7 Genetics

Question 1

genotype

Answer 1

genetic make up of an organism

describes all the alleles that an organism has

Question 2

phenotype

Answer 2

the observable or biochemical characteristics of an organism.
it is the result of the interaction between the the expression of the genotype and the environment

Question 3

gene

Answer 3

a length of dna that normally code for a particular polypeptide
genes exist in two or more different forms called alleles
the position of a gene on a particular dna molecule is known as the locus

Question 4

allele

Answer 4

one of the different forms of a gene
only one allele of a gene can occur at the locus of any one chromosome
Diploid organisms have two alleles (one on each chromosome).
There may be many alleles of a single gene.
Alleles may be dominant, recessive or codominant.
In a diploid organism, the alleles at a specific locus may be either homozygous or heterozygous.

Question 5

homozygous

Answer 5

if the allele on each of the homologous chromosomes is the same

Question 6

heterozygous

Answer 6

if the two alleles on the homologous chromosomes are different

Question 7

homologous chromosomes

Answer 7

in diploid organisms the chromosomes occur in pairs called homologous chromosomes
there are therefore two loci that each carry one allele of a gene

Question 8

dominant

Answer 8

the allele of the heterozygote that expresses itself is said to be dominant

Question 9

recessive

Answer 9

the allele of the heterozygote that doesn’t express itself is said to be recessive

Question 10

homozygous dominant

Answer 10

a homozygous organism with two dominant alleles

Question 11

homozygous recessive

Answer 11

a homozygous organism with two recessive alleles

Question 12

diploid organism

Answer 12

organism with 1 set of homologous chromosome, i.e., organism has two copies of same gene in the system. humans for example has 23 pairs of homologous chromosomes, with 2 copies of each gene on two chromosomes

Question 13

codominant

Answer 13

when two alleles both contribute to the phenotype
the phenotype is either a blend of both features or both features are presented
both alleles are equally dominant

Question 14

monohybrid inheritance

Answer 14

the inheritance of a single gene

e.g. gregor mendel and colour of pea plants

Question 15

example of monohybrid inheritance- inheritance of pod colour in peas

Answer 15

if pure-breeding green-pod plants are crossed with pure-breeding yellow-pod plants, all the offspring, known as the first filial, or F1, generation, produce green pods. this means that the allele for green pods is dominant to the allele for yellow pods, which is therefore recessive
when the heterozygous plants (Gg) of the F1 generation are crossed with another (=F1 intercross), the offspring (F2 generation) are always in an approximate ratio of three plants with green pods to one plant with yellow pods
these observed facts led to the formation of a basic law of genetics- the law of segregation

Question 16

law of segregation

Answer 16

states that 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

Question 17

dihybrid inheritance

Answer 17

a cross between two different genes that differ in two observed traits.
e.g. gregor mendel and pea seed shape and colour

Question 18

example of dihybrid inheritance- inheritance of pea seed shape and colour

Answer 18

• round shape dominant to wrinkled shape
• yellow colour dominant to green colour

he carried out a cross between two pure breeding types of plants:

• one always producing round shaped, yellow seeds
• one always wrinkled and green

in the F1 generation he obtained plants all of which produced round yellow seeds, dominant features.
he raised the plants from these seeds and crossed them
produced 4 types of gamete (RG Rg rG rg) as gene for seed colour and gene for seed shape are on separate chromosomes
ratio of 9:3:3:1
observations led him to formulate his law of independent assortment

Question 19

law of independent assortment

Answer 19

each member of a pair of alleles may combine randomly with either of another pair

Question 20

multiple alleles

Answer 20

where there are more than two alleles, of which only two may be present at the loci of an individual’s homologous chromosomes

Question 21

codominance example- snapdragons

Answer 21

one allele codes for an enzyme that catalyses the formation of a red pigment in flowers
the other allele codes for an altered enzyme that lacks this catalytic activity and so does not produce the pigment
three colours are found:
-in plants that are homozygous for the first allele, both alleles code for the enzyme, and hence pigment, production
-in plants that are homozygous for the other allele, no enzyme and hence no pigment is produced. these plants have white flowers
-heterozygous plants, with their single allele for the functional enzyme, produce just sufficient red pigment to produce pink flowers

we cannot use upper and lower case letters for the alleles
C for colour
C^R for red
C^W for white

Question 22

multiple alleles example- inheritance of ABO blood groups

Answer 22

there are three alleles associated with the gene I (immunoglobulin gene), which lead to the presence of different antigens on the cell-surface membrane of red blood cells:

• allele I^A, which leads to the production of antigen A
• allele I^B, which leads to the production of antigen B
• allele I^O, which does not lead to the production of either antigen

only two alleles can be present in an individual at any one time, as there are only two homologous chromosomes and therefore only two gene loci
alleles I^A and I^B are codominant where as I^O is recessive to both
possible blood groups are A, B, AB or O

Question 23

sex inheritance in humans

Answer 23

sex chromosomes are X and Y

• as females have two X chromosomes, all the gametes are the same in that they contain a single X chromosome
• as males have one X and one Y, they produce two diff types of gamete- half have an X chromosome and half have a Y

Question 24

sex linkage

Answer 24

any gene that is carried on either the X or Y chromosome is said to be sex-linked
however, the X chromosome is much longer than the Y chromosome
this means that, for most of the length of the X chromosome, there is no equivalent homologous portion of the Y chromosome.
those characteristics that are controlled by recessive alleles on this non-homologous portion of the X chromosome will appear more frequently in the male
as there is no homologous portion on the Y chromosome that might have the dominant allele, in the presence of which the recessive allele does not express itself

Question 25

sex linkage example- haemophilia

Answer 25

an X-linked genetic disorder the blood clots only slowly and there may be slow and persistent internal bleeding, especially in the joints almost entirely confined to males one of a number of causes is a recessive allele with an altered sequence of DNA nucleotide bases that therefore codes for a faulty protein which doesn't function X^H = dominant allele for production of the clotting protein. linked to X chromosome) X^h = recessive allele for the non-production of clotting protein. linked to X chromosome) as males can only obtain their Y chromosome from their father, it follows that their X chromosome comes from their mother if the mother doesn't suffer from the disease, she may be heterozygous for the character (X^HX^h) such females are called carriers because they carry the allele without showing any signs of the disease in their phenotype. this is because these carriers possess one dominant H allele and this leads to the production of enough functional clotting protein as males pass the Y chromosome to their sons, they cannot pass haemophilia to them however they can pass it to their daughters, who would then become carriers

Question 26

pedigree charts

Answer 26

a way to trace the inheritance of sex-linked characters, such as haemophilia - a male is represented by a square - female by a circle - shading within a shape indicated the presence of a character in the phenotype

Question 27

autosomal linkage

Answer 27

when two or more genes are carried on the same autosome autosomes are the remaining 22 chromosomes that aren't sex chromosomes any two genes that occur on the same chromosomes are said to be linked all the genes on a single chromosome form a linkage two genes A and B with heterozygous alleles are on diff chromosomes, there are four possible combinations of the alleles in the gametes (AB Ab aB ab) however if the two genes are linked (A and B on same chromosome) and provided there is no crossing over, there are only two possible combinations of alleles in the gametes ( AB ab).

Question 28

autosomal linkage example- fruit fly

Answer 28

two linked genes of fruit fly: -one determines body colour -one determines wing size there are two alleles for body colour: one produces a grey body and is dominant to the other which produces a black body there are two alleles for wing size: one produces normal sized wings and is dominant to other that produces tiny wings) if the alleles for body colour and wing size are not linked (on sep chromosomes) then an individual that is heterozygous for both characters would produce 4 diff gametes (GN Gn gN gn) rather than just 2 (GN gn)

Question 29

epistasis

Answer 29

arises when the allele of one gene affects or masks the expression of another in the phenotype e.g. mice with several genes that determine coat colour

Question 30

epistasis example- mice coat colour

Answer 30

- gene A controls the distribution of a black pigment called melanin in hairs and therefore whether they are banded or not. the dominant allele (A) of this gene leads to hairs that have black bands while the recessive allele (a) produces uniform black hairs when it is present with another recessive allele (a) (homozygous=aa) - gene B controls the colour of the coat by determining or otherwise, the expression of gene A. the dominant allele (B) leads to the production of melanin while the recessive allele (b) leads to no pigment and any hair will therefore be white when it is present with another recessive allele (homozygous =bb) agouti mice (brown)- have hair with black bands (AABB AABb AaBB AaBb) black mice- uniform black hairs (aaBB aaBb) albino mice- hairs lack melanin (AAbb Aabb aabb) if an agouti mous with genotype AABB is crossed with an albino mouse with genotype aabb, then the offspring are all agouti. if individuals from F1 generation are crossed to produce F2 generation the following ratio is produced 9 agouti 4 albino 3 black

Question 31

chi squared test

Answer 31

test used to test the null hypothesis the null hypothesis is used to examine the results of scientific investigations and is based on the assumption that there will be no statistically significant difference between sets of observation, any difference being due to chance alone. a means of testing whether any deviation between the observed and the expected numbers in an investigation is significant or not.

Question 32

what criteria must be met for the chi squared test

Answer 32

- the sample size must be relatively large, over 20 - the data must fall into discrete categories - only raw counts and not percentages, rates etc., can be used - it is used to compare experimental results with theoretical ones, e.g. in genetic crosses with expected mendelian ratios

Question 33

chi squared formula

Answer 33

chi squared= sum of ([observed numbers - expected numbers]^2 / expected numbers) x^2 = Σ ([O-E]^2 / E) value obtained is then read off a chi squared distribution table to determine whether any deviation from the expected results is significant or not to do this we need to know the number of degrees of freedom

Question 34

number of degrees of freedom

Answer 34

the number of classes (categories) minus one, that is, if a human can have blood group A or B or AB or O, there are four classes and three degrees of freedom in this case

Question 35

using the chi squared table

Answer 35

e.g. X^2= 1.0 heads and tails decide how many classes of results there are (e.g. 2). this corresponds to one degree of freedom. on table look at row showing two classes for our calculated value of 1.0. this lies between 0.45 and 1.32 so our value lies between 50% and 25%. in the chi squared test the critical value is p=0.05 this is the attribution to chance accepted by statisticians, that is, 5% dues to chance if the probability that the deviation is due to chance is equal to or greater than 5%, the deviation is said to be not significant and the null hypothesis is accepted