homozygous.

Aa x Aa

Child 1: AA (homozygous, dominant)

Child 2: Aa (heterozygous, dominant)

Child 3: Aa (heterozygous, dominant)

Child 4: aa (homozygous, recessive)

Males carry an X-chromosome and a Y-chromosome, while females carry two X-chromosomes. A son, by necessity, must inherit a Y-chromosome from the father and an X-chromosome from the mother. In this case, the affected father cannot possibly transfer his allele to the son because the condition is linked to the X-chromosome. We also know that the trait is X-linked dominant and that the mother is unaffected; since the trait is dominant, she cannot possible be a carrier, and must have two copies of the wild type allele. As a result, the son must inherit a Y-chromosome from the father and a wild type X-chromosome from the mother. The son cannot possibly be affected.

The cross between two heterozygous plants (Tt and Tt) will result in 25% TT, 50% Tt and 25% tt.

This is a simple Mendelian recessive trait. It must be recessive if the father is a carrier but never affected. 

We are unsure if our yellow pea is homozygous dominant (CC) or heterozygous (Cc). Crossing it with a homozygous recessive (cc) green pea will yield only yellow peas if it is homozygous dominant, or a mix of green and yellow if it is heterozygous. We could also cross it with a known heterozygote, as we would see the same pattern as if we had crossed with a homozygous recessive.  

We are told that the mother is homozygous for blood type A (AA) and the father has blood type AB (AB). Using a simple punnett squre we can find that 50% of the offspring would be blood type A and 50% would be blood type AB.