AQA A2 Biology B17 Inherited Changes Kerboodle Answers

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C17.1 Studying Inheritance AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 156

The genetic composition of an organism is called the genotype (1) and any change to it is called a mutation (2) and may be inherited by future generations. The actual appearance of an organism is called the phenotype (3). A gene is a sequence of nucleotides/bases (4) along a section of DNA that determines a single characteristic of an organism. It does this by coding for particular polypeptides (5) that make up the enzymes needed in a biochemical pathway. The position of a gene on the DNA of a chromosome is called the locus (6). Each gene has two or more different forms called alleles. If the two alleles on a homologous pair of chromosomes are the same they are said to be homozygous (7), but if they are different, they are said to be heterozygous (8). An allele that is not apparent in the phenotype when paired with a dominant allele is said to be recessive (9). Two alleles are called codominant (10) where they contribute equally to the appearance of a characteristic.

17.2 Monohybrid inheritance AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 159

1 2 b

17.3 Probability and genetic crosses AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 161

1 a Homozygous dominant (GG) b We cannot be absolutely certain because if the unknown genotype were heterozygous (Gg) the gametes produced would contain alleles of two types: either dominant (G) or recessive (g). IL is a matter of chance which of 1hese gametes fuses with those from our recessive parent -all these gametes have a recessive allele (g). Tt is just possible that, in every case, it is the gametes with the dominant allele that fuse and so all the offspring show the dominant character. Provided the sample of offspring is large enough, however, we can be reasonably sure that the unknown genotype is homozygous dominant. 2 a Heterozygous (Gg) b We can be certain because 7 of the offspring display the recessive character (in our case yellow pods). These plants arc homozygous recessive and must have obtained one recessive allele from each parent. Our unknown parental genotype must therefore have a recessive allele and be heterozygous (in our case Gg). 11 is theoretically possible that the: plants with yellow pods were due to a mutation but this is most unlikely. The unexpectedly low number of plants with yellow pods is the result of random fusion of the gametes. c 50% d. 7.29%

17.4 Dihybrid inheritance AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 164

1 Red eyes and nom1al wings are dominant because these characteristics are expressed in the F1 generation while pink eyes and vestigial wings are not expressed in the F1 generation and so these are recessive. Also red eyes and normal wings appear around 3 times more often in the F2 generation than pink eyes and vestigial wings. 2 R for red eyes and r for pink eyes, N for normal wings, n for vestigial wings. 3

17.5 Codominance and multiple alleles AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 167

1 The man is not the father. Reasons -child has blood group AB and therefore has alleles I^AI^B. The mother is blood group A and therefore either I^AI^O or I^AI^A. In either case she could have provided the I^A alleles to the child but not the I^B allele. The I^B allele must have come from the real father. The supposed father is blood group O and therefore has alleles I^OI^O. He cannot provide an I^B allele and so cannot be the father. 2

17.6 Sex-linkage AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 171

1 E = XX F = XY 2 A= not colour blind/normal vision B =not colour blind/normal vision D= colour blind 3 4 0% -because sons inherit their X chromosome from their mother and she has only alleles for normal vision (X^R). 5 By mutation (of the R allele).

17.7 Autosomal linkage AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 175

1 ln sex-linkage the linked genes arc on the same sex chromosome (usually the X chromosome) whereas in autosomal linkage they arc on any chromosome other than the sex chromosomes. 2

17.8 Epistasis AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 178

1 a Mouse 1 = albino(white); mouse 2 =agouti b AABb; AaBb; AAbb; Aabb; AaBb; aaBb; Aabb; aabb c 4 albino : 3 agouti : I black 2 a AaBb purple seeds b c i AABB; AABb; AaBB; AaBb; AABb; AaBb; AaBB; AaBb; AaBb. ii They all possess at least one dominant allele A and one dominant allele B. d The production of anthocyanin uses a biochemical pathway that requires two functional enzymes each coded for by the dominant allele of both genes A and B. If either gene is represented by two recessive alleles the enzyme it codes for is non-functional and the pathway cannot be completed. This is an example of epistasis because it affects the other gene in that, even if it is functional and produces its enzyme, its effects cannot be expressed because no pigment can be manufactured. Banner 2

17.9 The chi-squared (X²) test AQA A2 Biology B17 Inherited Changes Kerboodle Answers : Page No. 181

1. There is no significant difference between the observed and the expected results. 2 Three degrees of freedom. 3 Chi-squared value = 9.11. 4 The value of 9.11 lies between 7.82 and 9.84, which is equivalent to a probability of 0.05 (5%) and 0.02 (2%) that the deviation is due to chance alone. 5. This deviation is significant and we must reject the null hypothesis. Banner 3

Practice questions: Chapter 17: Page No. 182-183

1 (a) (Genes/loci) on same chromosome. (b) 1. GN and gn linked.

2. GgNn individual produces mainly GN and gn gametes.
3. Crossing over produces some/few Gn and gN gametes.
4. So few(er) Ggnn and ggNn individuals.

(c) (Grey long : grey short : black long : black short) =1:1:1:1 (d) Expected all 773.75, Chi2 is 579.3; P is much less than 0.001/0.1%, Observations differ significantly from the expected. 2 (a) (i)

1. Animal 2 / 5 has hair but offspring do not.
2. So 2 / 5 parents must be heterozygous/carriers.

OR

3. 4/7/8 are hairless but parents have hair.
4. So 2 / 5 must be heterozygous/carriers.

(ii) Hairless males have fathers with hair / 4 is hairless but 1 is hairy / 7 and/or 8 are hairless but 6 is hairy / only males are hairless. (b)

1. Parental genotypes X^HX^h and X^HY

Gametes X^H X^h X^H Y;

2. Genotypes of offspring X^HX^H, X^HY, X^HX^h , X^hY;
3. Phenotypes of offspring female with hair male with hair male hairless; 4. 0.25 / ¼ / 1 in 4 / 25%;

3 (a)

1. Large number of eggs/offspring/flies (therefore) improves reliability / can use statistical tests/ are representative / large sample (size) / reduces sampling error.
2. Small size / (breed) in small flasks / simple nutrient medium (therefore) reduces costs/easily kept/stored.
3. Size / markings / phenotypes (therefore) males/females easy to identify.

(b) (i) (3 marks) 4 (ii) Fertilisation is random / fusion of gametes is random / small/not large population/sample /selection advantage/disadvantage / lethal alleles. (c)

1. Males have one allele.
2. Females need two recessive alleles / must be homozygous recessive / could have dominant and recessive alleles / could be heterozygous/carriers.

4 (a) (i)

1. Parents are heterozygous.
2. Kittens receive white allele from parents /black cat.

(ii) 1:1. (b)  (i) 2 chocolate kittens 5 cinnamon kittens. Black, Chocolate, Black. (ii) (iii) Explain why the actual numbers were different from those expected.

1. Offspring ratios are a probability/not fixed/arise by chance.
2. Gametes may not be produced in equal number.
3. Fertilisation /fusion of gametes is random.
4. Small sample

(iv)

1. Possible if parents homozygous/ bb.
2. Don’t know genotype of chocolate cat / chocolate cat

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