Student Textbook Answers - Chapter 1 Genetics & Biotechnology - SCNSW 2014 PDF

1 of 16 STUDENT TEXTBOOK Answers Chapter 1 Genetics and biotechnology Experiment 1.1 Extracting DNA Results 1 DNA was extracted from the nucleus of banana (or other) cells. 2 Responses will vary, but substance should be fragile and mil...

1 of 16 STUDENT TEXTBOOK Answers Chapter 1 Genetics and biotechnology Experiment 1.1 Extracting DNA Results 1 DNA was extracted from the nucleus of banana (or other) cells. 2 Responses will vary, but substance should be fragile and milky white in appearance. Discussion 3 Responses will vary. The purpose of each step is listed in the method. 4 Yes. Factors to consider: surface area exposed and amount of liquid in the original tissue (why wheat germ is such a good, concentrated source). 5 Adequate answers repeat information provided in the method; excellent answers will discuss the mechanisms of how detergents work (such as formation of micelles and recognition that cell membranes contain lipids). 6 Alcohol is less dense; adding it carefully demonstrates this. Adding it without care still precipitates the DNA, but the solution may look cloudy and it will be hard to lift out. 7 Differences could arise because there could be technical issues (spillage), or some materials (particularly watery fruit) may have proportionally less DNA. The important aspect is that students demonstrate efforts of reflection to obtain their answer, even though it is likely to be overestimated, because the ‘DNA’ they extracted is mostly histone (refer to Figure 1.8 on page 13). 9 8 50 3 1 000 000 3 3 3 21 meals 5 3.15 310 or 3 150 000 000 km. 9 The use of the buffer compacts the DNA, which is then precipitated. The extracted ‘DNA’ is not a single strand, and a lot of it is histone. Conclusion 10 The important aspect is that students demonstrate efforts of reflection to obtain their answer, even though it is likely to be overestimated, because the ‘DNA’ they extracted is mostly histone (refer to Figure 1.8 on page 13). © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 1 26/04/14 4:38 PM  2 of 16 Questions 1.1 What have you learnt? Understanding 1 Genetics is the study of heredity. 2 a Deoxyribonucleic acid b A nucleotide c A sugar, a phosphate and a (nucleic acid) base. d A labelled diagram of a double helix (two twisted spiral strands). 3 Adenine only pairs with thymine; guanine only pairs with cytosine. Applying 4 Franklin was notorious for not being careful near X-rays, even as the dangers were becoming known, yet it would always be difficult to prove this was the cause of her cancer. Evaluating 5 Responses will vary. An important consideration is whether we should judge actions of the past by the ethics we are familiar with today (even when this history is relatively recent). Franklin’s own reaction to the Watson–Crick model is reputed to have been delight; she felt it didn’t matter how the problem was solved, only that it had been solved. 6 Responses will vary. Many scientific discoveries are likely to be the work of more than one person and over time. Had Franklin lived, she may have been eligible for the Nobel Prize in a different category. See ‘More information: Rosalind Franklin’. Creating 7 Responses will vary. Questions 1.2 What have you learnt? Understanding 1 The process of semi-conservative replication involves the DNA molecule splitting, and a new complementary strand being made using the original as a template. The outcome is that the two copies are genetically identical. 2 Name of the molecule Where does the activity take place? What is the name of the process? DNA (triplet) Nucleus Transcription mRNA (codon) Cytoplasm Translation tRNA (anticodon) Cytoplasm Protein synthesis © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 2 26/04/14 4:38 PM  3 of 16 Applying 3 a TAT AAC CCG CGG TTC TGA b UAU AAC CCG CGG UUC UGA c Six. Each codon (group of three bases) in this chain is unique. There are six codons, so there will be six different amino acids. Analysing 4 Responses will vary. Reflecting 5 Responses will vary. Activity sheet Sex as a game of ‘chance’ Method 1 X only 2 X or Y 3 Male 4 Female 1 5 2 1 6 2 7 Coin toss outcomes will vary. Results 1 Student responses will vary – some will have the same ratio. 2 No, they should not, because these are independent events. 3 Yes, the resulting ratio is likely to approach 50 : 50. 4 By chance, every time they have a child it could be the same sex. This activity highlights the importance of chance in determining ratios. Sometimes students assume that, because they understand how the Mendelian ratio comes about, any family with four children should therefore express the full range of possible outcomes. However, by chance, all four children may be homozygous recessive. Genes or chromosomes both segregate randomly. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 3 26/04/14 4:38 PM  4 of 16 Activity sheet Values and issues 1 Responses will vary. 2 The following are suggested answers only. a Answers could include the couple’s desire to have a particular family format. However, there could be risks – some sexes are more prone to genetic disease. b Medium consequences. If the choice results in an imbalance of sexes, (such as that which has occurred in China and India in the last few decades), the majority sex may have difficulty dating or finding a partner. c Some human societies practise polygamy (multiple wives) and, more rarely, polyandry (multiple husbands). In some cases, these traditions have come about because of different sex ratios or wealth distribution. 3 Responses will vary. Questions 1.3 What have you learnt? Understanding 1 A gamete is a sex cell – sperm in males and ova in females. It contains half the number of chromosomes. A zygote is a fertilised egg cell, formed by the fusion of a male and a female gamete, and therefore contains the full set of chromosomes. 2 Each gamete contributes DNA from the respective parent, and each DNA molecule contains many genes placed end to end. These carry code for the characteristics to be expressed in the new individual. 3 Genes are on the chromosomes (Figure 1.12 on page 18). Chromosomes are in the nucleus (Figure 1.10 on page 16). More astute students might observe that during cell division, chromosomes (and hence genes) move out of the nucleus onto the spindle fibres. 4 Chromosomes come from our parents. Each of their sex cells (sperm or egg) would have half the chromosomes in their own cells. 5 a The total number of chromosomes contributed by both parents. b 11 c 1 d 12 6 a Half: 26 b 26 c 26 d 26 e Sperm, as a gamete, contains half the total number (52) of chromosomes. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 4 26/04/14 4:38 PM  5 of 16 f No, because of random assortment. Maternal and paternal chromosomes are sorted differently (randomly). Also, because of crossing over during prophase. g 52 h The cells will be genetically identical. Applying 1 7 a 50% or 0.5 or 2 1 b 50% or 0.5 or 2 1 1 1  1 3 1 c 3 3 5  5 2 2 2  2 8 1 1 1 1 d There are two combinations: boy then girl 5 3 5 ; or girl then boy 5. Add the two 2 2 4 4 1 1 1 possibilities together: 1 5. 4 4 2 Analysing 8 This is metaphase, showing the chromosomes on the centre plate. It is probably during mitosis as there is no evidence of homologues paired up (therefore, not metaphase I, meiosis). 9 Mitosis Meiosis a What happens to the number of Exactly the same Halved chromosomes? b How many cell divisions are 1 2 involved? c How many daughter cells are 2 4 produced? d What is the purpose of this kind To keep the information identical; To halve the information in the sex of division? e.g. in growth and repair, in sexual cells reproduction (binary fission) Evaluating 10 a Identical twins form from a fertilised egg dividing by mitosis. Therefore, the chromosomes in their nuclei are exactly the same. b No, because sex is genetically determined. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 5 26/04/14 4:38 PM  6 of 16 Activity 1.3 Variable offspring – ‘If … then’ statements 1 Examples of answers (pre-evolutionary thinking, anticipating Chapter 2) below: If only poorly adapted organisms are produced, then these will be disadvantaged if a resource becomes critical (therefore they may not be able to reproduce). If only well-adapted organisms are produced, they will be able to take advantage of resources better than others. 2 Brainstorming responses will vary. Note: it is only vital if the environment changes. There are many organisms; for example, extremophyles and prokaryotes seem to have changed little in the past 3.8 billion years. This is because they survive in conditions suspected to resemble those of the early Earth; for example, hot springs and black smokers in ocean depths. Questions 1.4 What have you learnt? Understanding 1 A mutation is a change in DNA. 2 Changes can result from environmental factors, including radiation (UV, X-rays and gamma rays) and harmful chemicals. Changes can also occur from a copying error during mitosis and meiosis (estimated, on average, to be five point mutations in 3 billion per mitotic cycle in human cells). 3 DNA mutations can result in: (a) no changes in the protein, because the amino acid it codes for is the same; (b) no effect, because it does not disadvantage the organism in any way; or (c) harmful effects, because the change damages a vital protein or causes cancer. 4 ‘Silent’ mutations are changes in DNA that do not change the protein. This is because the majority of amino acids can be translated from more than one sequence of three base pairs (DNA triplet). 5 DNA mutation rates can be increased by exposure to UV radiation, high-energy radiation such as X-rays, or mutagenic chemicals. Applying 6 Student responses will vary and will draw on information on page 22. Activity 1.4 Mendel’s results 1 Replicates increase the statistical probability that the results match the average of the general population. For example, if Mendel tested only the outcomes in four offspring, it is possible that the ratio would be wildly different from the 3 : 1 ratio he observed. 2 In the F2 all the results are in a 3 : 1 ratio. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 6 26/04/14 4:38 PM  7 of 16 Activity sheet The Mendelian ratio 1 a RR 3 rr F1 5 Rr F2 5 1RR : 2Rr : 1rr (phenotypes – 3 red : 1 white) b Rr 3 rr F1 5 1Rr : 1rr (phenotypes – 1 red : 1 white) 2 Ww 3 Ww F1 5 1WW : 2Ww : 1ww (phenotypes – 3 long-winged : 1 short-winged) Activity sheet Punnett squares 2 2 1 a Rr : rr in 1 : 1 ratio ( round : wrinkled) 4 4 2 2 b GG : Gg in 1 : 1 ratio, but all will look dark green ( : ) 4 4 2 2 c yy : Yy in 1 : 1 ratio ( bright green : yellow) 4 4 4 d 100% Yy – heterozygous yellow ( ) 4 e Students could suggest their own symbol for this trait. Example: TT 3 tt 100% 4 Tt – heterozygous tall ( ) 4 2 2 2 The outcome becomes like a backcross: Tt : tt in 1 : 1 ratio ( tall : short). 4 4 3 This phenotype requires two recessive types of information. 4 a The correct ratios are: 1 : 2 : 1; 3 : 1; 1 : 1; 1. b The ratio can be expanded and then add up to 1 (that is, all the offspring). Example: 1 2 1 1:2:1 5 1 1 5 1 4 4 4 3 1 3:1 5 1 5 1 4 4 2 2 1:1 5 2:2 5 1 5 1 4 4 4 15 4 © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 7 26/04/14 4:38 PM  8 of 16 Activity sheet Dissecting scientific language 1 Word Meaning Anaerobic Not needing oxygen or air Anoxic Without (oxygen) Archaean Ancient (organism) Autotroph Self (nutrition) Biochemistry Life (living) chemistry Biodiversity Life (living) diversity Chlorophyll Green leaf Cytokinesis Cell (division) Endosymbiosis Inside symbiosis Genetics Birth/origin (of) Heterotroph Different nutrition Heterozygous Different (from zygote) Homology Same study of Homozygous Same (from zygote) Palaeontologist Someone who studies old things Photosynthesis Light making Polymer Many (parts) Polymorphic Many forms Symbiosis Together life/living Triphosphate Three phosphates 2 Meaning Term The study of genes Genetics An instrument to look at tiny things Microscope Something that can kill bacteria Bacteriocide A blood protein Haemoglobin An illness involving the stomach Gastritis 3 Responses will vary. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 8 26/04/14 4:38 PM  9 of 16 Questions 1.5 What have you learnt? Understanding 1 a Allele – form of a gene. b Homozygous – both alleles are the same. c Heterozygous – the alleles are different. d Pure-breeding – every generation is the same for a particular trait, indicating the trait has homozygous alleles. e Phenotype – the appearance of an organism. f Genotype – the genetic composition of an organism. 2 a Any three of the following: (a) round seed form; (b) yellow/orange seed colour; (c) brown seed coat; (d) inflated ripe pods; (e) dark green unripe pods; (f) flowers along stem; (g) tall stems. b These traits are all dominant because they were three times as common in the F2 generation. 3 Reginald Punnett was a geneticist who devised the ‘Punnett square’ method for determining the ratio of genotypes in offspring. Applying 4 Let the genotype of the homozygous woman be WW. All her gametes will have the W allele. The genotype of the heterozygous father is Ww. Half his gametes will have the w allele and, therefore, half of his children will be heterozygous for widow’s peak. 5 a Autosomal genes are found on autosomes. ‘Dominant’ means one copy of the allele affects the phenotype; all carriers eventually develop Huntington’s disease. b Let the allele causing Huntington’s be H, and normal alleles be h. The shaded squares (50%) indicate the probability that the child has the condition. Alleles in gametes of heterozygous individual H h Alleles in gametes of h Hh hh unaffected individual h Hh hh Analysing 6 By considering each trait as a single variable, Mendel was able to observe consistent patterns. Later, Mendel investigated pairs of traits, and discovered they behaved independently. (He was observing random assortment, but did not know it.) Evaluating 7 Mendel’s observations are essentially of independent assortment. The two genes that are on the same chromosome were highly likely to have a crossing over occurring between them, and therefore behaved as if they were being randomly assorted. When genes are on one chromosome, the traits would often be inherited together depending on whether they were on parts of the chromatids being exchanged. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 9 26/04/14 4:38 PM  10 of 16 Activity sheet Australian cattle dogs – a genetic case study This activity sheet should be used as extension material. 1 ee 2 kk 3 bb y t 4 There are black hairs sprinkled through the red coat of an A A dog. 5 a No b Yes. Students may like to reflect on the motto of a dog obedience school: It is the deed, not the breed, that counts. 6 All the alleles are homozygous. For example, black and tan is common, and this is homozygous. Brown dogs are considered undesirable, and b is a rare allele. 7 The suggestion in the table is that it is a dominant gene. This could be tested by breeding dogs with and without the feature, and observing phenotypes of offspring. Note that the frequency of the mark does not prove its dominance: O blood group in humans is the most common and it is a recessive phenotype. 8 German shepherds, kelpies or any breed that normally has a black saddle. 9 50% chance each animal is a carrier; therefore, a 25% (0.5 3 0.5) likelihood the cross is like an F1 cross. 10 25% of random matings will produce 25% affected homozygous recessive puppies. 11 You can get dogs genetically tested (students may not be aware of this). You can ensure that you breed from dogs that have no evidence of the disease in their ancestors (but they may still be carriers). You can also monitor the health of puppies: one puppy with the trait is evidence that both parents were carriers and, therefore, you would never breed from them again. You can ensure only to breed from older animals that have had eye checks. This will eliminate homozygous animals certain to pass on the disease. 12 Merle is a lethal recessive – the Merle animals are all Mm. Solid-coloured puppies are MM. Homozygous puppies mm are miscarried. Students may be able to show this with a Punnett square. Activity sheet Pedigrees 1 Hh. This is an autosomal dominant gene. His father is homozygous recessive (hh). 2 hh. This individual does not express dominant phenotype. 3 Individual II1 is hh; her husband is hh. None of their children will be affected (shaded boxes). Gametes produced by II1 h h Gametes produced by h hh hh unaffected man h hh hh © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 10 26/04/14 4:38 PM  11 of 16 4 Here is an example of how an extension of II1 might be shown, if the couple had two sons: Husband II1 5 I1 5 homozygous recessive (see key). I2 5 has at least one dominant allele – homozygous dominant or heterozygous. II1 5 must be heterozygous; II2, II3, II4 and II5 must all be heterozygous because their mother is homozygous recessive. III1 5 homozygous (see key); III2, III3 and III4 all have at least one dominant allele. 6 Haemophilia is a recessive allele because the disease is only apparent in homozygous sufferers. 3 7 a Both individual III2 parents are heterozygous, therefore of the offspring will not have the sickle 4 2 cell phenotype. Of these, will carry the allele for sickle cell anaemia. 3 If III2 is homozygous dominant, all offspring will have at least one dominant allele, and there is zero chance the first born, or any of his children, will have sickle cell anaemia. If III2 is heterozygous, the child has a 25% chance of having sickle cell anaemia (see shaded box in Punnett square). Gametes produced by III2 if he is heterozygous S s Gametes produced by S SS Ss heterozygous woman s Ss ss 2 Combining the probabilities discussed: (probability III2 is heterozygous 5 ) 3 (probability child is 3 1 homozygous 5 ) 4 2 1 2 1 3 5 5 3 4 12 6 b In this suggested example (section shown), IV1 is a daughter without sickle cell anaemia. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 11 26/04/14 4:38 PM  12 of 16 h h 8 I1 is homozygous X X. I2 is a normal XY male. All the mother’s gametes carry the haemophilia allele on h the X. h II1 and II3 are heterozygous daughters X X, the normal X chromosome coming from the father. h II2 and II4 are haemophiliac sons X Y, the Y chromosome coming from the father. 9 Every son produced by the parents will be a haemophiliac because both the mother’s X chromosomes carry the haemophilia allele. 10 a Gametes produced by II3 X h X Y XY h XY Gametes 25% haemophiliac son 25% normal son produced by normal male X XhX XX 25% carrier daughter 25% normal daughter b The pedigree can be extended as follows (the example given is the normal son): Activity sheet Karyogram Results 1 Person X is male. 2 47 chromosomes: XXY Discussion 3 Responses will vary. Negative: risk of spontaneous abortion, false negatives (and positives). Positive: emotional distress of having a severely disabled child, possibly with a fatal genetic disease that causes pain and suffering. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 12 26/04/14 4:38 PM  13 of 16 Activity sheet Genetically modified food 1 To stop fruits over-ripening (banana), protection from fungal rot (banana) and to make fish grow bigger. 2 Some genes can be found naturally in the organism, but are switched off; or they can come from other unrelated organisms such as radishes, onion or salmon. 3 Student responses will vary, but they should understand that every domestic crop or animal is the product of ‘human interference’ with nature. 4 Genetic modification is changing the genome of the organism by introducing genes from species that are very different (remotely related genetically); for example, a bacterium that has been engineered to produce human insulin to help diabetics. Hybridisation and selection works with the genes that occur naturally within the organism. Activity sheet Human Genome Project 1 To understand how our body works, how illness can be overcome, and how to produce better and more tailored drugs. 2 Currently, medical information is private because it could be misused to disadvantage individuals. It may not be in the interests of individuals to know their ‘genetic future’ if it predicts a chronic disease or low life expectancy. 3 Ethical issues: if commercial companies are able to select who they will insure (inevitably those deemed at lowest risk), this does not spread the risk or benefits through the entire population (a fairer way to support a community). Because genetic illness is not a choice, this type of discrimination counters ‘human rights’. It is similar to discriminating against a person because of sex or race (which are likewise genetically determined). An alternative answer includes the right of a company to be profitable. National insurance schemes replace private schemes in some countries. 4 a Responses will vary. b The company provides an exclusive service for all patients with this type of cancer, and this is potentially profitable. Without competition, the company can set its charges. The company may argue its monopoly provides quality assurance. c By limiting access to its scientific data, the company’s intellectual property is quarantined, hampering the development of improved, alternative tests. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 13 26/04/14 4:38 PM  14 of 16 Questions 1.6 What have you learnt? Understanding 1 Something else. Extension: intermediate dominance would be expected to produce plants of average height. Co-dominance reveals both alleles (an example is AB blood groups), but short alleles are not apparent. This is a phenomenon called ‘hybrid vigour’ and is often used in agriculture to boost yields. The effect in pea plants is so small that it is often overlooked and given as a simple Mendelian ratio. 2 Sex-linkage (X-linkage) is when the gene is on the X chromosome. Analysing 3 It shows the child’s chromosomes from largest to smallest. The sex chromosomes are shown last. The child is male and has 47 chromosomes, with three copies of chromosome 21, a condition called trisomy 21 or Down syndrome. Evaluating 4 Responses will vary. Advantages: solving crime, connecting people to families (for example, adoptees, refugees, people who have amnesia). Disadvantages: privacy issues, particularly if exploited for commercial gain, enabling life insurance companies to ‘cherry pick’ individuals unlikely to be at risk of common diseases. Creating 5 Student responses will vary, but they should be developing a growing understanding of the complexity of genetic influence. Questions 1.7 What have you learnt? Understanding 1  Removal of eggs from the mother’s ovaries. External fertilisation in a laboratory. Implantation in the mother’s uterus. Applying 2 Benefits: (1) may help infertile couples conceive a child; (2) allows the parents to select the sex of the child; and (3) allows selection of embryos without identified genetic disorders. Risks: (1) surplus embryos, which may be perfectly viable, may be killed; (2) as an invasive procedure it may introduce health risks to the mother; and (3) is very expensive, but may not be successful. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 14 26/04/14 4:38 PM  15 of 16 Analysing 3 Some possible answers: Natural reproduction IVF Internal fertilisation Laboratory fertilisation Often emotionally fulfilling Clinical Single egg fertilised (usually) Multiple eggs fertilised Requires no invasive medical procedures Invasive medical procedures Unsuccessful for infertile couples May be successful for infertile couples Inexpensive Expensive May involve ethical dilemmas May involve ethical dilemmas Evaluating 4 Responses will vary. 5 Responses will vary. 6 Responses will vary. Chapter review Remembering 1 Responses will vary. An example is IPMAT PMAT. Understanding 2 Responses will vary. 3 a Self-fertilisation. This meant lines naturally became ‘pure’. Mendel was also lucky that the traits he chose to study were mostly on different chromosomes, sorted independently. b Transparent eggs 4 Transcription: the DNA information is copied using a different type of base. (DNA base sequence replaced with RNA base sequence – essentially the same form, so you can ‘read’ it back to the original sequence.) Translation: the RNA information is changed into a protein. (RNA base sequence converted to an amino acid sequence – but the ‘mode’ has changed, therefore it cannot be used to find the original sequence.) 5 Here is a simple version: P 5 phosphate group, S 5 Deoxyribose sugar, B 5 Base. P S P S P S P B B B The diagram should show alternating phosphates and sugar (deoxyribose). If drawing the complementary strand, students may want to describe the A–T and C–G relationship, too. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 15 26/04/14 4:38 PM  16 of 16 6 DNA: the chemical (polymer) making up the gene. Nucleotide: a unit of the DNA (polymer). Base: the type of subunit in a nucleotide (such as A – adenine, T – thymine, C – cytosine, G – guanine). Gene: section of chromosome that codes for a protein. Chromosome: length of DNA within cell nucleus. 7 Anaphase I and II are both parts of the double cell division in meiosis. Anaphase I: the homologous pairs of chromosomes are separated (so the two daughter cells are haploid – each homologue consisting of two sister chromatids). Anaphase II: the sister chromatids of each chromosome are separated. Applying 8 Crossing two heterozygote plants should result in a Mendelian ratio in the F : 25% homozygous dominant, 50% heterozygous (phenotypically the same as the homozygous dominant plant) and 25% homozygous recessive. 9 Diploid: 28 3 2 5 56; Triploid: 28 3 3 5 84. 10 a In metaphase 1, the homologous pairs align in the middle of the cell. They then separate and move to opposite ends of the cell (Anaphase 1). Since the mule or hinny already has an unpaired chromosome, in the resulting cell division one chromosome would be unpaired. b Horse and donkey chromosomes don’t match up very well. To the extent that sex chromosomes are involved, the Y chromosome differences would produce different results in the male offspring. The mother’s physical size may also determine the size of the foetus. A horse supports a larger foetus. The preference could also be caused by animal behaviour. Female donkeys are much less likely to mate with a stallion than a mare with a male donkey; therefore, hybrids with donkey mothers are much rarer. In some species, for some genes, the sex that the allele originates from has an influence. This type of environmental influence on how genes are translated – in this case, the maternal environment – is called an epigenetic effect. Analysing 11 16 : 5 can be interpreted as approximately a 3 : 1 ratio (the Mendelian ratio) or a 2 : 1 ratio in which one of the possible phenotypes is missing (because the total numbers are small, there can be a large drift from the expected ratio – this is called a stochastic effect in genetics). Assuming the former, and that yellow is dominant to black, the Mendelian ratio is indicative of both parents being heterozygous (Cc). Evaluating 12 Responses will vary. 13 Responses will vary. Creating 14 Responses will vary. 15 Responses will vary. Reflecting 16 Responses will vary. 17 Responses will vary. © Cengage Learning Australia Pty Ltd 2014 SCNSW10AW00023 www.nelsonnet.com.au SCNSW10AW00023.indd 16 26/04/14 4:38 PM