Inheritance 10.2 HL

  2. 10 Genetics AHL
  3. Inheritance 10.2 HL
    4. 30'
Inheritance has been covered in the SL genetics topic already so this HL topic covers dihybrid crosses and the discovery of linkage by Morgan with Drosophila flies. The prediction of phenotype ratios using larger Punnett squares is required as is the understanding of how phenotype ratios of 9:3:3:1 or 1:1:1:1 ratios can appear. The identification of recombinant phenotypes is important and so is the ability to recognise signs of linkage.

Key concepts

Learn and test your biological vocabulary using these 10.2 Inheritance flashcards.

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Essentials

These slides summarise the essential understanding and skills in this topic. 
They contain short explanations in text and images - great revision.

Read the slides and look up any words or details you find difficult to understand.

These slides summarise the essential understanding and skills in this topic. 
They contain short explanations in text and images - great revision.

Read the slides and look up any words or details you find difficult to understand.

Summary

Summary list for 10.2 Inheritance

Linked genes

  • Genes are linked when they occur on the same chromosome.
  • Unlinked genes show independent segregation in meiosis.

Polygenic inheritance

  • Variation between living can be discrete or continuous.
  • Phenotypes controlled by many genes (polygenic) show continuous variation. They may also be influenced by environmental factors.
  • Chi-squared tests are used to calculate if differences between observed and expected frequencies are statistically significant.

Student skills and applications

  • A awareness of Morgan’s discovery of non-Mendelian ratios in Drosophila.
  • The ability to complete Punnett squares for dihybrid traits and analyse them.
  • Ability to calculate predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • The skill to identify recombinants in crosses involving two linked genes and to understand when they are shown as vertical pair in crosses of linked genes.
  • The skill to use of a chi-squared test on data from dihybrid cross.

Mindmaps

This diagram summaries the main sections of topic 10.2 Inheritance HL
Test if you can draw something like these concept maps from memory.

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Exam style questions

Exam style question about inheritance

Explaining the process of Mendelian inheritance is an important skill from this topic.

Answer the question below on a piece of paper, then check your answer against the model answer.

In the mustard plant Arabidopsis (Arabidopsis thaliana), there are two genes which affect plants: The tallness gene and the leaf shape gene. The allele for tall plants is T (dominant) and the allele for short plants is t. The allele for round leaves is R (dominant) and the allele for oval leaves is r.

a) Determine the phenotype of a plant with the genotype: Tt rr.       [1].

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b) Plants which are tall with round leaves are crossed with plants that are short with oval leaves.
    Surprisingly all the 1000 offspring were tall with round leaves.

   Deduce the genotype of the offspring.                                            [1]

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c) In another cross offspring from (b) were crossed and four different phenotypes were found in the offspring.
   Suggest what the phenotype ratio of offspring would be.               [2]

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d) Identify the recombinant phenotypes in this cross.                    [1]

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Click the + icon to see a model answer.

Extra exam question on polygenic inheritance.     Click to open.

Polygenic inheritance and continuous variation are important concepts in this topic.

Answer the question below on a piece of paper, then check your answer against the model answer.

Explain why traits determined by polygenic inheritance tend to demonstrate continuous variation. (4 marks).

Hint: Do this in italics, you may gain marks for the method of the calculation if your answer is incorrect.
Click on the "+" icon to see a model answer.

Extra exam question on dihybrid inheritance     - - - click to open.

Deriving genotypes and expected phenotypic ratios from the results of breeding is an important concept in this topic.

Answer the question on a piece of paper, then check your answer against the model answer below.

Antirrhinum flower

A pure-breeding red-flowered, short stem snapdragon (Antirrhinum majus) is crossed with a pure-breeding white-flowered long stem variety. The seeds are planted and the F1 generation are all pink-flowered and long stem.

i Derive and explain the genotypes of the parental generation. (4 marks).

ii Predict the result of breeding the F1 generation with a white flowered short- stemmed plant (assuming that the gene for flower colour and stem length are on different chromosomes). (5 marks).

iii Suggest two reasons why an achieved phenotypic ratios can significantly differ from the predicted ratio. (2 marks).

Hint: Use the correct terminology in responding to genetics questions.
 
Click on the "+" icon to see a model answer.
 

Test yourself

Test yourself - multiple choice questions

This is a self marking quiz containing questions covering the topic outlined above.
Try the questions to check your understanding.


START QUIZ!

Drag and drop activities

Test your ability to construct biological explanations using the drag and drop questions below.

You need to be able to work out parental genotypes from their offspring.

In humans, curly hair, C is dominant to straight hair, c.

The haemophilia gene is a recessive gene on the X chromosome denoted as Xh. Normal blood clotting is represented as X.

Habib’s mother, Amal and father, Omer both have curly hair and neither have haemophilia. Their son, Habib has straight hair and is a haemophiliac.

Deduce, with reasons, the genotype of Habib, Amal and Omer by dragging and dropping the correct genotype into the box beside the name.

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XYhcc XhYCc XhYhcc XXhCc XhYCc XYCc XXhCC Xhc XXCc

Amal

Omer

Habib

Explanation: Habib inherited the haemophilia gene from his mother who is a carrier. Both parents are heterozygousfor curly hair.

Extra drag and drop explanation on dihybrid inheritance  click to open.

You need to be able to work out parental genotypes from their offspring. You need to know the genetics of blood type.

In humans, white forelock is a rare trait, shown in the image.

Both Anielle and Virat have white forelock and are A blood group. Their daughter, Christianna does not have the white forelock trait and is O blood group

Explain the genotypes of Anielle. Virat and Christianna by dragging the correct genotype into the box next to the name and complete the explanation.

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iiww phenotype dominant heterozygous IAiWw recessive OOww AOWw IAiWw genotype

White forelock is a trait. Anielle and Virat are both for the white forelock trait and their daughter, Christianna is homozygous , not exhibiting the trait.

Anielle and Virat both have the recessive gene for O for blood group as Christianna has the O .

Anielle's genotypes for these traits could be denoted and Christianna's as

Use the clues in the data given to work back to the parental genotype.

Examiner hint: Ensure that you represent the blood group genes correctly as IA, IB, i.

A, B, AB and O are phenotypes.

Extra drag and drop explanation on linked genes  click to open.

The concept of linked genes is an important concept in this sub topic.

Drag and drop the correct term into the gap to outline the concept of autosomal gene linkage on chromosomes (see also the next activity).

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locus chromosome do ratios chiasmata many recombinants do not a different sequence few alleles independently linked same sequence

Genes are always found in a certain (position) on a . Chromosomes with the of genes are homologous. As genes may have one or more , homologous chromosomes do not necessarily carry identical genetic information.

During meiosis, genes on the same chromosome independently assort, they are linked and inherited as a group on a chromosome, not as individual genes. Normal Mendelian predicted phenotypic for genetic crosses are valid for genes that are assorted - when the ratio differs greatly from a predicted Mendelian ratio, it is very likely that the genes being investigated are on the same chromosome.

If genes are linked, (genotypes different from the parent genotype) are produced by and are generaly in number.

There are two forms of linkage - sex-linked and autosomal. The explanation of this is in another drag and drop activity.

Extra drag and drop explanation on autosomal and sex-linked genes  click to open.

The concept of autosomally linked genes is an important concept in this topic. Sex-linked genes are considered in Topic 3.

Drag and drop the correct word or phrase into the gap to summarise the differences between autosomally linked and sex-linked genes.

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autosomes non genderosomes carrier male different dominant identical heterosomes loci single one copy heritable disorder homozygous heterozygous

Sex-linked genes are in on the gender-determining chromosomes ( ) whereas autosomally linked genes are located on the gender determining chromosomes, ( ).

If genes are sex linked, there will be phenotypic outcomes in humans as males have a X chromosome and females have two. A recessive on the X chromosome (such as red-green colour blindness), if present, will always be expressed in the phenotype but not in the female if the normal gene is present. A heterozygous female would be a of the trait.

If genes are autosomally linked, there will be phenotypic outcomes in either gender. Both a male and a female could be a carrier of a trait if they possess of the recessive gene for the inherited disorder.

Sex-linked genes are inherited with the X chromosome. The Y chromosome determines gender (in humans) but does not have genes corresponding to those on the Y chromosome.

Just for fun

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