A large part of the grade 11 university genetics unit involves predicting the genetic makeup of the offspring resulting from a cross between two individuals. Various types of genetics problems are covered below. See the image to the left of each picture to view a guided solution of the problem. Teachers can find a SMARTnote file under the activities resources section entitled "Genetics Problems". This file contains blank punnet squares which can be used to do genetics problems on the SMARTboard.
Monohybrid Cross
Example 1: In pea plants, smooth peas (S) are dominant to wrinkled peas (s). What are the genotypic and phenotypic ratios in a cross of two plants that are both heterozygous for the seed shape trait?
Answer:
- Considers only one trait
Example 1: In pea plants, smooth peas (S) are dominant to wrinkled peas (s). What are the genotypic and phenotypic ratios in a cross of two plants that are both heterozygous for the seed shape trait?
Answer:
- Plants that are heterozygous for a trait have one of each allele (one S and one s).
- The use of a punnet square enables us to determine the genotypic and phenotypic ratios of the offspring. See image for full solution.
Dihybrid Cross
Example 1: In pea plants, S is the allele for the dominant, smooth shape and s is the allele for the recessive, wrinkled shape. Also, Y is the allele for the dominant, yellow pea and y is the allele for the recessive, green pea. What is the resulting phenotypic ratios of a cross between two individuals that are heterozygous for both traits?
Answer:
- Considers two traits (whose alleles segregate independently) simultaneously
Example 1: In pea plants, S is the allele for the dominant, smooth shape and s is the allele for the recessive, wrinkled shape. Also, Y is the allele for the dominant, yellow pea and y is the allele for the recessive, green pea. What is the resulting phenotypic ratios of a cross between two individuals that are heterozygous for both traits?
Answer:
- The genotypes for both parents is SsYy since they are both heterozygous for both traits.
- We keep the law of independent assortment in mind when we determine the gametes of the parents.
- Determining the gametes of the parents can be done using the FOIL method. This is when you take the First, Outer, Inner, and Last pairs of alleles in the parent's genotype (see the image for further clarification.
- Using these gametes in a punnet squares enables us to determine phenotypic ratios. See images for full solution.
Test Crosses & Unknown Parents
Example 1: In humans, brown eye colour (B) is dominant over blue eye colour (b). A man has brown eyes and his wife’s eyes are blue. They have a child with blue eyes. What is the father’s genotype?
Answer:
- The cross of an organism with a dominant phenotype (genotype could be AA or Aa) with an organism that is homozygous recessive for that trait (aa).
Example 1: In humans, brown eye colour (B) is dominant over blue eye colour (b). A man has brown eyes and his wife’s eyes are blue. They have a child with blue eyes. What is the father’s genotype?
Answer:
- Since we know that the wife's eyes are blue (the recessive characteristic), we also know her genotype to be bb.
- The fact that the father has brown eyes (the dominant characteristic) tells us that one of his alleles is B, while the other is unknown (represented by a box in the picture).
- Since we know that they have a child with blue eyes, we know that the second allele from the dad must be b because in order for the blue eye phenotype to be expressed, there must have been one b contributed from each parent. See image for full solution.
Example 2: When considering peas, tall plants (T) are dominant over short ones (t). Two tall pea plants are crossed, producing 105 tall plants & 32 short plants. What are the genotypes of both parents?
Answer:
Answer:
- We know that one of each parent's alleles for this trait is T (since they are both tall).
- The other allele in each parent is unknown.
- The total number of plants in the F1 generation is 137.
- 76.6% of the F1 generation is comprised of tall plants, while the remaining 23.4% are short plants.
- We can therefore take the phenotypic ratio to be 3:1 tall to short plants.
- The punnet square is filled out accordingly, and the empty boxes are filled in with t alleles. The genotypes of both parents are Tt. See image for full solution.
Codominance
Answer:
- In this type of genetics problem, two genes share dominance and are therefore both expressed.
Answer:
- Since the mother's blood type is O, we know that her genotype is ii.
- Since the child's blood type is A, we know that their genotype must include at least one A allele. This means that the father's genotype must include one A allele. The empty boxes in the picture represent an unknown allele.
- Since the father's genotype must only include one A allele, we know that he can be blood type A or AB. See image for full solution.