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Pat of Inheritance
| Question | Answer |
|---|---|
| What is a gene? | A length of DNA (sequence of nucleotide bases) that usually determines a single characteristic of an organism. |
| What is a locus? | The position of a gene on a chromosome. |
| Define the term allele | One of the different forms of a gene. |
| State the definition of a dominant allele | The allele that expresses itself in the phenotype in a heterozygote. |
| What is codominant? | Condition in which two alleles both contribute to the phenotype. Eg. pink flower in snapdragons from red and white parents |
| What is phenotype? | Observable characteristics of an organism. Attributed to the interaction between the genotype and the environment which can modify organism's appearance. |
| What is a genotype? | The genetic make-up of organisms. All the alleles that an organism contains. |
| What is homozygous? | Condition in which both chromosomes (maternal and paternal) contain the same allele on their locus. |
| What is heterozygous? | Condition in which both chromosomes (maternal and paternal) contain different alleles on their locus. |
| Define the term mutation | Any change to the genotype as a result of a change in DNA. |
| Define the term modification | Any change to the phenotype that does not affect the genotype and is not inherited. |
| What is recessive? | The allele that is not expressed in a heterozygote. |
| What are multiple alleles? | Organism has more than two possible alleles for a characterisic of which only two may be present at the loci of an individual's homologous chromosomes (e.g. human ABO blood system) |
| What does F1 generation stand for? | First filial generation |
| What does F2 generation stand for? | Second filial generation. |
| What type of genotype is pure-bred? | Homozygous |
| If pure-breeding green-pod plants are crossed with pure-breeding yellow-pod plants and the F1 generation turn out to be only green, what conclusion can be made? | The allele for green-pod plants is dominant and the allele for yellow-pod is recessive |
| When green-pod plants of the of the F1 generation were crossed with each other, ratio of F2 offspring were 3 green pod plants : 1 yellow pod plant.What Mendelian law arose because of this? | Law of segregation (Mendel's 1st law) : In diploid organisms, characteristics are determined by alleles that occur in pairs. Only one of each pair of alleles can be present in a single gamete. The F1 generation were heterozygous. |
| Why did Mendel choose to work with pea plants? | 1) They were easy to grow 2) Large amounts of offspring could be produced 3) They possessed many contrasting features that could be easily observed. |
| How did Mendel accurately manage his experiments? | He carefully pollinated his plants by transferring pollen from one plant to another with a paint bush. He also ensured plants were pure breeding for each feature by self pollinating them for many generations. |
| What is monohybrid cross/inheritance? | A single characteristic controlled by one gene is passed from one generation to the next. |
| What is dihybrid cross/inheritance? | Two characteristics determined by two genes located on non-homologous chromosomes are inherited. |
| What is test cross? | A test to determine the genotype of an individual that displays a dominant characteristic. It is carried out between an individual with the unknown genotype and an individual with the homozygous recessive genotype. |
| If the unknown genotype is heterozygous, what would the result of a test cross be? | At least one of the offspring will display the recessive characteristic. |
| If the unknown genotype is homozygous dominant, what would the result of the test cross be? | All offspring will display the dominant characteristic. |
| Why are actual results from experimental crosses often not identical to theorectical expected ratios? | Each fertilization event is independent of the next. It's like tossing a coin. It is random whether it will land on heads or tails. |
| The four blood types in the ABO system | A, B, AB and O |
| What are the two possible genotypes for blood group A | AA or AO |
| The two possible genotypes for blood group B | BB or BO |
| The only possible genotype for blood group AB | AB |
| The only possible genotype for blood group O | OO |
| Epistasis | A gene interaction where one gene interferes with the expression of another. E.g. an allele that produces a defective enzyme early in a biochemical pathway blocks the remainder of the pathway. |
| Mendel's dihybrid crosses led to the discovery of what is today known as his second law. What does it state? | Law of Independent Assortment (Mendel's 2nd law) : For genes that are on separate, non-homologous chromosomes, each member of a pair of alleles may combine randomly with either of another pair. |
| F2 phenotypic ratio from a monohybrid cross. (Parents were homozygous dominant and homozygous recessive. F1 were all heterozygous.) | 3:1 |
| F2 phenotypic ratio from dihybrid cross. (Parents were homozygous recessive and homozygous dominant. F1 were all heterozygous) | 9:3:3:1 |
| Offspring phenotypic ratio from a dihybrid test cross with unknown being heterozygous | 1:1:1:1 |
| F2 from a cross involving sex linkage where female parent is homozygous recessive and the male in hemizygous dominant | 1:1:1:1 |
| Offspring phenotypic ratio from a monohybrid test cross with unknown being heterozygous | 1:1 |
| F2 from a monohybrid cross with codominance (the parents were homozygous) | 1:2:1 |
| Autosome | Homologues that appear the same in males and females. Non-sex chromosomes. |
| Heterosome | Sex chromosomes - different in males and females. |
| Unlike all other characteristics, sex is determined by chromosomes rather than genes. Explain. | Females have two X chromosomes (they are homogametic). Males have one X and one Y chromosome (they are heterogametic). Males always pass their Y chromosome to their son and their X to their daughter. |
| Sex-linked gene | Any gene that is carried on either the X or Y chromosome. Very few genes are carried on Y therefore MOST sex-linked conditions are 'X'-linked. |
| List a few sex-linked conditions | Haemophilia, eye-colour in Drosophila fruit flies etc |
| Why can a male never be a 'carrier' for a sex-linked condition? | He only has one X and one Y. He cannot have the recessive condition since he cannot have two alleles. He either has the condition or he does not. |
| Males are hemizygous. What does this mean? | They have one allele of each of the genes on the X chromosome. These alleles are always expressed as there is no chromosome homologous to their X chromosome. |
| Why can females be 'carriers' for sex-linked conditions? | They have two X chromosomes. They can have the allele without showing signs of the condition in their phenotype since most conditions are found on the recessive allele. |
| Why can a male never pass on haemophilia to his sons? | Haemophilia (is sex-linked) and is carried on the X chromosome. Since males pass the Y chromosome to their sons, they cannot pass haemophilia to them. They can pass it to their daughters though. |
| Define 'carrier'. | An individual who is heterozygous for a condition which is located on the recessive allele. |
| How are the symbols on a pedigree chart interpreted? | Square for males. Circle for females. Shading indicates the presence of a characteristic in the phenotype. Half-shading represents a normal phenotype but carrier of the allele for the condition. |
| Chi-squared test | A statistical test used to test the null hypothesis. It determines whether any deviation between the observed and expected numbers in an investigation is significant or not. |
| Null hypothesis | The assumption that there will be no statistically significant difference between sets of observations (experimental and theoretical), any difference being due to chance alone. |
| Degrees of freedom | The number of classes (categories) minus 1 e.g. if there are 4 classes then 4-1 = 3 degrees of freedom. |
| Chi-squared formula | = sum of ((observed - expected numbers) sq. / expected numbers) |
| Critical value accepted by statisticians | p = 0.05 ie. 5% due to chance |
| What should be done after calculating your chi-square value? | After calculating chi-sq value, calculate the degrees of freedom then use the given chi-sq data table to determine the probability that your results are due to chance. |
| Accept null hypothesis if | If the probability that the deviation is due to chance alone is greater than the critical value 0.05 (5%) |
| Reject null hypothesis if | If the probability that the deviation is due to chance alone is less than 0.05 (5%) |
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CAPE Biology Unit 1
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