C6.1 -- inheritance 1

Question 1

when these leaves were self-pollinated, explain why only the white-flowered, three-leaved plants bred true. [2]

Answer 1

1. white-flowered, three-leaved plants are double homozygous recessive/ homozygous recessive at BOTH gene loci (Reject if no mention of both or double)
2. the other plants are heterozygous at either one or both gene loci

Question 2

describe how pure-breeding, red-flowered, 5-leaved clover plants (RRLL) could be produced using the offspring of the original cross [3]

Answer 2

1. select red-flowered, 5-leaved offspring from the cross and allow it to self-pollinate
2. 1/16 of the offspring would be double homozygous dominant
3. a test cross can be used to test the red-flowered 5 leaved offspring to pick out the double homozygous dominant from the double heterozygotes

formula:
1. do the cross (rough)
2. % proportion of F2 offspring showing the phenotype
3. how to distinguish the plants of the same phenotype

Question 3

suggest why the IA and IB alleles are dominant over the i allele [3]

Answer 3

1. IA and IB represent the allele for production of antigen A and antigen. B on surface of red blood cells
2. I represent the allele that does not produce any antigen on surface of red blood cells
3. in heterozygous individuals IAi or IBi, only the IA or IB allele will determine the blood group of the individual

Question 4

suggest where the rhesus (Rh) factor gene may be found given it’s inherited independently from blood group. explain. [2]

Answer 4

1. located on a chromosome other than chromosome 9
2. allows for independent assortment of homologous chromosomes at metaphase I of meiosis

Question 5

Explain the coat colour of labrador retrievers with the genotype Eebb. [3]

Answer 5

1. Eebb = brown coat colour
2. Allele E of one gene codes for a functional enzyme E, which converts the yellow coat
   pigment to brown coat pigment
3. With Ee, there is still one functional allele to convert yellow to brown
4. If Allele B of another gene is present, it codes for enzyme B, which converts the brown coat
   pigment to black coat pigment
5. Since the genotype is bb, there is no functional enzyme B thus resulting in Eebb having
   brown coat colout

Question 6

Within a struggle for existence with surrounding cells, discuss whether the cell
lineage that has the mutator phenotype and the mutation shown as a black
triangle has a selective advantage compared to cell lineages without these
features. [5]

Answer 6

1. The cell lineage with both mutator phenotype and black triangle mutation will have a
   selective advantage;
   Mutator phenotype has an increased rate of mutation
2. The mutator phenotype has an increased rate of mutation which resulted in all three
   types of mutations, passenger, driver, and chemotherapy resistance mutations;
3. Mutator phenotype will also result in a higher rate of mutation giving rise to greater
   variation;
4. And higher chance of forming favourable alleles/phenotypes for the cell e.g. faster
   growth, ability to gain resources;
   Black triangle represents chemotherapy resistance mutation
5. When exposed to chemotherapy, the cell is able to survive and continue to divide
   while other cells without this mutation will die;

Question 7

One of the ears of maize resulting from self-pollination of the F1 plants had 216 smooth
purple grains, 65 smooth yellow grains, 79 wrinkled purple grains and 21 wrinkled yellow
grains.
A chi-squared test was carried out on these results to compare the expected number of
each phenotypes with the observed number.
The calculated chi-squared value was 1.73.
The critical chi-squared value for these results at a probability of p = 0.05 is 7.81.
State and explain what can be concluded about inheritance of grain colour in maize from
the chi-squared value of 1.73. [3]

Answer 7

1. The inheritance of grain colour in maize follows normal Mendelian monohybrid
   inheritance;
2. Since the ꭕ2

= 1.73, it is less than the critical ꭕ2

value of 7.81,

3. and we do not reject the null hypothesis;
4. There is no significant difference between the observed and expected phenotypes
   in the offspring generation and any difference is due to chance only;

Question 8

Suggest how pure-breeding maize with smooth yellow grains can be obtained from the F2 grains by selection and breeding. [3]

Answer 8

1. Perform a cross between the smooth, yellow grain from the F2 generation with
   homozygous recessive* wrinkled yellow grains;
2. If all the offspring from the cross are smooth, yellow grains, then the original parent
   grain was a pure-breeding smooth, yellow grain;
3. If the offspring from the cross produced a ratio of 1 smooth yellow grain : 1 wrinkled
   yellow grain, then the original parent grain was not pure-breeding (i.e. Aabb);

Question 9

None of queen Victoria’s ancestors suffered from haemophilia. Suggest in whom and what organ a mutation occurred that accounts for haemophilia in her body [2]

Answer 9

1. Queen Victoria
2. Occurred in her ovaries

Question 10

Explain how the environment determines whether fertilised eggs develop into queen bees or worker bees. [3]

Answer 10

1/ ref to after hatching, all larvae from fertilised eggs, are fed a diet of royal jelly
2. Ref to larvae destined to be worker bees are switched to a diet containing honey and pollen, larvae destined to be queen bees stay fed with royal jelly
3. Ref to high protein content of royal jelly stimulates production of the female reproductive system in queen bees and diet is an environmental factor

Question 11

Use genetic diagrams to explain how the person knew that only one of the two original rabbits was not pure-breeding and describe how he could find out whether it was the male or female black Rex rabbit that was not pure-breeding. [20]

Answer 11

1. Since both parents were black Rex rabbits, they would be homozygous recessive at the r locus/ have rr genotype
2. Thus the parents would only be able to have variation at the B locus / ref to being homozygous or heterozygous at B locus

Max 5m for scenario 1
3. Scenario 1: if one parent is homozygous and the other is heterozygous at the B locus (BBrr x Bbrr) all F1 offspring would be black Rex rabbits (BBrr or Bbrr)
4. [genetic diagram]

5. Hence it would be possible to mate F1 generation (Bbrr x Bbrr) to produce F2 generation with 3 black Rex:1 brown Rex rabbits.
6. [genetic diagram]
7. Thus presence of brown offspring in F2 generation explains how the person knew that one parent was not pure-bred in this scenario

Max 5m for scenario 2 and 3
8. Scenario 2: if both parents are homozygous at the B locus (BBrr x BBrr), all F1 offspring would be homozygous BBrr
9. Hence would not produce any F2 offspring with the brown/ bb genotype, since both parents are pure breeding, thus reject this scenario
10. Scenario 3: if both parents are heterozygous at the B locus, the F1 offspring would have a ratio of 3 black Rex: 1 brown Rex
11. Genetic diagram

12. The person would know that both parents were not pure breeding even without crossing the F1 generation thus reject this scenario

Max 5m to identify if the female or male is not pure-bred
13. Male or female black Rex rabbits cold be (B_rr)
13. A test cross were performed for each parent
15. With a pure breeding brown Rex rabbit (bbrr)
16. Of opposite gender
17. If the parent were pure breeding (BBrr), there would only be the presence of black Rex (Bbrr) rabbits

Question 12

explain the term epistasis in this context [3]

Answer 12

1. 2 gene loci are involved in pigment synthesis/ colour of flower/ fruit
2. Expression of a dominant allele of the epistatic gene, inhibits the expression of alleles of the other/hypostatic gene, and will result in white fruit
3. homozygous recessive condition at either / both gene loci will result in white flowers/ green fruit
4. double heterozygote condition will result in purple flowers
5. when the epistatic gene locus is homozygous recessive, together with the presence of a dominant allele at the other/ hypostatic gene locus, will result in yellow fruit

Points 3, 4, 5 required for full credit. note that the answer given in here is modified from 2 sets of answers asked in 2 qs, phrased in the same way.

Question 13

explain how the chi-squared calculated value of 1.60 / 2.88 supports the statement that /(dominant) epistasis is the correct explanation of these results [3]

Answer 13

1. since the chi-squared calculated value of 1.60 is smaller than the chi-squared critical value of 3.84 for level of significance of 5%
2. the probability that chance alone is the reason for the difference between observed results and expected F2 phenotypic ratio of 9:7 is greater than 0.05/ 5%, the deviation is not significant
3. hence, do not reject H0 and the results support the 5% level of significance that epistasis is responsible for the colour of flowers observed

OR
1. probability that chance alone is the reason for the deviation from the 9:7 ratio is greater than 5%/ 0.05
2. deviation between the observed results and expected F2 phenotypic ratio of 9:7 is not significant
3. hence, do not reject H0 and the results support the 5% level of significance that epistasis is responsible for the colour of flowers observed

/
1. At p = 0.05, degree of freedom = 2; since chi-squared calculated = 2.88 < chi-squared critical = 5.99
2. the probability that chance alone is the reason for the deviation between the observed results and the expected F2 phenotypic ratio of 12:3:1 is greater than 0.05/ 5%, the deviation is not significant
3. hence, do not reject H0 and the results support the 5% level of significance that dominant epistasis is responsible for the colour of fruits observed

Question 14

explain what is meant by codominance [3]

Answer 14

1. codominance is observed where the 2 different alleles, IA and IB, each give rise to distinct functional products
2. and the effect of each allele can be observed as they are expressed equally in the phenotype of the heterozygote
3. giving rise to the simultaneous presence of surface antigens A and B on red blood cells in AB blood group

Question 15

explain why a man with haemophilia cannot pass haemophilia to his son but may pass to his grandson [3]

Answer 15

1. as the man with haemophilia will only pass the Y chromosome to his son, his son will not receive the recessive allele
2. however the man can pass the recessive allele on X chromosome to his daughter
3. who has a 50% chance of passing the recessive allele to her son

Question 16

explain how a farmer could use a breeding programme to find out the genotype of a male chicken with blue, barred feathers [3]

Answer 16

1. the farmer could do a test cross
2. between a male chicken and a female chicken with non-barred feathers
3. if the ratio of offspring with barred feathers to offspring with non-barred feathers is 1:1, the male chicken is heterozygous for the gene for barred feathers OR if all offspring have barred feathers, the male chicken is homozygous for the gene for barred feathers

Question 17

describe how you would determine the unknown genotype of a female mouse with a bent tail [2] !!!!

Answer 17

1. carry out a test cross by breeding with a normal male of genotype X^aY
2. if all offspring have bent tails, then the female must be homozygous dominant XAXA
3. if there are 50% with normal tails, then the female would be heterozygous XAXa

Question 18

suggest why mutations are usually recessive [2]

Answer 18

1. most mutations result in a loss of function in the protein coded
2. in diploid organisms, a single functioning copy (dominant allele) of a gene will result in production of sufficient functional protein to mask the loss of function (recessive) allele
   (there are 2 copies of the same gene)

Question 19

describe how structures P (sister chromatids) differ from structures Q (a pair of homologous chromosomes) [2]

Answer 19

1. structures P (sister chromatids) are genetically identical/ contains the same genes and same alleles, while structures Q are not genetically identical/ contains the same genes but may not have the same alleles for each gene
2. P are separated during anaphase II of meiosis while Q are separated during anaphase I of meiosis
3. P are derived from a single parent while Q are derived from both parents (paternal and maternal)

Question 20

account for the different test cross results obtained from plants A and B [2]

Answer 20

1. in A, dominant alleles H and F are linked on one chromosome and the recessive alleles h and f are linked on the other homologue
2. in plant B, its chromosomes have dominant allele linked to a recessive allele, ref to H and f, h and F
   OR
   ref to coupling arrangement in plants A/ repulsion arrangement in plant B
   OR having different allelic arrangements on their chromosomes/ produces gametes with different allelic combinations

(basically offspring of test cross with both plants do not give 1:1:1:1 ratio, hence both have incomplete linkage, not complete linkage as 1:1 will be expected if complete linkage. then this is explanation)

Question 21

explain how the distance between 2 linked genes on a chromosome can affect the products of meiosis [2]

Answer 21

1. ref to when 2 linked genes are located far apart, higher chances of crossing over between non-sister chromatids/ chiasmata formation
2. to produce recombinant gametes/ new combination of alleles/ recombinant chromatids in gametes
   OR
3. ref to when 2 genes are closely linked, no crossing over
4. producing only parental gametes

Question 22

suggest why the observed numbers recorded for plant A do not exactly match the expected proportions [1]

Answer 22

1. sample size is too small
2. random fertilisation of gametes
3. some seeds could have died before counting was done
4. ref to adverse environmental conditions
5. ref to chance events

Question 23

distance between 2 loci on a chromosome

Answer 23

distance between 2 loci on a chromosome = [total number of recombinants/ total number of progeny] x 100%

if the distance between the 2 loci on a chromosome is given, let’s say it’s 32.6 map units. this means 32.6% of offspring will be expected to be recombinant. hence 32.6%/2 =0.163 which will be the expected proportion for each recombinant phenotype (rmb there’s 2 recombinant phenotypes)

Then the other 2 phenotypes’ frequency will be expected to be (100%-32.6%)/2

• recombinants data will be the total number of plants/ variable that has a different genotype than their parents
• progeny will be the sum of all 4 genotypes’ offspring sampled.

Question 24

explain the meaning of the terms locus and allele [4]

Answer 24

locus:
1. ref to specific location of a gene on a chromosome
2. it may contain alternate forms of a gene

allele:
1. it is an alternative form of a gene at a particular gene locus
2. it is responsible for determining contrasting traits of the same character

Question 25

explain why pea plants are good experimental organisms for carrying out such crosses [2]

Answer 25

1. pea plant have many observable characters with contrasting traits 2. pea plants are easy to handle/ fertilised 3. pea plants produce large number of offspring/ progeny (hence data) in relatively short periods of time

Question 26

explain why there is a range of phenotypes (continuous variation) for this characteristic in F2 offspring [4]

Answer 26

1a. ref to independent assortment of homologous chromosomes during metaphase I of meiosis/ crossing over between non-sister chromatids of homologous chromosomes during prophase I of meiosis in F1 individuals 1b. ref to production of a variety of genotypes/ allelic combinations in F2 offspring upon random fertilisation 2a. ref to combined effect of polygenes/ multiple genes that act on the phenotype in an additive manner 2b. ref to each gene contributing a small amount to/ having little overall effect on the phenotype produces a great variety of phenotypes 3. the cumulative effect of varying environmental factors acting on different F2 genotypes further increases phenotypic variation in F2

Question 27

suggest how the corolla tube length of offspring resulting from a cross between the first generation (F1) of offspring and the parental variety with long-tubed corollas will be. EYA. [3]

Answer 27

any 1 from 1. the offspring corolla tube length will show a larger continuous range of variation, between 30 to 90mm 2. skewed to the range of the long-tubed pure-breeding parental form/ no short-tubed corolla phenotype/ no plants with corolla length below 30mm will be observed any 2 from 3. independent assortment of homologous chromosomes, during metaphase I of meiosis, resulting in new combination of alleles 4. crossing over between non-sister chromatids of homologous chromosomes, during prophase I of meiosis resulting in new combination of alleles 5. ref to multiple gene loci/ genes with multiple alleles controlling corolla tube length 6. ref to production of a variety of recombinant genotypes that give rise to a range of intermediate tube lengths in the offspring

Question 28

explain the conclusion that may be drawn from your probability value. [4]

Answer 28

1. no. of degrees of freedom = 3 - 1 = 2; probability that deviation of observed results from expected ratio due to chance is greater than 0.05 2. since the probability alone for the reason from deviation from 1:2:1 ratio is greater than 5%, deviation between the observed results and the expected F2 phenotypic ratio of 1:2:1 is not significant 3. therefore, the results support 5% level of significance that there is a pair of segregating alleles at a single locus/ corolla colour exhibits monohybrid inheritance 4. with one allele incompletely dominant over the other/ ref to incomplete dominance

Question 29

with reference to figure, predict and explain the most likely mode of inheritance of G6PD deficiency. [3]

Answer 29

1. sex-linked recessive 2. sex-linked: all affected individuals in the family are males such as II-1, indicating that the males display disease phenotype more often than the females as males are homozygous/ have only 1 copy of the sex-linked gene 3. recessive: unaffected parents such as I-1 and I-2 can produce affected offspring such as II-1, indicating that the mothers must have a dominant allele to mask the effect of the recessive allele

Question 30

deduce if the observed results follow the expected phenotypic ratio of 1 blood group A: 2 blood group AB: 1 blood group B. EYA. [3]

Answer 30

1. no 2. at p = 0.05, degree of freedom = 2, since chi^2 calculated = 7.22 > chi^2 critical = 5.99 3. there is less than 5% probability that the difference between observed and expected results is due to chance alone, indicating that the deviation is significant

Question 31

explain the type of gene interaction observed in this context [3]

Answer 31

1. epistasis (duplicate dominant) 2. allele A is epistatic to alleles B and b/ gene B/b, and allele B is epistatic to alleles A and a/ gene A/a 3. the presence of at least 1 dominant allele of either gene A or B causes the conversion of precursor X to product Z hence allowing the formation of heart-shaped fruits 4. if the genotypes at both loci are homozygous recessive, the conversion of precursor X to product Z does not occur, resulting in formation of narrow fruits

Question 32

describe how you could identify Capsella sp plants with heart-shaped fruits, which are homozygous dominant in at least one gene locus [2]

Answer 32

1. perform a test cross by crossing the plants with heart-shaped fruits with plants with narrow fruits/ double homozygous recessive 2. if the plants with heart-shaped fruits are homozygous dominant in at least one gene locus, all offspring produced will only have heart-shaped fruits (why at least one gene locus: continuous variation -> you can have 2 or more genes controlling the same phenotype, and correspondingly more alleles controlling the phenotype expressed too)

Question 33

Vestigial wing and ebony body are caused by artificially induced mutations in laboratory bred fruit flies. Suggest why such mutants are unlikely to be found in natural populations. [2]

Answer 33

any 2 1. Flies with vestigial wings would be disadvantaged in the wild, as they cannot fly far ; 2. Flies with ebony bodies are more visible to predators, therefore unlikely to survive as they are more likely to be eaten ; 3. Gene mutation, is a rare occurrence ; 4. Such alleles not passed on to offspring, because mutants are unlikely to reproduce ; 5. Such characters are coded for by recessive alleles, which are therefore masked by dominant alleles ;

Question 34

Explain why it is unlikely that any daughters fathered by R would suffer from haemophilia or colour blindness. [2]

Answer 34

1. Since haemophillia and colour blindness are sex-linked diseases, and females have 2 copies of each sex-linked gene ; 2. Daughters must possess 2 recessive alleles / be homozygous recessive, for haemophilia or colour blindness to inherit the disease ; 3. Thus the mother would need to be a carrier (heterozygous) or be affected (homozygous recessive) for either disease ;

Question 35

Suggest how the genotype of S has arisen. (X^H Y X^b Y) [2]

Answer 35

1. Genes for colour blindness and haemophilia are incompletely linked on X chromosome ; 2. Crossing over during prophase I, result in mutual exchange of segments of non-sister chromatids in P (thus causing son S to inherit only colour blindness but not haemophilia) ;

Question 36

define locus [2]

Answer 36

Refers to the specific location of a gene on a chromosome ; 2. May contain alternate forms of the gene ;

Question 37

define linkage in this context [2]

Answer 37

Genes such as red eye and normal wings or purple eye and vestigial wings located on the same chromosome ;

Question 38

Suggest how similar breeding experiments with many different pairs of characters could be used to map the position of genes on the chromosomes of fruit flies. [3]

Answer 38

1. The chance for crossing over occurring between two linked genes is reduced if they are close to each other / ORA ; 2. Via numerous breeding experiments, cross-over value (COV) may be determined by calculating number of individuals showing recombination / number of offspring x 100% ; 3. Thus, by analyzing the relative distance between different pairs of genes, the exact order of various genes on a chromosome can be determined which is used to generate a chromosomal map ; 4. If the expected phenotypic ratio is obtained, there is no linkage present ;

Question 39

Explain how different genotypes give rise to a white phenotype (homozygous recessive). [3]

Answer 39

1. Genotype rr / homozygous recessive condition for gene R/r is epistatic over the E/e locus masking the expression of the alleles for red and purple pigmentation ; 2. Dominant allele R codes for an enzyme that converts its colourless substrate to give a red product ; 3. Dominant allele E codes for a second enzyme that subsequently converts the red product to give a purple product ; 4. With genotype rr, the phenotype remains white as the first enzyme to form the red product will not be produced, thus there will be no substrate for the second enzyme to convert to the purple product even with the genotype E_ ;

Question 40

Explain how you could determine that a plant grown from a white grain was homozygous recessive at both loci. [3]

Answer 40

1a. Cross with a true-breeding red grain plant with genotype RRee ; 1b. If offspring of this cross are all red, then the white grain plant is homozygous recessive at both loci ; OR 2a. Cross with a red grain plant with genotype Rree ; 2b. If offspring of this cross are 50% red grain and 50% white grain, then the white grain plant is homozygous recessive at both loci ; 3. If any purple grain plants are seen (from either cross), then the original plant grown from a white grain must have at least one allele E at the E/e locus, and hence is not homozygous recessive at both loci ;

Question 41

Explain how the allele B affects the gene M/m. [4] (B is dominant epistatic to gene M/m)

Answer 41

1. Allele B is epistatic to gene M/m OR Gene M/m is hypostatic to allele B ; 2. Allele B suppresses / inhibits the expression of the banding gene ; 3. A single copy of the allele B is sufficient in blocking / suppressing / inhibiting the pathway for banding ; 4. No banding occurs even in the presence of one dominant M allele ;

Question 42

Define gene

Answer 42

Gene: A unit of inheritance located at a particular locus of a chromosome./ A specific DNA nucleotide sequence which codes for RNA (mRNA, tRNA or rRNA) or a polypeptide.

Question 43

Define genotype

Answer 43

Genotype: The complete genetic makeup / allelic composition of an organism. The term is also commonly used in reference to the paired alleles carried by an organism that give rise to a phenotype.

Question 44

Define phenotype

Answer 44

Phenotype: The physical manifestation of a genetic trait that results from a specific genotype and its interaction with the environment.

Question 45

Define homozygous and heterozygous

Answer 45

Homozygous: Condition in which the alleles of a gene pair in diploid condition are identical, e.g. TT or tt, where T is the dominant allele for tall stem and t is the recessive allele for short stem. An organism with this condition is known as a homozygote and referred to as true or pure breeding. All gametes produced by this organism will carry the same allele. Heterozygous: Condition in which the alleles of a gene pair in diploid condition are different, e.g. Tt. An organism with this condition is known as a heterozygote. 50% of its gametes will carry one allele while the other 50% will carry the other allele.

Question 46

Define dominant and recessive

Answer 46

Dominant alleles: Produce their effects in both homozygous and heterozygous condition i.e. one copy of the allele is sufficient to cause the organism to express the phenotype it encodes, e.g. allele T causes the plant to be tall in either TT or Tt condition. A dominant allele masks the influence of the recessive allele. An organism homozygous for a dominant allele is known as homozygous dominant. Recessive alleles: Produce their effects only in homozygous condition, e.g. allele t causes the plant to be short only in the tt condition. An organism homozygous for a recessive allele is known as homozygous recessive.

Question 47

Linkage

Answer 47

Linked genes: Genes that control different characters and situated on the same chromosome at different loci

Question 48

Define codominance

Answer 48

Codominance: a phenomenon in which both alleles are equally expressed in the phenotype of the heterozygote. The heterozygote simultaneously expresses the phenotypes of both types of homozygotes.