Tonight, we’re going to talk about the effects of heterozygous and homozygous individuals and their dominant and recessive alleles. Firstly, we should know by now that homozygous means having two of the same alleles of a gene, and heterozygous means having two different alleles of a gene. Between the four possible combinations between dominant and recessive alleles (AA, Aa, aA, and aa), the dominant allele would be the one that determines the phenotype of the organism. The phenotype is the physical characteristics of an organism as opposed to the genotype which are the alleles posses by an organism. It’s important to know that though a trait can be seen in an organism, it’s occasionally hard to tell whether the gene is heterozygous or homozygous even if the phenotype shows the dominant trait. The way to find out would be to do a test cross, which crosses the unknown genotype with a homozygous recessive, and if all the offspring only show the dominant trait, obviously the unknown genotype was homozygous but if half of the offspring show the recessive trait, then obviously the unknown genotype was heterozygous.
There are also co-dominant alleles that are pairs of alleles that both affect the phenotype when in a heterozygote. The best example is the Mirabilis jalapa, which is a flower that displays this pattern of inheritance. The flower has an allele for having a red colour as well as an allele for having a white colour, but when both alleles are present in a heterozygous individual, the flower is pink. The specific alleles are thus: C^wC^w = White flowers, C^rC^r = Red flowers, C^rC^w / C^wC^r = Pink flowers.
The ABO blood group system in humans is also another example of co-dominance where there are two dominant alleles that always code for blood types A and B, but blood type O can only be coded for by a homozygous recessive gene.
Alright, so, remember when we previously discussed that meiosis results in variation? Natural selection requires variation so that species can constantly evolve along with the changes in the environment (think of the Red Queen analogy that keeps coming up in class – in biology!). New alleles must then be formed to always maintain variation amongst species, and this is done through gene mutations, which are changes to the base sequence of a gene (they’re usually very few base changes, one or at least a very small number). These mutations are either neutral or harmful because the alleles that already exist have been selected for by evolution – by natural selection itself! Some mutations are therefore lethal, especially if they end up in the gametes, which will then pass into the offspring and cause genetic disease.
Finally: sex linkage. It is the association of a characteristic with gender, because the gene controlling the characteristic is located on a sex chromosome. Females, because of the two x-chromosomes, have two copies of sex-linked genes, while males, who only have one x-chromosome, hold only one copy of sex-linked genes. Hemophilia is an example of a sex-linked characteristic.
- Define the terms gene and allele and explain how they differ. 4 marks
- Describe the consequences of a base substitution mutation with regards to sickle cell anemia. 7 marks
- Outline the formation of chiasmata during crossing over. 5 marks
- Explain how an error in meiosis can lead to Down syndrome. 8 marks
- 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
- Compare the processes of mitosis and meiosis. 6 marks
- Outline one example of inheritance involving multiple alleles. 5 marks
- 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
- Outline sex linkage. 5 marks
- Explain, using a named example, why many sex-linked diseases occur more frequently in men than women. 9 marks
DATA BASED QUESTIONS
Page 146, deducing genotypes from pedigree charts
It isn’t possible to investigate human genotypes with test crosses! Sometimes a person’s genotype can be deduced from a pedigree chart. (Roman numerals indicate the generations and Arabic numbers are used for individuals in each generation.)
1. Explain, using evidence from the pedigree, whether the condition is due to a recessive or a dominant allele.
The condition is due to a recessive allele, as seen in parents I 1 and I 2. Neither are affected but they both produce a daughter and a son who are affected, which means that they received recessive alleles from both parents that just didn’t show in the parents’ generation. Similarly, parents I 3 and I4 are unaffected and don’t show the trait, but they both produce a daughter and a son who are affected and do show the trait in the next generation.
2. Explain what the probability is of the individuals in generation V being:
- a) homozygous recessive: 4/4
- b) heterozygous: 0/4
- c) homozygous dominant: 0/4
3. Deduce, with reasons, the possible genotypes of:
- a) 1 in generation III: Hh and hh
- b) 13 in generation II: HH, Hh, and hh
4. Explain which sets of parents in the pedigree are genetically similar to a test cross.
Generation II 13 and 14, because the rest of the couples have parents who only offer one recessive allele between the two of them, and only 13 and 14 have the possible individual (13) who could possibly have a homozygous dominant gene for this trait.
Page 147, the two spot ladybird
Adalia 2-punata is a species of ladybird. In North America ladybirds are called ladybugs. The commonest form of this species is known as typica. There is a rarer form called annulata.
1. Compare the typica and annulata forms of Adalia 2-punata.
Typica has only two dots on each wing (?) while annulata has a more intricate design all over its shell (?). Typica has a more yellow-covered head while annulata’s head is less yellow.
2. The differences between the two forms are due to a single gene. If male and female typica are mated together, all the offspring are typica. Similarly, the offspring produced when annulata forms are mated are all annulata. Explain the conclusions that can be drawn.
Because only typica ladybirds are produced when typica ladybirds mate and vice versa for annulata ladybirds, we can conclude that the single gene holds the same alleles among all of the individuals.
3. When typica is mated with annulata, the F1 hybrid offspring are not identical to either parent. Distinguish between the F1 hybrid offspring and the typica and annulata parents.
The F1 hybrid offspring have more complex and different designs from their original parents. It isn’t just slightly yellow heads or very yellow heads or simply two dots on each wing, it’s a combination of everything, not just a combination of yellow head + two dots, or a combination of not-so-yellow head + intricate shell design.
4. If F1 hybrid offspring are mated with each other, the offspring include both typica and annulata forms, and also offspring with the same wing case markings as the F1 hybrid offspring.
- a) Use a genetic diagram to explain this pattern of inheritance. —> I’m sorry, I’m not going to do a genetic diagram because it wouldn’t be efficient at this point in February, but in any case: this pattern of inheritance is called co-dominance, wherein pairs of alleles both affect the phenotype when in a heterozygote. In this case, the heterozygote’s genotype would be L^TL^A (L^T = typica, L^A = annulata) for example. This genotype codes for the mixed ladybird while the homozygous version of each ladybird codes for that ladybird (e.g. L^tL^t can only code for typica). Whenever the alleles are heterozygous, they will code for the mixed ladybird, while any combination of homozygous alleles will code for either original ladybird.
- b) Predict the expected ratio of phenotypes. —> In this generation (F2), the expected phenotypes are 1 L^TL^T : 2 L^TL^A : 1 L^AL^A
Page 148, distribution of ABO blood groups
1. State the frequency of I^B in:
- a) Northern India: 20%-25%
- b) Japan: 15%-20%
- c) North America: 10%-15%
2. The frequency of the allele I^A in South America is less than 5%. Calculate the frequency of the allele i in South America.
I^B frequency = 10%-15%, I^A frequency < 5%, therefore i frequency (10 or 15 + 5 – 100) = 80%-85%
3. Suggest an explanation for gypsies in Europe having a much higher frequency of I^B than other indigenous groups in Europe.
Perhaps the gypsies developed a system that produced anti-A antibodies efficiently in all their travels around Europe. Their bodies probably had to build up a stronger system which involved that production of Anti-A antibodies.
4. Use the data in the map to evaluate Thor Heyerdahl’s theory that Pacific islands were colonized by groups from South America sailing west. The frequency of I^B on these islands is between 10% and 25%.
That honestly doesn’t sound accurate because South America had 5-10% frequency of I^B while the Pacific islands had more than 10%. This means they had a higher frequency than the Pacific islanders so maybe we could say that the Pacific islanders spread their anti-B antibodies into the Pacific islanders, but they didn’t have much of it to start off with.