Back to homeBiology · 5th Year OL · Change subject
Start for free
5th Year OL · Biology
Molecular genetics · Mendelian genetics

Mendelian crosses

In 5th Year Biology, genetics questions go beyond a single gene: in the Leaving Cert you will be asked to use the notation fluently, deduce genotypes from observed ratios and solve dihybrid crosses. A farmer in Kilkenny who wants a pure breed of cattle needs exactly these skills, and we will come back to her cattle at the end of the page.

Gene, allele and genetic notation

When two yellow pea plants are crossed, 300 yellow seeds and 100 green seeds are obtained: how can a colour absent from the parents reappear if inheritance does not blend hereditary factors?

Pair of homologous chromosomes, one maternal and one paternal, with the same locus at the same height: the maternal one carries allele A and the paternal one allele a, so the genotype is Aa, heterozygous. pair of homologous chromosomes locus allele A allele a maternal chromosome paternal chromosome genotype: Aa (heterozygous)
Practice★☆☆Not started
In pea plants, the allele for smooth seeds (R) is dominant over that for wrinkled seeds (r). Two heterozygous plants are crossed. State the genotypes of the parents, the…

In pea plants, the allele for smooth seeds (R) is dominant over that for wrinkled seeds (r). Two heterozygous plants are crossed. State the genotypes of the parents, the gametes each one produces and the expected genotypic and phenotypic ratios in the offspring.

Solve
Practice★☆☆Not started
In dogs, short hair (L) is dominant over long hair (l). State all the possible genotypes of a short-haired dog and the genotype of a long-haired dog, and explain why…

In dogs, short hair (L) is dominant over long hair (l). State all the possible genotypes of a short-haired dog and the genotype of a long-haired dog, and explain why there is no doubt about the genotype when the phenotype is recessive.

Solve

Law of Segregation: the monohybrid cross

Practice★☆☆Not started
In tomato plants, red fruit (R) is dominant over yellow (r). A homozygous red-fruited plant is crossed with a yellow-fruited plant. State the genotype and the phenotype…

In tomato plants, red fruit (R) is dominant over yellow (r). A homozygous red-fruited plant is crossed with a yellow-fruited plant. State the genotype and the phenotype of the whole F1 and reason whether any yellow-fruited plant can appear in it.

Solve
Practice★☆☆Not started
In rabbits, a black coat (N) is dominant over a brown one (n). A heterozygous male is crossed with a brown-coated female. State the ratios expected in the offspring and…

In rabbits, a black coat (N) is dominant over a brown one (n). A heterozygous male is crossed with a brown-coated female. State the ratios expected in the offspring and calculate how many rabbits of each phenotype are to be expected in a litter of 40 young.

Solve
Practice★☆☆Not started
Two pure-breeding lines of an ornamental plant, one with red flowers and one with white flowers, are crossed, and the whole F1 has red flowers. State the genotypes of…

Two pure-breeding lines of an ornamental plant, one with red flowers and one with white flowers, are crossed, and the whole F1 has red flowers. State the genotypes of the parental generation and of the F1, and the phenotypic ratios expected in the F2 obtained by self-fertilisation of the F1.

Solve

The test cross and probability in genetics

Product rule: the probability that each child is aa is 1/4, and the probability that both children are aa at the same time is 1/4 × 1/4 = 1/16. 1st child: P(aa) = 1/4 2nd child: P(aa) = 1/4 1/4 × 1/4 P(both aa) = 1/4 × 1/4 = 1/16
Practice★☆☆Not started
What is a test cross and what is it used for? Design the cross that would make it possible to find out whether a black-coated mouse (a dominant trait) is homozygous or…

What is a test cross and what is it used for? Design the cross that would make it possible to find out whether a black-coated mouse (a dominant trait) is homozygous or heterozygous, stating what results you would expect in each case.

Solve
Practice★★☆Not started
When two purple-flowered plants are crossed with each other, the offspring consists of 37 purple-flowered plants and 13 white-flowered ones. a) Deduce the genotypes of…

When two purple-flowered plants are crossed with each other, the offspring consists of 37 purple-flowered plants and 13 white-flowered ones. a) Deduce the genotypes of the parental plants and justify your answer from the observed ratios. b) Explain which cross you would carry out to find out whether a purple plant from the offspring is homozygous or heterozygous, stating what results would be expected in each case.

Solve
Practice★★☆Not started
Polydactyly (having extra fingers) is due in humans to an autosomal dominant allele (P), while the normal number of fingers depends on the recessive allele (p). A…

Polydactyly (having extra fingers) is due in humans to an autosomal dominant allele (P), while the normal number of fingers depends on the recessive allele (p). A heterozygous polydactylous man has children with a woman with the normal number of fingers. a) State the genotypes of both. b) Calculate the probability that their first child will be polydactylous. c) Calculate the probability that their first two children will have the normal number of fingers.

Solve
Practice★★★Not started
In cattle, the polled trait (hornless, P) is dominant over the presence of horns (p). A farmer acquires a polled bull with the commercial guarantee that it is homozygous…

In cattle, the polled trait (hornless, P) is dominant over the presence of horns (p). A farmer acquires a polled bull with the commercial guarantee that it is homozygous dominant (PP).

  1. a) Design a test cross to empirically verify the bull's homozygosity. Specify the genotype and phenotype of the cows with which it should be crossed and justify why this choice minimizes the number of crosses needed.
  2. b) The farmer crosses the bull with a group of horned cows and obtains 7 calves in a row, all of them polled. Calculate the probability of having obtained this offspring if the bull were actually heterozygous (Pp) and reason whether the farmer can certify 100% that the bull is genetically pure.
  3. c) Assuming that the bull is indeed homozygous (PP), in another cross this same bull mates with a polled cow whose mother was horned, giving birth to a polled calf. The farmer claims that this calf has a 0.5 probability of being homozygous. Demonstrate, by calculating the conditional probability, whether this statement is correct or false.
Solve

Law of Independent Assortment: the dihybrid cross

Mendel crossed dihybrid plants with yellow round seeds and obtained in the offspring $315$ yellow round, $101$ yellow wrinkled, $108$ green round and $32$ green wrinkled: why do combinations absent from the grandparents appear, and what mathematical rule fixes that $9:3:3:1$ ratio?

Four-by-four Punnett square for the dihybrid cross AaBb × AaBb: sixteen genotypes coloured according to their phenotype, in a 9:3:3:1 ratio. AaBb × AaBb: 16 equally likely combinations AB Ab aB ab AB Ab aB ab AABB AABb AaBB AaBb AABb AAbb AaBb Aabb AaBB AaBb aaBB aaBb AaBb Aabb aaBb aabb 9/16 A_B_: two dominant characteristics 3/16 Abb: A dominant, b recessive 3/16 aaB_: a recessive, B dominant 1/16 aabb: two recessive characteristics _ = either allele (A or a; B or b) Phenotypic ratio 9 : 3 : 3 : 1

From observed ratios to genotype

A breeder in Cork gets from one pair of guinea pigs $28$ black, curly-coated offspring, $31$ black with straight hair, $9$ white curly and $11$ white with straight hair: how do you deduce the exact genotype of the parents by breaking those figures down one characteristic at a time?

Three Punnett squares coloured by phenotype: Aa × aa gives 1:1, Aa × Aa gives 3:1 and AaBb × AaBb gives 9:3:3:1, each with its parental genotype. observed ratio → parental genotype Aa aa Aaaa Aaaa 1:1 Aa × aa Aa Aa AAAa Aaaa 3:1 Aa × Aa ABAbaBab AB Ab aB ab 9:3:3:1 AaBb × AaBb = Aa × Aa and Bb × Bb each characteristic is compared with its own Mendelian ratio
Practice★★☆Not started
When two agouti-coated mice, heterozygous for the colour trait, are crossed with each other, a litter of 96 young is obtained. a) State how many young of each phenotype…

When two agouti-coated mice, heterozygous for the colour trait, are crossed with each other, a litter of 96 young is obtained. a) State how many young of each phenotype would theoretically be expected. b) In the actual litter, 68 agouti mice and 28 non-agouti mice are born: reason whether these results are compatible with the expected Mendelian ratio and what explains the small deviations from the theoretical values.

Solve
Practice★★★Not started
In a genetic counseling session, a family's history is analyzed regarding oculocutaneous albinism, an autosomal recessive condition (allele a versus normal pigmentation…

In a genetic counseling session, a family's history is analyzed regarding oculocutaneous albinism, an autosomal recessive condition (allele a versus normal pigmentation A).

  • Seán and Aoife have normal pigmentation, but Seán's father and Aoife's mother were albino.

  • Seán and Aoife have two healthy children: Cian and Niamh.

  • Cian plans to have offspring with his first cousin Saoirse (daughter of a healthy sister of Aoife and a healthy man with no family history of albinism).

  1. a) Draw the family pedigree using standard genetic symbols and indicate the definite or probabilistic genotype of each of the individuals mentioned.
  2. b) A family member claims: "Since there was never albinism in Saoirse's paternal branch and Cian does not have it, it is impossible for their child to be born albino." Refute this argument and calculate the exact probability that Cian and Saoirse's first child will be albino.
  3. c) If Cian and Saoirse were to have a first albino child, calculate and justify what the probability would be that a second child would also exhibit albinism. Explain the reason for the difference with respect to the probability calculated in the previous section.
Solve

Summary

See all the exercises in this section