predicting-inheritance-monohybrid-crosses

Predicting inheritance: monohybrid crosses

Monohybrid inheritance

• Monohybrid inheritance looks at how the alleles for a single gene are passed on from one generation to the next

• Known information about the genotypes, phenotypes and the process of meiosis are used to make predictions about the phenotypes of offspring that would result from specific breeding pairs

• When two individuals sexually reproduce there is an equal chance of either allele from their homologous pair making it into their gametes and subsequently the nucleus of the zygote

  • This means there is an equal chance of the zygote inheriting either allele from their parent

• Genetic diagrams are often used to present this information in a clear and precise manner so that predictions can be made

  • These diagrams include a characteristic table called a Punnett square

• The predicted genotypes that genetic diagrams produce are all based on chance

  • There is no way to predict which gametes will fuse so sometimes the observed or real-life results can differ from the predictions

Worked example: genetic diagram

• One of the genes for the coat colour of horses has the following two alleles:

  • B, a dominant allele produces a black coat when present

  • b, a recessive allele produces a chestnut coat when present in a homozygous individual

• In this example a heterozygous male is crossed with heterozygous female

Parental phenotype: black coat x black coat

Parental genotype: Bb Bb

Parental gametes: B or b B or b

• Predicted ratio of phenotypes in offspring – 3 black coat : 1 chestnut coat

• Predicted ratio of genotypes in offspring – 1 BB : 2 Bb : 1 bb

Codominance

• When working with codominant alleles the genetic diagrams can be constructed in a similar way, however the genotypes are represented using a capital letter for the gene and superscript letters for the alleles (e.g. I^AI^A)

• There will be more possible phenotypes and so the predicted ratios will be different

Worked example: codominance

• The gene for blood type has three alleles:

  • A, a dominant allele produces blood type A

  • B, a dominant allele produces blood type B

  • O, two recessive alleles will produce blood type O

• In this example a blood type A person is crossed with a blood type B person

Parental phenotype: Blood type A x Blood type B

Parental genotype: I^AI^O I^BI^O

Parental gametes: I^A or I^O I^B or I^O

Predicted ratio of phenotypes in offspring – 1 Blood type AB : 1 Blood type A : 1 Blood type B : 1 Blood type O

Predicted ratio of genotypes in offspring: 1 I^AI^B : 1 I^AI^O : 1 I^BI^O : 1 I^OI^O

Sex-linkage

• Sex-linked genes are only present on one sex chromosome and not the other

• This means the sex of an individual affects what alleles they pass on to their offspring through their gametes

• If the gene is on the X chromosome males (XY) will only have one copy of the gene, whereas females (XX) will have two

• There are three phenotypes for females – normal, carrier and has the disease, whereas males have only two phenotypes – normal or has the disease

Worked example: sex-linkage

• Haemophilia is a well known sex-linked disease

• There is a gene found on the X chromosome that codes for a protein called factor VIII. Factor VIII is needed to make blood clot

• There are two alleles for factor VIII, the dominant F allele which codes for normal factor VIII and the recessive f allele which results in a lack of factor VIII

• When a person possesses only the recessive allele f, they don’t produce factor VIII and their blood can’t clot normally

• The genetic diagram below shows how two parents with normal factor VIII can have offspring with haemophilia

Parental phenotypes: carrier female x normal male

Parental genotypes: X^FX^f X^FY

Parental gametes: X^F or X^f X^F or Y

Predicted ratio of phenotypes in offspring – 1 female with normal blood clotting : 1 carrier female : 1 male with haemophilia : 1 male with normal blood clotting

Predicted ratio of genotypes in offspring: 1 X^FX^F : 1 X^FX^f : 1 X^FY : 1 X^fY