4.3c Genetics

06/02/2013 § 1 Comment

Last but not least – genetic diseases, carriers, and the sex genes! We are aware of what genetic diseases are and how they are caused. Are we aware, though, of those individuals that pass down the disease into their offspring even if they themselves don’t show it? Most genetic diseases are caused by the recessive allele of the gene, therefore three out of the four possible alleles (AA, aA, and Aa, for example) will not result in the disease. The last possibility though (aa) will in fact result in the genetic disease, and this is shown in the offspring of carriers. Carriers are individuals that have one copy of a recessive allele who can then pass that allele down to their offspring, potentially giving them the genetic disease (if the other allele turns out to also be recessive).

As we’ve mentioned before, sickle-cell anemia is the most common genetic disease in the world. The base substitution mutation that changes the codon GAG to GTG is what transcribes as GUG instead of the appropriate GAG and the cell would then produce the amino acid valine instead of glutamic acid. Hemoglobin molecules in the cell are then altered to stick together instead of letting blood flow freely. The amount of red blood cells that are lost are done so rapidly – too fast to be replaced – and the individual develops anemia.

Now, there’s a thing called sex linkage (because I’m so great at transitioning) which is a pattern of inheritance where there are differences in genotypes and phenotypic ratios between males and females. Obviously the sex genes are “linked” or important in this idea. If a gene is carried on the X chromosome (of which females have two), then the pattern of inheritance is different for males and females. Because females have two X-chromosomes and males have only one, sex-linked genes are almost always located on the X-chromosome. Females would then have two copies of the sex-linked gene while males have only one copy.

Last but not least, the gender of a child is determined by the chromosome it receives from its father. Half of a male’s gametes carry Y-chromosomes, and the other half carry the X-chromosome. (It doesn’t matter for the mother because all gametes hold X-chromosomes.) If the offspring receives the Y-chromosome, it will be a boy; if the offspring receives the X-chromosome, it will be a girl.

Okay, now I have a test to study for and a talent show to run. You better be there! Yes, you!

Essay Questions

  1. Define the terms gene and allele and explain how they differ. 4 marks
  2. Describe the consequences of a base substitution mutation with regards to sickle cell anemia. 7 marks
  3. Outline the formation of chiasmata during crossing over. 5 marks
  4. Explain how an error in meiosis can lead to Down syndrome. 8 marks
  5. Karyotyping involves arranging the chromosomes of an individual into pairs. Describe one application of this process, including the way in which the chromosomes are obtained. 5 marks
  6. Compare the processes of mitosis and meiosis. 6 marks
  7. Outline one example of inheritance involving multiple alleles. 5 marks
  8. Describe the inheritance of ABO blood groups including an example of the possible outcomes of a homozygous blood group A mother having a child with a blood group O father. 5 marks
  9. Outline sex linkage. 5 marks
  10. Explain, using a named example, why many sex-linked diseases occur more frequently in men than women. 9 marks


Page 153, hemophilia in Queen Victoria’s family

1. Explain why there are no male carriers of the allele for hemophilia in this pedigree chart.

There are no male carriers for hemophilia in this chart because hemophilia is a sex-linked gene, meaning the allele for the disease can only be found in x-chromosomes. The males in the chart should then only have half a chance of having the chromosome and even then, the x-chromosome they get might be one that doesn’t have the allele.

2. Explain why there are no females with hemophilia in this pedigree chart.

Females have a higher chance of having both or one of each allele, therefore the dominant allele that they all appear to have cancels out the recessive, skipping over hemophilia entirely.

3. Explain the evidence from the pedigree chart for hemophilia being a sex-linked condition.

Queen Victoria was apparently a carrier, which means that her x-chromosome that held the allele for hemophilia is the one that gave Maurice, Leopold, Alfonso, Viscount Trematon, Fredrick William, Henry of Prussia, and the other hemophiliac males the condition. Although Prince Albert wasn’t hemophiliac, his heterozygous genes, combined with Victoria’s X-chromosome with the recessive allele are what caused Leopold I to have hemophilia. Victoria’s X-chromosome is the main factor.

4. Deduce the possible genotypes and phenotypes of the daughters of Nicholas II of Russia and Alexandria. Show your working and indicate the probabilities.

Alexandra was a carrier female, meaning her chromosome pair must have been heterozygous (Hh). Nicolas II of Russia was a normal male, indicating that he was either homozygous dominant (HH) or heterozygous (Hh). In that case, the possible outcomes would be: HH, Hh, OR hh. The chances of HH are 5/12, the chances of heterozygous Hh are 6/12, and the chances of the homozygous hh is 1/12.

5. Deduce the possible genotypes and phenotypes of the second and third sons of Alfonso XIII of Spain and Victoria Eugenia. Show your working and indicate the probabilities.

Alfonso XIII of Spain and Victoria Eugenia are in the same situation as Alexandria and Nicholas II of Russia, therefore the chances of their children’s genotypes are the same. 5/12 chance – HH, 6/12 (½) chance – Hh, and 1/12 chance – hh.


Chapter 12 Questions

1. Distinguish between:

  • a) genotype and phenotype – The genotype is the combination of alleles that codes for the trait, and the phenotype is what you see on the individual (the physical characteristics of the gene).
  • b) dominant and recessive alleles – Dominant alleles are expressed no matter if the gene is heterozygous or homozygous; recessive alleles can only be expressed if the genes are homozygous recessive
  • c) homozygous and heterozygous – Homozygous chromosomes contain the same allele (dominant or recessive) and heterozygous chromosomes contain one of each allele (both dominant and recessive)
  • d) gene and genome – a gene is a portion of the chromosome that codes for a certain trait (it is a heritable factor controlling a specific characteristic) and the genome is the entire set of chromosomes of that organism that codes for the entire organism
  • e) haploid and diploid – states of ploidy: haploid is n, diploid is 2n; haploid is one chromosome of each type, diploid is two chromosomes of each type

2. A three-leaf clover plant was crossed with a four-leaf clover plant. All the offspring were four-leaf. When these F1 hybrid plants were crossed together, there were both three-leaf and four-leaf plants in the F2 generation.

a) State the genotype of the F1 plants, using suitable symbols for the alleles. Include a key to the symbols in your answer. – The genotype of the F1 plants is Ll (L = dominant, l = recessive).


  • i) Deduce the expected ratio of genotypes and phenotypes in the F2 generation, showing your working. – The expected ratio of genotypes – 1 LL : 2 Ll : 1 ll. The expected ratio of phenotypes – 3 four-leaf clovers : 1 three-leaf clovers. I used a Punnett square when I crossed the F1 generations with genotypes of both Ll.
  • ii) Discuss whether this is the expected ratio of three-leaf and four-leaf clover plants in wild populations of clover. – This is probably not the expected ratio of three-leaf and four-leaf clover plants in wild populations because there could be other species of clovers that might have five or two leaves instead of only three and four. In wild populations, there is usually more variety than, say, here where we live in Rokko Island, so the ratio would be different and spread out amongst the different species. 

c) Explain how homozygous varieties of three-leaf clover and four-leaf clover could be developed, starting with wild plants. – Homozygous varieties of three-leaf clover and four-leaf clover could be developed when different species are crossed with each other. If there are heterozygous individuals amongst the wild plants (and chances are that there are heterozygous individuals somewhere in the wild), then crossing those heterozygous organisms would result in at least 1 out of 4 individuals that are homozygous dominant or homozygous recessive. // Three-leaf clovers will produce more three leaf clovers when crossed together and we can identify the homozygous four-leafed plants by test crossing; if four leaf clovers only produce four leaf clovers during a test cross, then three leaf clovers are homozygous.

3. The pedigree in Figure 23 shows the ABO groups of three generations of a family.

a) Deduce the genotype of each person in the family.

  • – In order, the genotypes on generation I are: (1) I^AI^B, (2) I^Bi, (3) ii, (4) I^Bi. 
  • Generation II: (1) I^BI^B or I^Bi, (2) I^Ai, (3) I^Bi, (4) ii, (5) ii. 
  • Generation III: (1) ii, (2) I^Ai, (3) I^BBi, (4) ii, (5) ? Wait, give me a second.

b) Deduce the possible blood groups of individual III 5, with the percentage chance of each. – Blood group AB  = 25%, blood group A = 25%, blood group B = 25%, blood group O = 25%.

c) Deduce the possible blood groups and the percentage chance of each blood group:

  • i) of children of individual II 1 and his partner who is also in blood group O – (Because II 1 and his partner are both in blood group O, and the genotype for O is ii, then) the children will be 100% in blood group O, too (unless they go through a mutation). 
  • ii) of children of III 2 and her partner who is in blood group AB. – 50% blood type A, 25% blood type AB, 25% blood type B.


4. [Figure 24] shows two chromosomes and eight genes on them.

  • a) State one allele that is recessive, one that is dominant and one that is co-dominant. – Recessive: h, dominant: D, co-dominant: C^WC^R.
  • b) State one gene for which the chromosomes are homozygous and one for which they are heterozygous. – Homozygous: FF, heterozygous: Ee
  • c) Explain whether the chromosomes are identical, homologous, or different. – The chromosomes are not IDENTICAL because the alleles are DIFFERENT but they are HOMOLOGOUS because they have the same genes, even if the alleles are DIFFERENT. WE CAN CONCLUDE THAT THE CHROMOSOMES ARE HOMOLOGOUS.
  • d) Explain how the diagram shows that B and b are different alleles of the same gene. – The diagram shows this because the chromosomes are lined up (the centromere between C^WC^R and Dd indicates alignment) and the alleles b and B are lined up one on top of the other, which suggests that they are different alleles for the same gene.

5. A mother and father have two sons, one of whom has hemophilia and the other does not. They also have a daughter. The son who has hemophilia was treated with factor VIII, the protein that is missing from his blood, but the factor VIII was contaminated with HIV and the son has developed AIDS. Hemophilia is sex-linked. – I HAVE A QUESTION: Is the dad hemophiliac or not? That is the question.

  • a) Explain how the son inherited hemophilia. – The son inherited hemophilia because of his innocent mother. With his one X-chromosome and one Y-chromosome, the chances of him getting hemophilia or not were 50-50 when he became a boy (?). His mother had a genotype of X^HX^h and he just so happened to receive the recessive X-gene with the sex-linked allele for hemophilia (while his brother got the dominant gene that did NOT code for hemophilia). Since the Y-chromosome doesn’t hold any of these hemophiliac genes and came from his father, the son with hemophilia got the disease from one of his mother’s X-chromosomes.
  • b) Explain the advice for the daughter and non-hemophiliac son about the chance of them passing on hemophilia to their children. – Good afternoon Mr. and Ms. Smith, I’m sorry to hear about your brother (thank you, they say), but if only to cut to the chase, there is good news and bad news. (Okay, what’s the good news) Well, Mr. Smith, there is very little chance, almost 0%, that you’ll be passing hemophilia to any of your future children. (Oh, that’s great! But why is that, he asks) See, you received your Y-chromosome from your father, which made you a male. The X-chromosome you received from your mother could have either been a dominant gene, in terms of hemophilia, or a recessive gene. The dominant gene, the one you inherited, does not code for hemophilia, hence why you don’t have it, but your brother, bless his soul, received the recessive gene that DOES code for hemophilia, which is how he got it. That being said though, you don’t have the recessive gene of hemophilia within your genome and unless a mutation occurs in the future amongst your children, there’s a very, very small chance you’ll be passing it down to them. (Then what’s the bad news, they ask) The bad news is that, Ms. Smith, we’ve deduced the possible genotypes of your hemophiliac gene based on whether your father has hemophilia or not – and we still don’t know because he’s missed his doctor’s appointment for the fourth time this month (I’m so sorry about that, I’ll talk to him right away, she says). Nevertheless, there is a 25% chance your genotype is X^HX^H (which means no hemophilia), a 50% chance you have X^HX^h (also no hemophilia, but your children could have it), and a 25% chance you have X^hX^h (which means hemophilia). (Oh, no, Dr. Medalla, what does this mean! she exclaims) We’ll test you later to determine your actual genotype to be certain but this means there is a 75% you can most definitely pass hemophilia to your children in the future, especially if you yourself are hemophiliac, and if your partner is heterozygous for the gene or hemophiliac themselves. (What should I do? she says desperately) There’s nothing you can do, ma’am, I’m sorry. If it matters that much, I’d suggest choosing a partner who is homozygous dominant to avoid the occurrence entirely, or if you’re a risk-taker, a heterozygous individual. GOOD LUCK.
  • c) Suggest possible social effects for the family of the hemophilia and AIDS. – Especially now that the family has AIDS through the brother who has hemophilia, they would be receiving a lot of pity and some more insensitive individuals would probably try to avoid the family, especially the brother. At the same time, knowing the family has a history of hemophilia or a potential of passing down hemophilia may or may not reduce the chances of the brother without hemophilia or the sister with/without hemophilia ever landing a partner. // Also: anxiety/guilt for passing on hemophilia, ill health and hospital visits, loss of employment/income, grief after death of relative (brother) who developed AIDS.


  • a) A female magpie moth, Abraxas grossulariata, was discovered that had paler wings than normal. It was female and was crossed with a normal dark-winged male. All the offspring had dark wings. Explain what this shows about the inheritance of wing colour in this species. – This could suggest that the female moth’s pale wings are coded for by recessive alleles and the dark wings are coded for by dominant alleles.
  • b) The dark-coloured offspring were crossed with each other, to produce the F2 generation. Explain, using a Punnett grid, what the expected ratio of dark and light offspring is in the F2 generation. – 3 dark : 1 light
  • c) The actual F2 results were: male: all dark winged; females: 1:1 ratio of dark and pale winged. Explain what these results show about the inheritance of wing colour in this species. – It means that the inheritance of wing colour in this species is sex-linked.
  • d) Sex determination in the magpie moth is not the same as in humans. Deduce the sex chromosomes of magpie moths from experimental results. – So apparently the males have two X chromosomes and the females have one X and one Y chromosome. THIS WAS A TRICK QUESTION WOW RUDE.
  • e) Predict the result of crossing a dark female moth with a pale male moth. Use a genetic diagram to show how you reached your answer. – 2 dark males : 2 light females, 1:1 ratio of dark males and light females, woah.

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§ One Response to 4.3c Genetics

  • cafergy says:

    IB 1-7

    Grade 7 A consistent and thorough understanding of the required knowledge and skills, and the ability to apply them almost faultlessly in a wide variety of situations.


    Data-based questions: hemophilia in Queen
    Victoria’s family
    page 153
    1 The condition is sex-linked
    if a male possesses the allele
    then they will be affected.
    2 The female would need two copies of the allele
    which would require a carrier mother and an affected father.
    This is a rare combination.
    3 Only males are affected
    inheritance in males appears to be from maternal line
    Carrier female and unaffected male produces affected male.
    4 Alexandria is a carrier and Nicholas is unaffected.
    Daughters could be carriers/ XHXh
    or possess nocopies of the allele / XHXH
    5 The normal male symbol in the pedigree suggests that they are XHY,
    though without reference to the pedigree there would a 50% chance that they could be XhY.

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