Top2-Ch3-P48-71End

Question 1

Explain the Mendel pea experiments.

Answer 1

He crossed homozygous round seeds and homozygous wrinkled seeds as P generation, which produced all round seeds.

He then self-fertilised the F₁ generation of round seeds and they produced the F₂ generation of 5474 round and 1850 wrinkled seeds, which was 3/4 round and 1/4 wrinkled seeds.

He also conducted reciprocal crosses where he crossed male round with female wrinkled in P generation and vice versa but it got exactly the same result. This showed that round seeds were dominant. Although F₁ plants display the phenotype of one parent, both traits are passed to F₂ progeny in a 3:1 ratio. The wrinkled seeds are recessive.

Question 2

What are the four conclusions that Mendel drew from his pea experiments?

Answer 2

1. F₁ plants possess phenotype of only one parent but must inherit genetic factors from both parents because they transmit both phenotypes to the F₂ generation. Wrinkled and round factors can only be explained only if F1 inherited both traits from P generation. The genetic factors are now called alleles.
2. Two alleles in each plant separate when gametes are formed and one allele goes into each gamete. When two gametes (one from each parent) fuse to produce a zygote the male and female allele unite to produce genotype of offspring.
3. Concept of dominance over recessive.
4. Two alleles of an individual plant separate with equal probability into the gametes in F1. When F1 produce gametes they pair randomly to produce the following genotypes in equal proportion RR, Rr, rR, rr. Because R is dominant end up with three R for every r pea.

Question 3

Principle of segregation is?

Answer 3

Mendel’s first law which is two alleles separate during gamete formation in equal proportions.

Question 4

Concept of Dominance is?

Answer 4

When two alleles are present in a genotype the dominant allele is observed in the phenotype

Question 5

Chromosome theory of heredity is?

Answer 5

genes are located on chromosomes

Question 6

What is a backcross?

Answer 6

between an F₁ genotype and either of the parental genotypes.

Question 7

Punnet square is?

Answer 7

see picture

Question 8

Multiplication rule is?

Answer 8

probability of two or more independent events occuring together is calculated by multiplying their independent probabilities. Words “and”

Ie probability of getting a four on two sequential rolls of the dice are ¹/₆ and ¹/₆ each time. So it is ¹/₆ x ¹/₆ = ¹/₃₆ is chance of rolling two fours in a row.

Question 9

Addition rule is?

Answer 9

the probability of getting any one of two or more mutually exclusive events is calculated by adding the probabilities of the events. Words either , or.

Question 10

Binomial expansion form is?

Answer 10

(p + q)ⁿ where p equals probaility of one event and q represents probability of alternative event. N equals the number of times the event occurs.

Question 11

Give formula of one family with albinism with two children and how to work out probability of five children with albinism.

Answer 11

p = ¹/₄ - probability of one child having albinism

q = ³/₄ probability of one child having normal pigmentation

(p + q)⁵ = p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵

p⁵ chances of five children with albinism

5p⁴q is four with albinism and one without

So finding chances of two with albinism is;

10p²q³ = 10(¹/₄)²(³/₄)³ = ²⁷⁰/₁₀₂₄= 0.26

Question 12

Explain how to work out how to expand binomial (p + q)ⁿ

Answer 12

Number of terms is n + 1

start with pⁿ then p^(n-1)q¹, p^(n-2)q², until n + 1 terms are done.

If n = 5 then;

(p + q)⁵ = p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵

To work out coefficient the first term is always 1. The second term is the power it is raised ie for above example that is 5. Then third term is using numbers from second term so (5x4)/2 = 10 where two is number on sequence. For fourth term (10x3)/3 = 10 and continue until the end. Ie for;

(p + q)³ = p³ + 3p²q + 3pq² + q³

Question 13

Testcross is when?

Answer 13

one individual of unknown genotype is crossed with another individual with a homozygous recessive genotype for the trait in question.

Ie for tallness. So the known is tt.

if the unknown is TT then (TT x tt –> all Tt) but if unknown is Tt then (Tt x tt –> ¹/₂Tt and ¹/₂tt). Of course not in book but if unknown is tt then all will be tt.

Question 14

Table of Phenotype ratios in off spring give info on parental genotype

Answer 14

Question 15

Genetic symbols explain

Answer 15

• lowercase - recessive alleles
• uppercase - dominant alleles
• Can use more than one letter with front letter upper or lower case depending on its nature.
• “Wild typ” alle signified by +
• Superscripts or subscripts can be used ie Lfr₁
• Slash may be used to distinguish alleles present in an individual genotype. ie El⁺ /El^R
• If genotypes at more than one locus are presented together a space separates the genotypes; Ie El⁺ /El^R G/g

Question 16

Table of genotypic ratios for simple genetic crosses (crosses for a single locus)

Answer 16

Question 17

dihybrid cross is?

Answer 17

a cross that differs in two characteristics, ie round, yellow seeds RR YY or wrinkled green seeds rr yy

Question 18

Mendel’s second law is and what is it about?

Answer 18

principle of independent assortment which states that alleles at different loci separate independently of one another.

Question 19

What is an important qualification of the principle of independent assortment?

Answer 19

it applies to characters encoded by loci located on different chromosomes because it is based wholly on behaviour of chromosomes in meiosis and each pair of homologous chromosomes separates independently of each other in anaphase I of meiosis.

Genes located on same chromosome will stay together unless crossing over takes place and sometimes if the genes are located far enough apart crossing over might take place every meiotic division.