Chapter 3.4 Inheritance

Question 1

Describe Gregor Mendel’s experiment

Answer 1

Experiment subject: different varieties of pea plants, each of which reliabily had the same characteristics when grow independently

Method: hybridization (transferring the male pollen from one variety to the female parts of flowers of another variety), the resulted pea seeds were grown to find out their characteristics – one cross

Each cross was repeated with thousands of plants –> ensure accuracy and reliability –> statistically significant results at high confidence level, anomalous results are also less likely to distort the whole set of data

7 crosses were tested in total

Question 2

Name of the scientist who did inheritance experiments with pea plants

Answer 2

Gregor Mendel

Question 3

Genotype

Answer 3

The combination of alleles of a gene carried by one organism in a diploid cell.

Question 4

Phenotype

Answer 4

The physical expression of the alleles of a gene processed by an organism. It is the resulted feature of one’s genotype.

Question 5

Dominant

Answer 5

An allele that is expressed whether its paired allele is identical or different (i.e. regardless if the organism is heterozygous/homozygous of the gene in question).

Masks the effects of recessive alleles.

Question 6

Recessive

Answer 6

Alleles that are only expressed if two copies of the same recessive allele are found in the genotype (i.e. homozygous recessive).

Question 7

Codominant

Answer 7

Pairs of alleles that are both expressed when present (e.g. blood type A and B alleles) –> joint effects.

Question 8

Loci

Answer 8

Specific locations on chromosomes where a gene is located.

Question 9

Centromere

Answer 9

Joins two chromatids together during cell division (mitosis and meiosis).

Question 10

Alleles

Answer 10

Different forms of the same gene.

Question 11

Carrier

Answer 11

A heterozygous individual carrying a recessive disease-causing allele, who does not personally suffer from the disease.

Question 12

Use of punnett grids

Answer 12

Predicting the outcomes of monohybrid (crossing a single trait) genetic crosses.

Question 13

Steps of constructing a punnett grid

Answer 13

1. Start with homozygous parents with different alleles (e.g. TT, tt).
2. Write down both the genotype and phenotype of the parents.
3. Write down the allele that would be contained in their gametes (e.g. T, t).
4. Write down the genotype and phenotype of the F1 generation (e.g. Tt, tall stem).
5. Construct a punnett grid (Tt * Tt) and write down the genotype and phenotypes for each grid.
6. Calculate the ratios between genotypes and phenotypes.

Question 14

Why blood testing is necessary to determine the blood group of a donor before a transfusion?

Answer 14

RBCs have agglutinogens (antigens) on their surfaces, which can be used to identify self/non-self cells by the immune system. The body of a patient who receives a wrong blood transfusion would react patally, as his lymphocytes would secrete antibodies which causes blood to clot.

Question 15

Blood AB is known as…

Answer 15

universal recipient (the recipient’s RBCs have both type A and B agglutinogens)

Question 16

Blood O is known as…

Answer 16

universal donor (no agglutinogens on type O RBCs)

Question 17

Why is ABO blood groups often used as a textbook example?

Answer 17

Only one gene determines one phenotype, whereas more than one gene is often necessary to determine a phenotype in many other scenarios.

Question 18

Why ABO blood groups is an example of codominance?

Answer 18

Both type A and B agglutinogens are present on RBCs for a IAIB person (type AB).

Question 19

An example of genetic disease caused by a recessive allele

Answer 19

Cystic fibrosis (CF)

Question 20

An example of genetic disease caused by a dominant allele

Answer 20

Huntington’s disease

Question 21

An example of genetic disease caused by a codominant allele

Answer 21

Sickle-cell anemia

Question 22

Examples of a sex-linked genetic disease

Answer 22

1. RG colour blindless

2. Hemophilia

Question 23

Cystic fibrosis

Answer 23

The P of a person being a CF carrier is relatively high, but since CF is caused by a recessive gene, both parents need to be carriers to have a 0.25 P of having a child with CF –> therefore overall P is low.

Caused by a recessive allele of the CFTR gene on chromosome 7.

The CFTR gene involves in the production of chloride ion channels that involve in secretions of sweat, digestive juices, and mucus.

The disease-causing allele results in production of faulty chloride ion channels –> instead of being a lubricant –> viscious sweat, digestive juices and mucus –> block tubes/ducts/air passages –> infections in lungs as sticky mucus accumulates + prevents digestive enzymes from reaching the duodenum from the pancreas.

Life expectancy: 35~50 years
C –> normal; c –> CF allele

Question 24

Huntington’s Disease

Answer 24

Caused by a dominant allele of the HTT gene on chromosome 4. The gene product is huntingtin (protein).

Causes degenerative changes in the brain –> changes in behaviour, thinking and emotions become increasingly severe –> patients will eventually need full nursing care and often suffer/die from heart failure and/or infectious diseases such as pneumonia (lung infection)

Symptoms usually show up when the person is between 30~50 years old –> late onset –> the diagonised patient may have already had children by that age

H –> HD allele; h –> normal

Question 25

Sickle-cell anemia

Answer 25

Hb(S) –> SC allele; Hb(A) –> normal

Three possible scenarios:

1. Hb(S)Hb(S): susceptible to malaria, severe anemia (WORST SITUATION!)
2. Hb(S)Hb(A): increased resistance to malaria, mild anemia
3. Hb(A)Hb(A): susceptible to malaria, not anemic

Question 26

Why sex-linked genetic diseases are more common in males?

Answer 26

Caused by recessive alleles on the sex-linkage: sex chromosomes are not homologous, there are some genes present on X but not on Y.

A male processing a disease-causing allele is affected, regardless whether the allele is dominant/recessive, as he does not have an alternative allele on the Y chromosome to combat the disease-causing allele on the X chromosome.

Question 27

RG colour blindness

Answer 27

Caused by a x-linked, recessive allele of the RG gene which codes for a photoreceptor protein in cone cells.

Locus of the gene is Xq28, the non-homologous section of the X chromosome.

XA –> normal
Xa –> colour blindness

Question 28

Hemophilia

Answer 28

Caused by a recessive x-linked mutation in hamophilia which causes one of the clotting factors not being produced.

The sufferer could bleed to death. X^X
Treatment: injecting commercial blood clotting factors produced by genetic modification. Restriction enzymes cutting out the genes which codes for blood-clotting factors –> inserted into a sheep embryo’s milk gene –> clotting factors produced along with sheep milk –> extraction and purify

XH –> normal
Xh –> hamophilia

Question 29

General rules when interpreting a pedigree diagram

Answer 29

1. Male – squares, female – circles
2. Generations are indicated by roman numerals (I, II etc.)
3. Individuals are indicated by letters (A, B etc.)
4. Sufferers are shaded, carriers are half-shaded –> KEY

Question 30

Uses of pedigree charts

Answer 30

Trace family history of genetic-diseases
Shows carriers, sufferers, non-sufferers
Predict the probability of a new-born child having a genetic disease.

Question 31

Causes of genetic mutations

Answer 31

1. Exposure of short WL radiations: UV, X-ray, gamma ray, alpha particles from radioactive isotopes
2. Chemicals: carcinogens in tobacco smoke, mustard gas (WW2)

Once a person dies, his mutations are eliminated from the human gene pool. Yet if that person has offspring, his genetic mutations would be passed on by gametes.

Question 32

Define autosomal genetic disease and give examples.

Answer 32

Genetica diseases caused by fatally genes from non-sex chromosomes (autosomes).

Examples: Hungtington’s disease, cystic fibrosis, sickle-cell anemia.