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GENETICS SL and HL Papers 1 and 2
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GENETICS

SL and HL

Papers 1 and 2

Part 1 – The Basics of Genetics

Assessment Statement

4.1.1 State that eukaryotic chromosomes are made of DNA and protein

4.1.2 Define gene, allele, and genome

4.1.3 Define gene mutation

4.1.4 Explain the consequences of a base substitution mutation in relation to the process of transcription and translation, using the example of sickle cell anemia

4.2.2 Define homologous chromosomes

4.2.4Explain that non-disjunction can lead to changes in chromosome number, illustrated by reference to Down Syndrome (Trisomy 21)

4.2.5 State that in karyotyping, chromosomes are arranged in pairs according to their size and structure

4.2.6 State that karyotyping is preformed using cells collected by chorionic villus sampling or amniocentesis, for prenatal diagnosis of chromosome abnormalities

4.2.7 Analyse a human karyotype to determine gender and whether non-disjunction has occurred

Question - What determines what you look like?

Genetics – the study of inheritance and of variation of inherited characteristics that chromosomes control.

Chromosomes hold the genetic blueprint, DNA (coded instructions) for the organization and activities of cells and for the whole organism in the form of genes.

Define Gene, Allele, Genome, and Gene pool

Eukaryotic chromosomes are made of DNA and Protein.

LOOK AT THE WORKSHEET, EUKARYOTIC CHROMOSOMES.

Basics of Chromosomes

When the cell divides, the chromosomes are thick, compact structures, coiled tightly. At all other times, the chromosomes are long, thin, uncoiled threads. The granular appearance at this point means the chromosomes are called chromatin.

Review of DNA

· double helix, paired strands

· DNA runs the full length of the chromosome, supported by protein (histones)

· 50% built of protein

· some proteins are enzymes involved in copying and repair

· bulk of chromosome protein has a support and packaging role for DNA

5 features of Chromosomes

1. Number of chromosomes per species

· In one species, the number of chromosomes is normally constant.

· For example, a mouse has 40 chromosomes per cell, onion has 16 and a human has 46

2. The shape of a chromosome is characteristic

· The chromosome is a long thin structure

· There is a narrow region called the centromere

3. Chromosomes occur in pairs in a cell, called homologous pairs.

· homologous means similar in structure

· one of each pair came from the other parent

· For example, humans have 46, 23 from one parent and 23 from the other

· homologous chromosomes resemble each other in structure, have the same sequence of genes, but come from different parents. (Ref to karyotyping)

4. Chromosomes hold the heredity factors – genes

· Gene can be at a specific region of a chromosome and a specific length

· A particular gene always occurs on the same chromosome in the same position, called a locus (plural – loci)

· Each gene has two or more alternate forms, called alleles

· Total of all genetic information is called a genome

5. Chromosomes copy themselves.

· Mitosis and Meiosis – later

Diagram of Chromosome

Karyotyping

Karyotyping is a “map” of chromosomes that have been paired up according to their structures.

The chromosomes can be dyed to show banding, and the arranging can be done easier.

Why is this done?

Often used to determine if there are any genetic abnormalities, like missing or extra chromosomes, or deformity, due to a non-disjuction (homologous pairs do not separate properly during mitosis or meiosis).

Examples are Down’s Syndrome, Turner (X), and Klinefelter (XXY)

Karyotyping is done following these steps:

· extract a cell sample from the placenta (amniocentesis)

· add colchine to stop cell division

· add water to burst cell

· apply stain to chromosomes

· create a photographic image of scattered chromosomes

· cut out and pair the chromosomes by their length and position of the centromere

· observe deviation from a “normal” chromosome set

MutationsChromosomal Mutations

As mentioned above, the chromosomal homologous pairs do not split evenly during mitosis or meiosis, creating more than the normal 46 or less.

Down’s Syndrome, is identified by an extra chromosome 21, giving them 47. The process for identifying the abnormality takes about 3 weeks.

Gene Mutations

A gene mutation is a permanent change in the DNA sequence that makes up a gene. Therefore, the sequence of base pairs for the production of amino acids, are changed, and the protein which is coded for may change. Some mutations can be beneficial or non-beneficial.

Gene mutations occur in two ways.

1. They can be inherited from a parent, called hereditary mutations. These mutations are present in a person, in virtually every cell in the body. Passed on through gametes.

2. They can occur in the DNA of a individual’s cells at some time in the person’s life. These are called acquired or sporadic mutations. These mutations can occur due to environmental factors or can occur if a mistake is made as DNA copies itself during cell division. These mutations that occur in cells other than egg or sperm, cannot be passed on to future generations.

Two types of Gene Mutations

Point Mutations – where only one organic base pair is changed, changing the genetic code, due to codons changing

This may have no effect because:

· It may involve part of the sense strand of the DNA which is not transcribed

· It involves part of the DNA which that particular cell does not use

· It changes the third (or second) organic base of a codon, and since the genetic code is degenerate, the same amino acid is still coded for.

Frame Shift Mutation - Insertion / Deletion – where a base is inserted or deleted from a gene

With insertion and deletion, the codon is definitely altered, as the position of the base is changed all the way down the strand. An extra base is deleted from a gene or is added. This causes the AA to change from this point on, which causes major changes in the protein.

An example of point mutation is Sickle Cell Anemia.

Haemoglobin is made up of 4 polypeptide chains, 2 alpha and 2 beta. Normal red blood cells are flexible and donut shaped (binoconcave). This allows them to carry four O2 molecules per RBC. They are said to have Hb, which is regular.

Sickle Cell is a base substitution in the gene coding for the 6th amino acid in the beta chain. This causes glutamic acid to be inserted instead of valine, resulting in abnormally shaped haemoglobin molecules. They are said to have HbS, or sickle cell haemoglobin.

The shape and rigidity of the sickle shape cause the RBC to bunch together, blocking the capillaries and the cells cannot carry as much oxygen. This can cause death due to lack of oxygen to major organs.

Example

But……. the gene has survived for years. Why?

The sickle cell trait is co-dominant with the gene coding for regular haemoglobin (meaning that it is not found more often in individuals, than regular haemoglobin). This means that individuals will be heterozygous (meaning the person will have both types of RBC), with more being normal. They will be mildly anemic.

A person who is homozygous (only contains the sickle cell trait), will die of anemia.

An advantage of having sickle cell anemia, is the sickle shape of the RBC cannot be infected by the protest, Plasmodium, that causes malaria. In areas where malaria is a problem, have HbS is an advantage. Natural selection has ensured that the sickle cell trait is more common among people living in malaria infected areas, such as West Africa.

Genetics Part 2 – Meiosis

Assessment Statement

4.2.1 State that meiosis is a reduction division of a diploid nucleus to form haploid nuclei

4.2.2 Define homologous chromosomes

4.2.3 Outline the process of meiosis, including pairing of homologous chromosomes and crossing over, followed by two divisions, which results in four haploid cells

10.1.1 Describe the behaviour of the chromosomes in the phases of meiosis

10.1.2 Outline the formation of chiasmata in the process of crossing over

10.1.3 Explain how meiosis results in an effectively infinite genetic variety in gametes through crossing over in prophase I and random orientation in metaphase I

Define Mitosis

Explain what mitosis is used for.

State the stages of mitosis and the process for each.

Define Meiosis

The purpose of meiosis is to produce gametes. It only occurs in diploid cells and reduces the number of chromosomes per cell.

The gametes contain half the number of chromosomes than the original cell. This means it goes from diploid to haploid. When the two gametes meet in the process of fertilization, the original number of chromosomes is restored.

Terms to know:

Define diploid, haploid, gametes and zygote

homologous chromosomes – also called a homologous pair. They are two chromosomes that look the same. They are the same size and show the same banding pattern in a karyotype. They carry the same genes but not necessary the same alleles. For example both chromosomes will carry the gene for eye colour, but one may have the allele for brown and the other may have the allele for blue.

Stages of Meiosis

Meiosis consists of two stages – Meiosis I and Meiosis II

Meiosis I

Interphase

· the DNA replicates

Prophase I

· Chromosomes condense

· Nucleolus becomes visible and the spindle forms

· Synapsis – when the homologous chromosomes are side by side ( now called a bivalent) and they cross over at a point called a chiasmata)

· Nuclear membrane disappears

Metaphase I

· Homologous pairs (tetrads or bivalents) move to the middle of the cell

Anaphase I

· Homologous pairs split up

· One chromosome goes to each pole

Telophase I

· Chromosomes arrive at poles

· Spindle disappears

Cell / Egg rests. There are now 23 Chromosomes in each, but sister chromatids

Meiosis II

Prophase II

· New spindle forms at right angles to the previous spindle

Metaphase II

· Chromosomes move to the equator

Anaphase II

· Chromosomes separate, chromatids move to opposite poles

Telophase II

· Chromatids arrive at poles

· Spindle disappears, nuclear membrane reappears

· Nucleolus reappears, chromosomes become chromatin

Summary – It takes one cell that is diploid and creates 4 haploid cells – this is a reduction division.

Meiosis Continued

Research - In what two ways does meiosis increase genetic variability?

The only way to get genes that are not identical is to have a mix of genetic material. This is done by crossing over, or synapsis.

Cross Over

During Prophase I, the chromatids of the bivalent are close together. During the coiling and shortening process, breakages of the chromatids occur frequently. Breakages are common in non-sister chromatids.

Process of Cross Over

Broken ends rejoin more or less immediately, but where it rejoins are between non-sister chromatids, swapping pieces of the chromatids. This is why it is called crossing over.

The point where the pieces join is called a chiasma (plural – chiasmata).

Every pair of homologous chromosomes forms at least one chiasma, and sometimes having two or more in the same bivalent is very common.

The new combinations are called recombinants.

Recombination is the re-assortment of genes or characters into different combinations from those of the parents.

As a result, meiosis results in an infinite number of variations in gametes during Prophase I and randon orientation in Metaphase I.

The number of different types of gametes produced by random orientation is represented by 2n, where n is the haploid number in a gamete. Throw in the affect of crossing over, and you have a lot of genetic variety.

For example, the human has 46 chromosomes. Therefore, the possible combinations of genetic material, would be 223. This works out to 8 388 608 possible combinations.

A pea plant has 8 chromosomes, so the possible combinations would be __________.

In order for a gamete to have exactly the same genetic make-up as the parents would occur every 246 or 7 x 10 13.

When the chromosomes separate during anaphase, this is called disjunction (junction – joined). If the chromosomes do not separate properly, there can be a problem with the amounts of chromosomes in the gametes, and as a result, more or less than the required 46 chromosomes in a human. This is called non-disjunction.

During each of the cell divisions, the chromosomes are pulled to opposite ends of the cells. Sometimes the chromosomes do not separate properly, leading to a condition known as aneuploidy.

Aneuploidy is either an extra or missing chromosome. This only happens in 9 cases, with Down’s Syndrome being the most common. In some cases, total non-disjunction takes place, which is called polyploidy.

Part 3A – Theoretical Genetics – Introduction and Monohybrid Crosses

Assessment Statement

4.3.1 Define genotype, phenotype, dominant allele, recessive allele, co-dominant alleles, locus, homozygous, heterozygous, carrier and test cross

10.1.4 State Mendel’s Law of Independent Assortment

10.1.5 Explain the relationship between Mendel’s law of independent assortment and meiosis

4.3.2 Determine the genotypes and phenotypes of the offspring of a monohybrid cross using a Punnett grid / square

Based upon how organisms look, we can deduce what genes the zygote developed from. In theory, each organism should have a set of genes from the male and a set of genes from the female. We all think this is new, done by scientists with computers and fancy equipment. Theoretical Genetics was actually done by a monk!! (You know, the chanting guys.)

Gregor Mendel was the first geneticist. The Austrian monk first looked at bees and noticed that there was some variation in how they looked. He wondered what controlled these characteristics, or traits.

He turned his curiosity to garden peas. He bred pea plants for several years until he found 7 traits that occurred generation after generation. This laid the groundwork for the Chromosomal Theory of Inheritance.

Some of the 7 Traits were:

· round vs. wrinkled

· yellow vs. green

· long stemmed vs. short stemmed

· red flowers vs. white flowers, etc.

Mendel and Theoretical Genetics

Define Trait, Expression, Genotype, Phenotype, Homozygous, Heterozygous, Dominant Allele, Recessive Allele, Co-dominant Allele, Multiple Allele, Carrier, Test Cross

Chromosomal Theory of Inheritance

Based upon the traits that Mendel saw in his pea plants, he labelled some dominant and some recessive.

Since he crossed the plants with each other, and saw the traits exhibited, he developed some laws to explain his findings.

Law of Segregation

The pea plants contain 2 heredity factors for each trait. When the gametes form, the two factors segregate and pass into separate gametes.

The gametes fuse, and the zygote will receive one of each factor from each parent. The factors remain distinct and re-segregate unchanged when new gametes are formed.

We now know that these factors are called genes, and each homologous pair has the gene on the same loci. Since the chromosomes go through synapsis (pairing) and separation (disjunction), one of each of the chromosomes ends up in a gamete.

Principle of Independent Assortment

If several genes are involved in a cross, they sort themselves out independent of one another. When they combine, any one of the pair can combine with one of the other pair.

For example:

Since any combination of chromosomes is possible in Metaphase I, any on of a pair of characteristics may combine with either one of another pair.

Eg. Pea Plantsgene: shape of peaalleles:wrinkled, round

gene: colour of peaalleles:yellow, green

When crossing, which we will go over in a bit, two plants that are heterozygous for both traits (genes), the offspring will show all combinations: green-round, yellow-round, yellow-wrinkled, and yellow-round. This shows that the genes for shape and colour inherit independently.

Test Crosses in Theoretical Genetics

How do we get the above? We need to cross the gametes and see what we get. We will use Mendel’s Monohybrid Cross for pea plants as an example.

Monohybrid – investigation of the inheritance of single contrasting characteristic

There are several parts to a test cross. As we saw earlier, in our definitions, the terms will be used now.

1. Constructing a Punnett Grid

· A Punnett Grid or Square, is used to find the ratio of the offspring, given parental phenotypes.

· It is like the multiplication tables you did way back when.

2. Figuring out the parental phenotypes and genotypes. (P)

· Plants showed two phenotypes, tall and dwarf.

· Mendel found that tall plants were dominant and dwarf plants were recessive, because when the plants were self-fertilized, they produced all tall plants. Therefore, the homozygous tall plant was either TT, or Tt (genotypes). The homozygous dwarf plants had to be genotype (tt). What would be the possible gametes?

3. First Filial Offspring (F1) from the cross of parental (P)

· Mendel found that the offspring were all tall plants. This lead him to conclude that tall was a dominant phenotype, but the plants also had to have some of the characteristics of the dwarf plants also.

· He concluded that the genotypes were Tt, based upon the original parental genotypes.

· Could Mendel have made an error here?

4. Second Filial Offspring (F2) from the self-pollination of the F1 generation.

· Mendel found that when the offspring were crossed, the majority of the F2 generation was tall, but there were some dwarf plants as well. The ratio was 3:1.

· He was able to conclude that the dominant trait was the tall, and those tall plants had a genotype of Tt or TT, depending on the gametes that fused. In the case of the dwarf plants, they cannot have T as a genotype, as the dwarf phenotype is recessive, or t. Therefore, all dwarf plants must be tt for their genotype.

Example of Mendel’s Test Cross

Mendel’s Conclusion from the Monohybrid Test Cross

1. Within an organism, there are breeding factors controlling characteristics such as “tall” and “dwarf”

2. There are two factors in each cell.

3. One factor comes from each parent

4. The factors separate in reproduction and either can be passed on to an offspring

5. The factor for “tall” is an alternative form of the factor for “dwarf”.

6. The factor for “tall” is an dominant over the factor for “dwarf”.

Steps to solving Genetics Problems 1 – Monohybrid Cross

1. Figure out the genotypes of the parents

2. Figure out the kinds of gametes the parents can produce

3. Set up a Punnett Grid for your mating

4. Fill in the babies inside the table by matching the egg allele at the top of the column with the sperm allele at the head of the row.

5. Figure out the genotypic ratio for your predicted babies

6. Figure out the phenotypic ratio for your predicted babies

7. Answer the questions you have been asked

The Problem

In gerbils, there is a gene for fur colour. One allele produces brown fur, the other produces black fur. The brown allele is completely dominant to the black. Honey and Toast, two brown gerbils, are mated. In their first litter, of babies, one was black. If they were to produce a large number of babies, what percentage would you expect to be brown?

The Solution

Practice Problems

1. What kinds of offspring result when pea plants with heterozygous green pods are crossed with pea plants with yellow pods?

1. In humans, having dimples (D) is dominant. If two parents are homozygous for dimples, what is the probability that their child will also have dimples?

2. In humans, straight hair is recessive (w) to wavy hair. What is the probability that two straight-haired parents will have a child with straight hair?

3. Using the same two parents as in #3, what is the probability that they will have a child with wavy hair?

4. A women with attached earlobes (f) mates with a man homozygous for free earlobes (F). What percent of their children would have free earlobes and what percentage would have attached earlobes?

5. In humans, long eyelashes are dominant (L) to short. If two parents are heterozygous for eyelash length, what are the possible phenotypes and genotypes of all offspring?

6. In squash, white fruit is dominant (W) to yellow fruit. What are the possible genotypes and phenotypes when a heterozygous white squash is crossed with a yellow squash?

7. Spotted coat is dominant in rabbits and the solid coat is recessive. What are the genotypes and phenotypes of offspring from a solid-coat rabbit and a heterozygous spotted-coat rabbit?

Theoretical Genetics – Part 2Co-dominance, Multiple Alleles, Sex Linked Genes and Pedigree Charts

Assessment Statement

4.3.3 State that some genes have more than two alleles (multiple alleles)

4.3.4 Describe ABO blood groups as an example of codominance and multiple alleles

4.3.5 Explain how the sex chromosomes control gender by referring to the inheritance of X and Y chromosomes in humans

4.3.6 State that some genes are present on the X chromosome and absent from the shorter Y chromosome

4.3.7 Define sex linkage

4.3.8 Describe the inheritance of colour blindness and hemophilia as examples of sex linkage

4.3.9 State that a human female can be homozygous or heterozygous with respect to sex-linked genes

4.3.10 Explain that female carriers are heterozygous for X-linked recessive alleles

4.3.11 Predict the genotypic and phenotypic ratios of offspring of monohybrid crosses involving any of the above patterns of inheritance

4.3.12 Deduce the genotypes and phenotypes on individuals in pedigree charts

We saw in the previous section that if we take one trait, with different alleles, and cross the two, we will get a 3:1 ratio in terms of the phenotypes and a 1:2:1 ratio for the genotypes.

These are very broad generalizations, and can be affected by numerous factors.

In some cases, there are multiple alleles, that can be expressed and some traits are co-dominant. Some traits are linked to sex (male or female). Using these as a guide, we can determine a pedigree, or the inheritance of several traits over several generations. We will explain these next.

Co-dominance

Co-dominance is when both alleles are expressed simultaneously rather than one being dominant and the other being recessive in the phenotype.

We saw in our test crosses that one trait will over ride another, if it is heterozygous. In the above, they are both seen in the organism.

Remember – they are not both dominant. One is expressed with the other.

For example:

For short horn cattle, the mating of a red bull (not the drink) and a white cow produces a calf that is described as roan. Roan is intermingled red and white hair. This is an example of co-dominance.

Predict the genotypic ratio and phenotypic ratio if a red bull mates with a roan cow. First, write out the genotypes and gametes. For co-dominance, the gene is the main letter and the allele is the superscript. Both are expressed in capital letters.

Then we go through the test cross just like a monohybrid cross before.

In the common garden flower Snapdragon, or Antirrhunum, when red flower plants are crossed with white flower plants, the F1 plants all are pink. The pink colouration of the petals occurs because both alleles are being expressed in the heterozygote and two pigment systems are present, rather than just the dominant allele only. When third type of phenotype appears, and blends the phenotype to produce a heterozygous offspring it is called incomplete dominance.

When the pink flower plants are crossed, the F2 plants are found to have the ratio of 1 red to 2 pink to 1 white. Why? Do the test cross.

Multiple Alleles

The genes introduced so far have two forms, or alleles. For example, tall or dwarf, red or white, and wrinkled or smooth. This is what Mendel identified.

When the gene for one trait exists as only two alleles, and the alleles play according to Mendel’s Law of Dominance, there are 3 possible genotypes and 2 possible phenotypes. We saw this with the Yellow vs. Green, with the yellow being the dominant phenotype in both the homozygous form (YY) and the heterozygous form (Yy).

If you remember from earlier, if there are only 2 alleles, but three possible phenotypes, there must be co-dominance or incomplete dominance occurring.

With some genes, there are more than two possible alleles. Then there are 4 or more possible phenotypes for a particular trait. These are called multiple alleles.

BUT, VERY IMPORTANT: There may be multiple alleles within a population, but individuals have only two of the alleles. WHY?

ANSWER: The individuals only have two biological parents. We inherit half our genes from one and the other half from the other, so we end up with two alleles for every trait in our phenotype.

Example of Multiple Alleles – Blood Type

As we discussed earlier, there are 4 blood types: A, B, AB, and O.

There are three alleles of the gene that determines blood type.

IA – codes for A blood

IB – codes for B blood

i - codes for O blood

(With multiple alleles, we choose a single capital letter to represent the locus at which the alleles may occur. The individual alleles are then represented by an additional single letter (usually a capital) in a superscript position.)

If you notice, the allele for O is recessive (i).

If you look at the three alleles, the possible combinations are as follows:

GENOTYPES

PHENOTYPES

IAIA

IAi

Type A

Type A

IBIB

IBi

Type B

Type B

IAIB

Type AB

ii

Type O

· There are 6 different genotypes and 4 different phenotypes.

· There are two genotypes for both A and B blood – either homozygous dominant or heterozygous

· There is only one genotype for O blood, homozygous recessive.

· What is the deal with AB blood? What is this an example of?

Answer the following questions below.

Multiple Alleles and Co-dominance Questions

1. For short horn cattle, the mating of a red bull and a white cow produces a calf that is described as roan. Roan is intermingled red and white hair. This is an example of co-dominance. Predict the genotypic ratio and phenotypic ratio of the following:

a. A roan bull and a roan cow

b. A roan bull and a white cow

2. A woman with Type O blood and a man who is type AB are expecting a child. What are the possible blood types of the kid?

3. What are the possible blood types of a child who’s parents are both heterozygous for B blood?

4. What are the chances of a woman with AB blood and a man with A blood having a child with type O?

5. Jill is type O. She has two older brothers with blood types A and B. What are the genotypes of her parents, with respect to blood type?

6. One busy night in an understaffed maternity ward, four children were born. The babies were mixed up and it was not certain who each baby belonged to. Baby Joe was type A, Baby Moe was type B, Baby Zoe was type AB and Baby Melvin was type O. The parents were the following:

Mr. and Mrs. Jones were A and B

Mr. and Mrs. Gerber were O and O

Mr. and Mrs. Lee were B and O

Mr. and Mrs. Santiago were AB and O

The nurses were able to figure out which child belonged to which family. Deduce how this was done.

Sex Linked Traits and Pedigree Charts

What genetically determines who is a boy and who is a girl?

That’s right, it’s the X and Y chromosomes!

In humans, gender is determined by specific chromosomes, known as the sex chromosomes. Each of us has one pair of sex chromosomes (either XX or XY) along with the 22 autosomal chromosomes.

What are the chances of having a boy? And who determines this chance?

Egg cells produced by meiosis all carry an X chromosome, but 50% of sperms carry an X and 50% carry a Y chromosome. At fertilization, an egg cell may fuse with a sperm carrying an X chromosome, which will produce a female offspring. The egg may fuse with a sperm carrying a Y chromosome, leading to a male offspring. The gender of the offspring is determined by the male partner. Also, due to the distribution of X and Y chromosomes, we would expect equal numbers of male and female offspring to be produced by a breeding population of humans, and other mammals, over time.

Evidence of 50:50 split.

Human X and Y Chromosomes and the Control of Gender

Initially, male and female embryos develop identically in the uterus. At the seventh week of pregnancy, the sexes are starting to be determined. This is shown by the development of male genitalia, only if a Y chromosome is present.

Why is XX female and XY male? On the Y chromosome is the prime male-determining gene. This gene codes for a protein, the testis-determining factor (TDF). TDF functions as a molecular switch; on reaching the embryonic gonad tissues, TDF initiates the production of a relatively low level of testosterone. The effect of this hormone at this stage is to inhibit the development of female genitalia, and to cause the embryonic genital tissue to form testes scrotum and penis.

In the absence of a Y chromosome, the embryonic gonad tissue forms an ovary. Then, partly under the influence of hormone from the ovary, the female reproductive structures develop.

In order for this to happen, the X and Y chromosomes have to pair up. We know that homologous chromosomes pair up during meiosis. Since the female has XX, the two are similar, containing the same genes, but maybe different alleles (more on this later).

With the X and Y chromosomes, only a very small part of the X and Y have complimentary alleles and pair up during Meiosis. The two only pair up at a very small portion of the chromosome.

In summary, the short Y chromosome carries genes specific for male sex determination, for coding for TDF and sperm production. The X chromosome carries an assortment of genes, very few of which are concerned with sex determination.

Sex linkage and Sex Linked Traits

Although most of our traits are carried on our autosomes, there are some traits that are located on the sex chromosomes – however, because the Y chromosome does not code for all the traits on the X, we have an interesting situation. Therefore, the genes present on the sex chromosomes are inherited with the sex of the individual. They are said to be sex linked characteristics or traits.

Sex linkage – is a special case of linkage occurring when a gene is located on a sex chromosome (usually the X chromosome)

The difference between inheritance of genes on autosomal chromosomes and sex-linked genes

The inheritance of sex-linked genes is different from the inheritance of genes on autosomes, because the X chromosome is much longer than the Y chromosome and many of the genes on the X chromosome are absent from the Y). (See diagram above)

In a female, if the X chromosome carries a recessive form of a gene (one allele), the pair X chromosome will often carry the dominant allele. As a result, the recessive allele will not be expressed.

In a male (XY), an allele present on the X chromosome is most likely to be on its own, and will be apparent even if it is recessive. In other words, it does not have a pair, dominant allele to “override” the recessive allele.

In conclusion, a human female can be homozygous or heterozygous with respect to sex-linked characteristics, whereas males will have only one allele.

An individual with a recessive allele of a gene that does not have an effect on the phenotype (ie. heterozygous) is known as a carrier. They carry the allele but it is not expressed. Therefore, female carriers are heterozygous for sex-linked recessive characteristics.

What does this mean for males?

Uh, oh!! The unpaired alleles of the Y chromosome deal mostly with male structures and male functions. But, some diseases are controlled by recessive genes found on the X chromosome. If a male human Y is paired with the recessive gene on the X, the disease-triggering allele will be expressed. A female must be homozygous recessive for a sex-linked characteristic for the allele to be expressed. Sorry guys!!!

Some examples of recessive conditions controlled by genes on the X chromosome are: Duchenne Muscular Dystrophy, red-green colour blindness, and haemophilia. The gene for hairy ears is on the Y chromosome, but it is more of an annoyance than a problem!!!

Examples of Genotypes for Sex Linkage

NB: A female has one of three possible genotypes for a sex-linked trait: homozygous dominant, heterozygous and homozygous recessive.

Human males cannot be heterozygous, since they only have one copy of the allele.

Heterozygous females are carriers.

Example 1 – Colour Blindness

Existing Alleles

XB for normal vision

Normal vision but a carrier

Xb for colour blindness

A female can be:

XB XB

XB Xb

Xb Xb

A male can be:

XBY

XbY

The homozygous recessive in a female is very rare.

Example 2 – Haemophilia

Existing Alleles

XH for normal blood clotting

Normal but a carrier

Xh for haemophilia

A female can be:

XH XH

XH Xh

Xh Xh

A male can be:

XHY

XhY

**** Xh Xh is homozygous lethal, meaning that the individual would not exist. Haemophilia is largely a male disease. The only way for a female to be a haemophiliac is to be homozygous recessive. When this happens, the condition is usually fatal in the uterus, typically resulting in a natural abortion. If the female was born with this condition, she would bleed to death during her first menstrual cycle.

Test Cross Examples

1. If a female carrier for red-green colour blindness had a child with colour blind male, what are the possible percentages of males and females and normal sight versus colour blindness in the F1 generation? (Give the phenotypes and genotypes)

2. If a female carrier for the Duchenne MD gene has children with man without Duchenne MD, what will be the genotypes and phenotypes for the F1 generation?

3. If each of the male offspring had children with a female who was heterozygous for the MD gene (a carrier), what would the phenotypes and genotypes be for the F2 generation?

4. A man and a woman both have normal vision, but a daughter has red-green colour blindness. The man files for divorce on the grounds of infidelity. Can genetics support his case or make him look foolish?

Pedigree Charts

A pedigree can be represented as a diagram showing the phenotypes of a trait that is inherited from generation to generation. The genotypes can then be determined or predicted from these charts. Look at the following example of a simple pedigree diagram.

The squares represent males and the circles represent females. A horizontal line joining a male and a female indicates that the couple is married or is capable of having children. A shaded square or circle means that this person has the trait being studied.

In this example, the parents had three children: one girl and two boys. The father exhibits the trait being followed (shaded) and none of the children show this trait (none are shaded).

The Roman numerals represent the generation. Each person in a generation can be numbered to show birth order (e.g., 1, 2, 3 and so on). For example, II-3 represents the third child born in the second generation. Moving up in the diagram shows ancestors; moving down shows descendants.

Try one!

Use a drawing program or pencil and paper to draw the pedigree chart for the following family:

One couple (first generation) has four children: two boys and two girls.

One of the girls gets married and has two boys.

Pedigree charts can be used to identify a genetic pattern of inheritance.

The trait being studied is earlobe attachment, which is an autosomal trait. Free lobes is dominant. Having attached earlobes is recessive.

The shaded individuals are homozygous recessive for the trait and have attached lobes.

Looking at the genotypes, can you see any assumptions that were made about certain individuals?

The diagram tells us that the female, I-2, has free earlobes. We are guessing that her genotype is heterozygous Ee, but it might also be homozygous dominant EE. Looking at her offspring, we see that she only had one child--a male with free lobes. That is not enough information to know her genotype for sure. If she has the genotype Ee, by chance all of her children could have free lobes (I-1 can only contribute the recessive gene because he is homozygous for attached lobes) just as they would if she had the EE genotype.

The male, II-2, must be heterozygous dominant Ee. He has free lobes, but we know that he is carrying the gene for attached lobes from his father. The female II-1 has attached lobes, so we know that she is homozygous ee.

The male, III-3, has the genotype Ee because he has free lobes. We know this definitely because we know with certainty his mother's and father's genotypes (see above).

Individual IV-1 has free lobes but cannot have genotype EE since her mother cannot give her a dominant allele. Therefore, her genotype has to be Ee.

What is the genotype of III-1? The pedigree chart shows it to be heterozygous Ee. Again, we don't have information to be sure. The genotype could be either EE or Ee. Either Ee or EE in III-1 would produce the child IV-1 with Ee. If we had information about the parents of III-1, we might be able to know, but as it is, we simply can't. You can do a simple Punnett square to show probabilities of offspring in any of these scenarios. In a pedigree, we look at phenotypes of actual offspring rather than possible combinations of genes.

Just keep these 5 Rules for Pedigree Analysis

1. Every gamete carries exactly one allele for every gene. The exceptions are sex linked genes in males.

2. Any individual with the recessive phenotype (appearance) must be homozygous recessive.

3. Any individual with a homozygous recessive offspring must have at least one recessive allele.

4. Any individual with a homozygous recessive parent must have at least one recessive allele.

5. If you don’t know, don’t guess. Use the Punnett Square.

Study the next pedigree to follow the inheritance of colour-blindness in humans, a sex-linked trait. Keep in mind that there are various types of colour-blindness and that this chart only shows one type.

Check your understanding of interpreting pedigrees and sex-linked traits by answering the questions below. From the pedigree chart above showing the inheritance of colour-blindness, explain how to determine the genotype of each of the following individuals: a) II-3b) IV-5c) V-1

Try this one.

1. From the pedigree chart above showing the inheritance of hemophilia (a sex-linked trait), determine the genotype of each individual listed below. Explain how you determined the genotype for each. a) I-2b) II-1c) What are the probabilities of various genotypes for the following: V-1, V-2, V-3, V-4?

Dihybrid Crosses, Recombinants, Gene Linkage and Chi-squared Test

Assessment Statement

10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes10.2.2 Distinguish between autosomes and sex chromosomes10.2.3 Explain how crossing over between non-sister chromatids of a homologous pair in prophase I can result in an exchange in alleles10.2.4 Define linkage group10.2.5 Explain and example of a cross between two linked genes10.2.6 Identify which of the offspring are recombinants in a dihybrid cross involving linked genes1.1.5 Deduce the significance of the difference between two sets of data using calculated values for t and the appropriate tables (Chi Squared Test)1.1.6 Explain that the existence of a correlation does not establish that there is a causal relationship between the two variables

Mendel also investigated the simultaneous inheritance of two pairs of contrasting characters, using the same garden pea plants. He referred to this as a dihybrid cross.

To test the theory, Mendel crossed pure breeding plants from round seeds with yellow cotyledons (seed leaves) with pure breeding plants from wrinkled seeds with green cotyledons. All of the F1 generation were round, yellow peas. When the F1 generation was allowed to cross, Mendel found that there was a ratio of 9 yellow round peas, 3 green round peas, 3 yellow wrinkled peas and 1 green wrinkled pea. The result from the monohybrid crosses was still valid. Yellow is dominant to green, and round / smooth is dominant to wrinkled.

Mendel’s Dihybrid Cross Example

Further Test Crosses – Drosophilia melanogaster (Fruit Fly)

Mendel died in 1884, in relative obscurity in the scientific community. Many did not know of his work or thought he did not understand his results. When his papers were rediscovered in 1900, Mendel’s results came to be confirmed and then extended by many others.

One was by Thomas Morgan in 1908. He used the Drosophilia as an organism to test Mendelian Genetics. For his work, he was awarded the Nobel Prize in 1933.

Morgan showed:· Mendel’s factors are linear sequences of genes on chromosomes (what is now known as the Chromosomal Theory of Inheritance)· discovered sex chromosome and sex linkage· demonstrated crossing over, and the exchange of alleles between chromosomes, resulting from chiasmata formed during meiosis

Why was the fruit fly chosen? It is quite common around rotting vegetable material and the female lays hundreds of eggs, which produce hundreds of offspring. The time from hatching to adult takes about 10 days. They can be anaesthetized, and sorted. They have four pairs of chromosomes. The characteristics, or phenotypes, are controlled by genes not on the sex chromosomes. Therefore, the characteristics are controlled by the autosomes, and are not sex-linked.

Morgan’s Test Cross

If a wild type fly is crossed with a mutant type, what will the offspring be. The wild type has a white body and normal wings. These are dominant traits. The mutant has a red body and vestigal wings (short). What will the F2 generation be? (The genes in both cases are not linked)

Recombinants

Recombination is the re-assortment of genes or characters into different combinations from those of the parents.

When cross over occurs (previous notes), we see that the genes on some of the chromosomes are kept the same, while others change. Recombination can be applied to non-linked genes but is often restricted to linked genes, which we will talk about later.

We can identify the recombinants in the above two examples. What are the parental combinations and what are the recombinants?

Examples

1. Two tomato plants were crossed. Tall plants (T) are dominant to short plants (t) and round tomatoes (R) are dominant to oblong fruit (r). If one parent with the genotype TTRr, is crossed with a parent homozygous recessive for both traits, what will the F1 generation phenotypes and genotypes be?

2. If the genotypes for the tomatoes above were TtRr, Ttrr, ttRr and ttrr, which two pairs would a plant breeder select in order to obtain a phenotype ratio of 1:1:1:1?

3. Frumpy plants have small flowers with green petals, which are dominant traits. Large white flowers are recessive, and the genes are not linked. A plant with small green flowers was crossed with one with large white flowers and the offspring had either small green flowers or large green flowers. Determine the genotypes of the parents. If the large green flowers were allowed to self-pollinate, determine the genotypes and phenotypes of the offspring.

Autosomal Gene Linkage

Autosomes are chromosomes, which are not sex chromosomes. They are the other 22 pairs, for example, in humans.

Sex chromosomes are those chromosomes, which help in determining the sex of an individual (XX or XY).

In our studies, we have studied that genes independently assort. In other words, they are located on different chromosomes, and have no effect on each other.

In reality, we only have 22 autosomes or homologous pairs, along with the sex chromosomes. There are many more genes (about 60,000!) in our genetic make-up. This means that some must be located on the same chromosome.

When cross over occurs in Prophase I, the alleles are exchanged, between non-sister chromatids of a homologous pair. But since some genes are on the same chromosome, when cross over occurs, the genes go together. This is called a linkage group.

Linkage group – a group of genes whose loci are on the same chromosome.

The Reason

Linked genes were discovered in 1906, by R. C. Punnett and W. Bateson. (the square guy!) When Mendel’s work was rediscovered, geneticists looked at Mendel’s ratios. Some worked out to the characteristic 9:3:3:1 ratio for the F2 generation. In some cases, using plants, the ratio for the F2 generation was 7:1:1:7! What happened?

Let’s go over an example.

In sweet peas, purple flowered plants are dominant to red, and long pollen plants are dominant to round pollen. When a homozygous purple and homozygous long plant is crossed with a red and round pollen plant, the resulting F1 phenotype is all purple and long, as you would expect.

What would be the expected ratio for the F2 generation, if two of the previous plants are mated?

The results were:

7 purple long: 1 purple round: 1 red long: 7 red round

This seemed to mean that more PL and pl gametes were produced than the Pl and pL gametes. What is the explanation?

If no cross over occurs, the phenotypes will have the parental characteristics and few recombinants.

Example

In fruit flies (Drosophila) body colour and wing length are linked. A tan body (b+) is dominant to a black body (b) and long wings (w+) are dominant to short wings. Cross a homozygous tan, homozygous long winged fly with a mutant black and short wing fly. What would be the F1 and F2 generations in terms of genotypes and phenotypes?

1. The snowy gull has white feathers and a yellow beak, but some birds have black feathers and an orange beak. Black and orange are recessive and the genes are linked. A pure breeding bird with white feathers and a yellow beak was mated with one with black feathers and orange beak. The offspring had white feathers and yellow beak. Two of these were mated and 75% of the offspring had the dominant phenotypes and the rest had the recessive. Explain.

2. Transparent wings and clubbed antennae are autosomely lined dominant alleles in the fairy fly. Grey wings and smooth antennae are the characteristics for being homozygous recessive for both alleles. A fly with grey wings and smooth antennae was mated with an unknown fly and all of the offspring had transparent wings and clubbed antennae. These offspring were all mated to flies, with the recessive traits and the total offspring were:

i. 227 transparent wings, clubbed antennae

ii. 10 transparent wings and smooth antennae

iii. 9 grey wings, smooth antennae

iv. 231 grey wings, clubbed antennae

Determine the genotypes of all the flies.

The Chi-squared Test

When you do a cross, generally you need a large amount of real life data to ensure that your results are significant. You would then compare the ratio with the expected ratio.

The chi-squared test is used to analyze the data produced from a mono or dihybrid cross, and determine if your results were simply by chance, or if the results are significant.

The formula is as follows:

A probability of less than 5% is considered to be significantly different than the expected value, and therefore the original hypothesis or expectations need to be re-examined. To put it another way, the greater the value of the chi-squared above the critical value, the greater the difference between the observed and expected values.

Once we calculate the chi-square, we have to look at our degrees of freedom.

Degrees of freedom are the number of observations minus the minimum number required to uniquely define the figure. In other words, if we are looking at two classes of a phenotype, we calculate the degrees of freedom by:

n - 1

where n is the number of classes. For example, if you are looking at red eyes vs. white eyes, the degree of freedom would be 1, as 2-1=1 or n-1.

For every degree of freedom, there is a critical value. Since in Biology, the probability is 5% or 0.05, we look at the probability (chi-square) vs. the degrees of freedom for 5%.See handout.

Basically, if your value is bigger than the critical value for 5%, the result is by chance and your original assumption (hypothesis) is correct. If the value is smaller, there is a less than 5% probability that the results are due to chance, so your original assumption may be flawed.

Confused? Let’s do an example.

Ex. In a hypothetical cross, we expect the phenotypic ratio to be 1:1. The actual results were 45:55, instead of 50:50. Is this deviation due to chance or is it significant?

Example 2

In a genetics experiment, tall pea plants were crossed with short pea plants. The resulting F1 was self-fertilized and F2 consisted of 787 tall plants and 277 short plants. Does this result confirm Mendel’s explanation?

Genetics Problems

1. List the possible gametes that would result from the segregation of the following parental genotypes.a. RRb. Mmc. AABBd. DdSse. gGHhRr

2. In a species of plants, flowers appearing at the end of the stem are dominant to flowers appearing in the middle of the stem. Predict the phenotype for this F2 generation, for a cross involving a homozygous dominant plant with recessive plant.

3. In garden peas, red flowers dominate white flowers and green pod colour is dominant to yellow pod colour.

a. Cross two heterozygous red flowered plants with green podsb. Cross a heterozygous red flowered plant with green pods with a plant that is completely recessive for all the traits.c. Cross two plants that are recessive for both traits.

4. In snapdragons, tallness (T) is dominant to dwarfness (t), while red flower (Cr) and white flower (Cw) show incomplete dominance. A dwarf red snapdragon is crossed with a plant homozygous for tallness and white flowers. Give the genotypes and phenotypes of the expected F1. If the F1 flowers were crossed, what would be the expected F2 genotypes and phenotypes.

5. A dominant allele, A causes yellow colour in rats. The dominant allele of another independent gene, R, produces black coat colour. When two dominants occur together (A_R_), they interact to produce grey. Rats of the genotype (aarr) are cream coloured. If a grey male and a yellow female, when mated, produce offspring in the following approximate ratio, 3/8 yellow, 3/8 grey, 1/8 cream and 1/8 black, what are the genotypes of the two parents?

6. What are the genotypes of a yellow male rat and a black female that, when mated, produce 46 grey and 54 yellow offspring? Does a chi-square test bear you out?

Polygenetic Inheritance

Assessment Statement

10.3.1 Define polygenetic inheritance

10.3.2 Explain that polygenetic inheritance can contribute to continuous variation using two examples, one of which must be human skin color

In the previous section, we saw that the height of a pea plant was determined by one gene with two alleles. This clear-cut difference is an example of discontinuous variation. This means there is no intermediate form and no overlap between the two phenotypes.

Actually, very few characteristics of organisms are controlled by a single gene. Most characteristics are controlled by a number of genes. Groups of genes which together determine a characteristic are called polygenes.

Polygenetic inheritance is the inheritance of phenotypes that are determined by the collective effect of several genes.

Genes that make up a polygene are often (but not necessarily always) located on different chromosomes. Any one of these genes has a very small or insignificant effect on the phenotype. The combined effect of all the genes on the polygene is to produce infinite variety among the offspring.

Many features are controlled by polygenes. Here are some examples.

Example 1 – Shape of the comb in Poultry

Different shape of comb exist:

Looking at the above, some generalizations can be made. The trait is controlled by a pair of genes that assort independently. If the genotype is P_rr, a pea comb is present. If the genotype is P_R_, a walnut comb is present. If the genotype is ppR_, a rose comb is present. If the genotype is pprr, a single comb is present.

What are the dominant alleles?

Cross a homozygous pea-combed chicken with a homozygous rose-combed chicken. What is the F1 and F2, if the F1 was allowed to mate?

Example 2 – Human Skin colour

The colour of the human skin is due to the amount of the pigment called melanin that is produced in the skin. Melanin synthesis is genetically controlled. It seems that three or four or more separately inherited genes control melanin production. The outcome is an almost continuous distribution of skin colour from very pale (no alleles coding for melanin production) to very dark brown (all alleles coding for melanin production).

The example below is for only two independent genes, but dark is dominant to white in all cases.

Some questions to really exercise your mind.

1. If A, B, C are the alleles for dark skin, and a, b, c, represent the alleles for light skin, what would the possible combinations be for two individuals mating who are heterozygous for all the alleles?

2. Seed colour in wheat is controlled by around 6 genes. Below are the combinations in the genotypes and the phenotypes.

If two plants which produce mid pink seeds are crossed, what are the ratios of the genotypes and phenotypes for the seed colours?

Extension 1 - Eye Colour

The exact color of the human eye is determined by the amount of a single pigment called melanin. When a lot of melanin is present, the eye will appear brown or even black. When little melanin is present, the iris appears blue.

Intermediate amounts of melanin produces gray, green, hazel or varying shades of brown. Eye color genes, through the enzymes they produce, direct the amount and placement of melanin in the iris. In general Caucasian babies are born with blue eyes because at the time of birth they haven't begun to produce melanin in their irises.

At one time scientists thought that a single gene pair, in a dominant/recessive inheritance pattern, controlled human eye color. At the present, three gene pairs controlling human eye color are known. Two of the gene pairs occur on chromosome pair 15 and one occurs on chromosome pair 19. The bey 2 gene, on chromosome 15, has a brown and a blue allele. A second gene, located on chromosome 19 (the gey gene) has a blue and a green allele. A third gene, bey 1, located on chromosome 15, is a central brown eye color gene.

Geneticists have designed a model using the bey 2 and gey gene pairs that explains the inheritance of blue, green and brown eyes. In this model the bey 2 gene has a brown and a blue allele. The brown allele is always dominant over the blue allele so even if a person is heterozygous (one brown and one blue allele) for the bey 2 gene on chromosome 15 the brown allele will be expressed.

The gey gene also has two alleles, one green and one blue. The green allele is dominant to the blue allele on either chromosome but is recessive to the brown allele on chromosome 15.

This means that there is a dominance order among the two gene pairs. If a person has a brown allele on chromosome 15 and all other alleles are blue or green the person will have brown eyes.

If there is a green allele on chromosome 19 and the rest of the alleles are blue, eye color will be green.

Blue eyes will occur only if all four alleles are for blue eyes.

This model cannot account for gray, hazel or multiple shades of brown, blue, green and gray eyes, or how eye colour changes over time.

Extension 2 – Other forms of Gene Interaction

Polygenetic inheritance adds to the range of phenotypes that may exist. But there are other mechanisms by which genetic variety is controlled.

Cases where characteristics are controlled by two or more genes, one or more of which masks or modifies the expression of other genes include:

· Plumage pigmentation in birds

· Agoutie coat colour in some mice

· Shell banding in the snail Cepaea nemoralis

There are some examples below. Other examples are controlled by factors in the environment. A tall plant may appear to be dwarf if it has consistently been deprived of essential nutrients.

Examples

1. In fruit flies, the allele for grey colour is dominant over black and straight wings are dominant over curly wings.

A heterozygous grey, straight winged fly was crossed with a black, curly winged fly. The offspring were as follows:

43 grey, straight

12 grey, curly

39 black, curly

10 black, straight

a. If these genes are not linked, which phenotypes would you expect for the F1?

b. Give the expected numbers of the phenotypes.

c. Based on these results, would you expect the genes to be linked?

2. In rabbits, having coloured fur is dominant over producing no pigment (albino). Grey fur is dominant over black. A homozygous completely recessive albino rabbit is mated with a homozygous grey rabbit. The F1 are allowed to interbreed. The genes are not linked.

a. What are the genotypes and phenotypes of the F1?

b. Predict the genotypes for the F2?

c. What are the expected ratios?

d. Predict the phenotypes of the F2?

e. What are the expected ratios?

Genetic Engineering and Other Aspects of Biotechnology

These are the objectives from the IB Curriculum (Assessment Statements 4.4.1 – 4.4.13). Research each area, following the action words required. Do the worksheets, keeping in mind the objectives. These will constitute your notes to use for the IB Exam.

1. Outline the use of polymerase chain reaction (PCR) to copy and amplify minute quantities of DNA.

2. State that in gel electrophoresis, fragments of DNA move in an electric field and are separated according to their size.

3. State that gel electrophoresis of DNA is used in DNA profiling.

4. Describe the application of DNA profiling to determine paternity and also in forensic investigations.

5. Analyze DNA profiles to draw conclusions about paternity or forensic investigations. To do this, complete the Murder Mystery by using DNA profiling. (Will be handed out)

6. Outline three outcomes of the sequencing of the complete human genome.

7. State that, when genes are transferred between species, the amino acid sequence of polypeptides translated from them is unchanged because the genetic code is universal.

8. Outline the basic technique used for gene transfer involving plasmids, a host cell (bacterium, yeast or other cell), restriction enzymes (endonucleases) and DNA ligase.

9. State two examples of the current uses of genetically modified crops or animals.

10. Discuss the potential benefits and possible harmful effects of one example of genetic modification.

11. Define clone.

12. Outline a technique for cloning using differentiated animal cells.

13. Discuss the ethical issues of therapeutic cloning in humans.