This is where things get a little complicated in terms of combinations and ratios.
So far, the genes and recombinations and generations we’ve been working with have invalid unlinked genes, which are clearly pairs of genes located on different types of chromosomes. These chromosomes, when paired up during meiosis, are called bivalents. Other times, pairs of genes can be located on the same chromosome, resulting in a ratio different from the 9:3:3:1 ratio found when crossing genes on different chromosomes. These genes are inherited together by the next generations – that’s what we call gene linkage.
Crossing Over and Recombination of Linked Genes
In this case, new combinations of alleles can emerge during prophase I in meiosis I in the process we know as crossing over, which involves exchanging DNA between the chromatids of two homologous chromosomes (not between the same chromatids, or else there’d be no difference) when they’ve synapsed. When crossing over has been done, then there is recombination amongst the genes because new combinations of alleles have been formed. Individuals born of this new combination of alleles (different from the parent) is called a recombinant. Genes that have their loci on the same chromosome type form a linkage group.
Also important to note are the type of chromosomes we see in a genome. Two chromosomes are the sex chromosomes that determine the gender of the individual. In humans, the two male chromosomes are the X and Y chromosomes. In females, the two X chromosomes. The rest of the chromosomes that are not the X and Y chromosomes are called autosomes. There are two types of linkage then. Sex linkage, involving the sex chromosomes, namely the X-chromosome, and autosomal gene linkage, involving genes located on the same autosome.
And this is what I mean about things getting a little complicated. Polygenic inheritance is when two or more genes affect the same characteristic. The quickest example would be Mendel’s white-purple flower plants. When he pollinated them, he expected the typical 9:3:3:1 ratio amongst the resulting daughter individuals. Instead, he was met with a range of colours, and the new ratio found was 1:4:6:4:1 – which should be familiar because that’s a line from Pascal’s triangle! This is where we get the distribution of the different results and the more the genes, the closer the distribution gets to a normal distribution.
Variation among polygenic inherited genes is continuous – gradual, if you will. Unlike the previous results we’ve been working with (tall or short, black or brown, wrinkled or smooth), there is a range of results in polygenic inheritance, and the best example is the variation of skin colour in humans.
Skin colour is affected by two main things: the amount of black pigment melanin (the more melanin, the darker your skin), and your environment. More sun makes a darker person and vice versa. I for example have darker skin than others but not so dark as, say, someone who comes from Africa. This just means I’m in the middle of the range of skin colour and have maybe the medium amount of melanin pigments in my skin.
- Calculate and predict; genotypic and phenotypic ratios of offspring of dihybrid crosses involving unlinked autosomal genes.
- Identify which of the offspring in dihybrid crosses are recombinants.
- Describe the methods and aims of DNA profiling.
- Outline a technique for transferring genes between species.
- Describe the technique for the transfer of the insulin gene using E. coli.
- Discuss the potential benefits and possible harmful effects of genetic modification.
- Discuss the ethical arguments for and against the cloning of humans.
- Outline the ethical issues of cloning humans.
DATA BASED QUESTIONS
Page 160, gene linkage in mutant tomatoes
Tomato plants usually have flowers in small groups and produce fruits that are round and smooth. Three mutations were found: peach (hairy fruits), oblate (flattened fruits), and clustered (with many flowers in a group).
1. Identify, with reasons, which tomato plants are recombinants in the F2 generations in:
- a) cross 1 – smooth oblate and peach round
- b) cross 2 – smooth clustered and peach few
2. Draw diagrams to show how recombinants were produced by crossing over in:
- a) cross 1 – diagram involves crossing over genes to create recombinants smooth oblate, and peach round
- b) cross 2 – diagram involves crossing over genes to create recombinants smooth clustered and peach few.
3. Calculate the percentage of recombinants in:
- a) cross 1 – total: 1000; total of recombinants: 467; 46.7%
- b) cross 2 – total: 1000; total of recombinants: 342; 34.2%
4. Discuss the reasons for the difference in recombination percentage between the two crosses.
There is a difference in the recombination percentage because of the distribution of these characteristics. These three genes are probably on the same chromosome, therefore the distribution would be different from a 9:3:3:1 distribution.
Chapter 13 Questions
- a) Distinguish between autosomes and sex chromosomes. —> The sex chromosomes are the X and Y chromosomes that determine the gender of the organism, and the autosomes are all the other chromosomes in the organism’s genome.
- b) Compare autosomal linkage and sex linkage. —> Autosomal gene linkage is on the same autosome and sex linkage involves genes located on sex chromosomes, mostly the X-chromosome.
- c) Distinguish between continuous and discontinuous variation. —> Continuous variation in a phenotype is the wide range of differences of that phenotype while discontinuous variation is quite the opposite – very few differences. Continuous variation is because of one gene but discontinuous is polygenic.
- d) Distinguish between polygenic inheritance and multiple alleles. —> Polygenic inheritance is when two or more genes affect a single phenotype, multiple alleles are when one gene (that controls one phenotype) can be coded for by … multiple alleles.
- e) State two examples of polygenic inheritance. —> Skin colour in humans, Mendel’s purple-flowered plants, height.
2. The diploid chromosome number of the pea plant, which Mendel used in his experiments, is 14.
a) How many linkage groups are there in peas? —> 7
b) Table 4 shows the chromosome location of the genes that affect each of Mendel’s character differences.
- i) State one pair of character differences that you expect to be unlinked. —> Colour and height.
- ii) State one pair of character differences that you expect to show autosomal linkage. —> Colour and flowers.
c) If pure-breeding plants with full green pods were crossed with pure-breeding plants with constricted yellow pods, and the F1 plants were allowed to self-pollinate, what ratio of phenotypes would you expect in the F2 generation? Use a genetic diagram to work out your answer. —> The F1 babies are all GgFf babies (phenotype: Green, full pods / zygote: heterozygous). Self pollination would use the gametes GF, Gf, gF, and gf, and the F2 babies would produce a ratio of 9: 3: 3: 1. Possible phenotypes are green full (9/16), yellow full (3/16), green constricted (3/16), and yellow constricted (1/16).
d) When dihybrid crosses are performed with plants differing in cotyledon colour and flower colour, the give results suggesting that the two genes involved are unlinked The two genes are located far apart from each other on chromosome number 1. Explain how this can cause them to behave as though they are in different linkage groups. —> Because the two genes are located far from each other on chromosome #1, the distance allows for a lot of variation and crossing over. The wide variation makes it seem like the genes were on different chromosomes the whole time (but that’s not true so don’t fall for that).
- a) Explain the relationship between Mendel’s law of independent assortment and the events of meiosis. —> Mendel’s law of independent assortment states that when genes are on different chromosomes (then they are unlinked), they can do independent assortment to add variation to the gene pool. During meiosis, the chromosomes have to line up at the equator at metaphases I and II and during metaphase I, they align at the equator randomly and independent of each other. There are multiple different ways the chromosomes can line up and the way they (randomly) do so creates variation in the offspring.
- b) Explain how recombination of linked genes occurs. —> Recombination occurs in linked genes when crossing over occurs. In prophase I of meiosis, the chromosomes synapse and the homologous chromosomes’s chromatids (that are NOT sisters) cross over, break off, and create new variations of DNA that way. The more distance between the two genes, the more variation there is.
- c) Explain the variation and inheritance of skin colour in humans. —> The variation of skin colour in humans ranges from dark tones (Africans!) to very light tones (Europeans! Caucasians!) and some light brown in the middle (Me! South-East Asians! Middle Easterns! Hispanics!). The phenotype of skin colour is polygenic which means two or more genes affect that characteristic, which is why there is such a large variation in skin colour. It is also very dependent on our ancestry and our environment, for example, people whose ancestors lived in the desert for the past five hundred years would still be dark today as that is where their genes are passed down from. Likewise, people living in the mountains would always wear thick coats therefore their skin wouldn’t be exposed to sun that much and they would then be much whiter than the people who had lived in the desert.
4. Chickens have a red fleshy crest on their head, called the comb. There are four shapes of comb: single, rose, pea, and walnut. LOL.
- a) If pure-breeding chickens with rose comb and single comb shapes are crossed, all the F1 hybrids have rose combs. Using A and a for the dominant and recessive alleles, state the genotypes of the parents and the F1 hybrids with rose comb. —> The parents’ genotypes are AA and aa, and the F1 hybrid’s (with rose comb) genotypes are all Aa.
- b) If pure-breeding chickens with pea comb and single-breeding chickens with pea comb and single comb shapes are crossed, all the F1 hybrids have pea combs. Using B and b for the dominant and recessive alleles, state the genotypes of the parents and the F1 hybrids with pea comb. —> Genotype of parents: BB and bb, F1 hybrid (with pe comb) genotype all Bb.
- c) If pure-breeding chickens with rose comb and pea comb shapes are crossed together, all the F1 have walnut combs. Deduce the genotype of the F1 hybrids with walnut comb. —> AaBb
- d) Explain, using a genetic diagram, the ratio of phenotypes you would expect in the F2 generation if F1 hybrids with walnut combs were crossed together. —> ratio = 9:3:3:1 // this is a dihybrid cross // 9 walnut : 3 rose : 3 pea : and 1 single
- e) Discuss whether comb shape in chickens is an example of polygenic inheritance. Comb shape is indeed polygenic because two genes (A/a and B/b) controls the shape. The dominant A codes for the rose comb while the homozygous recessive (aa) codes for single comb. The dominant B codes for the pea comb while the homozygous recessive (bb) codes for the single comb as well. Walnut is coded by both dominant A and B (therefore AB).
5. The scattergraph shows the results of a study by Newman, Freeman, and Holzinger of identical twins, who were reared apart. The intelligence of the twins was assessed by an IQ test and the differences in their education were also assessed using a numerical scale.
- a) Explain why identical twins reared apart are very useful in studies of inheritance. —> Twins have the same genetic makeup, therefore if placed in different environments (LIKE LAB RATS), we could observe a different pattern of inheritance among the individual twins and see how the environment influences them.
- b) Explain why identical twins reared together show a close correlation in intelligence. —> Because the twins are in the same environment and undertake the same influences, their intelligence would be affected the same way.
- c) Using the data in the scattergraph, state the smallest and largest difference in intelligence of the twins reared apart. —> Smallest difference: difference in education @ about 4. Largest difference: difference in IQ @ about 37 – 38.
- d) The scattergraph shows a positive correlation. Suggest a cause for this correlation. —> The environments the twins were placed in were very different from each other. Technically, the greater the difference in education, the greater the difference in intelligence.
- e) The points on the graph are widely scattered. Suggest reasons for this. —> The experiment used a wide variety of controlled variables and environmental factors that affected intelligence and had many different samples (trials).