Methods of Phonological Analysis

Date: 2015-10-07; view: 564.

List C

What does stabilizing selection mean?

What does disruptive selection mean?

a) selection against the middle form in two utmost forms favour

b) selection against the middle form in one utmost forms favour

c) selection against two utmost forms the in middle form favour

d) selection against two utmost forms and middle form

e) selection against one utmost forms and middle form

a) selection against the middle form in two utmost forms favour

b) selection against the middle form in one utmost forms favour

c) selection against two utmost forms the in middle form favour

d) selection against two utmost forms and middle form

e) selection against one utmost forms and middle form

1. Myoplegia inherit as dominant disease. What is the possibility of birth the child with myoplegia in the family, where mother is healthy, but father has myoplegia (heterozygous)?

2. In human being blue eye color is recessive to brown eye color. A brown-eyed man has a blue-eyed mother. What are the genotypes of this man and his mother? What are the possible genotypes of his father?

If this man get married a blue-eyed woman what are the possible genotypes of their offspring?

3. Tei-Sax disease inherit as recessive disease and, as rule, children die in 4-5 years. The first child died, when will be born the second child. What is the possibility, that the second child will be ill?

4. Glaucoma inherit of several ways: one of it^'s- dominant, others are recessive. Genes lie in different chromosomes.

What is the possibility of birth the child with glaucoma in family, where parents are diheterozygous?

What is the possibility of birth the child with glaucoma in family, where mother is diheterozygous and father has normal eyesight (homozygous for both pairs of the genes)?

5. In dogs, the barking trait is dominant over the silent trait and erect ears are dominant over drooping ears. What is the expected phenotypic ratio of the offspring when dogs heterozygous for both the traits are crossed?

6. In man brown eyes (B) are dominant to blue (b) and dark hair (R) dominant to red hair (r). A man with brown eyes and red hair, whose father was blue-eyed get married a woman with blue eyes and dark hair, whose mother was red-haired. They have four children. Give the genotypes and phenotypes of the parents and the children. What is the possibility of birth the child with blue eyes and red hair? penetance

7. In man polydaktilia, syndrome of short fingers and glaucoma are dominant characters. What is the possibility of birth the children without these characters in family, where parents are triheterozygous?

8. Retinoblastoma in children – dominant disease with penetance 90%. Mother in childhood have retinoblastoma, father is healthy. In family was born the child with retinoblastoma. What is the possibility of birth the second child with disease?

9. Podagra inherit as a dominant character. Penetance of the gene in man - 20%, in woman-0%. What is possibility be ill in podagra in family, where parents are heterozygous? What is possibility be ill in podagra in family, where mother is healthy and father ill in podagra (heterozygous)?

10. Rh-negative girl that has 0 group of blood and Rh-positive boy that has AB group of blood were married. The first their baby has B group of blood and Rh-negative. What is probability of birth children with A group of blood and Rh-negative?

11. Woman has myopia (dominant character), Rh-positive, and A group of blood. A child born by she has normal eyesight, Rh-negative, and 0 group of blood. Father has characters like mother's characters. What are the genotypes of mother and child? Define the genotype of the father.

12. It^'s necessary presence of two dominant genes D and E for normal hearing. Gene D determines the development of auditory receptors and gene E determines the development of auditory nervous. Parents are deaf, but their children have a normal hearing. What are the genotypes of all members of the family? What is the type of interaction of two dominant allelic genes?

13. Human skin color stipulate for dominant genes A and B, which interact for type of polygenic inheritance. People with genotype aabb have white color of skin. People with genotype AABB have black color of skin. Parents have one child with white color of skin and other child with black color of skin. Define the genotypes of parents? What are the possible genotypes of children?

14. Height of the people is controlled by several pairs of genes, which interact by type of polymeria (polygenic inheritance). In some population the shortest people have all recessive genes and height 150 sm. And the highest people have all dominant genes and height 180 sm. Define height of people, which are heterozygous for three pairs of genes.

15. Height of the people is controlled by several pairs of genes, which interact by type of polymeria (polygenic inheritance). In some population the shortest people have all recessive genes and height 150 sm. And the highest people have all dominant genes and height 180 sm. The woman with short height was married with man with middle height. In this family were born 4 children with height 150, 155, 160, 165 sm. Define the genotypes of parents and their height.

16. Mother and father have the 0 group of blood. Their daughter has B group of blood. Define the genotypes of parents and daughter. The phenomena of recessive epistasis take place.

17. Define length and mass of the gene, which code the protein with molecular mass 3200 Dalton.

18. How many aminoacids will have polypeptide, if molecula of mRNA has mass 20000 Dalton?

19. Gene has 25% of introns, it's molecular mass – 40000 Dalton. Define the quantity of aminoacids in polypeptide, which it codes?

20. Define length and mass of gene, if know, that it have 20% introns and it's polypeptide have mass 3600 Dalton.

21. The consecution of nucleotides in chain of DNA is AAAGGATTCCAACCTATCGTC. Build the consecutive of aminoacids in polypeptide, if know, that 7-11 and 15-17 nucleotides are introns.

22. Weight of polypeptide is 3000 Dalton. Weight of one region of chain DNA, in which the consecutive of aminoacids of its polypeptide code, is 93000. What part of introns is in this region of DNA?

23. Build the fragment of replicating DNA, if know, that chain contain the following nucleotides: AAAGTACCTTTCC.

24. Chain of DNA has the following nucleotides: CGTATCAAGTTTCGCCCAT. Build the consecutive of nucleotides in mRNA.

25. Molecule of DNA has 15% adenine nucleotides. What is percent of other nucleotides in it?

26. Hyperplasia of teeth enamel is inherited as dominant, X-linked character. Parents are ill, but son was born without illness. What are the genotypes of parents? What phenotype will have the second sun?

27. Hypertrichosis of ear is Y-linked character. What is possibility of birth the child with illness, if his father has this character?

28. Lack of sweat's glands is X-linked, recessive character. Albinism is recessive autosome linked character. Parents are healthy, but their son has two diseases. Indicate the possible genotypes of the parents. What is possibility of following birthing of sick and healthy children?

29.Woman has 0 group of blood and normal vision. She was married to man with color blindness (daltonism) and A group of blood. What must be the genotypes of parents, if the child has daltonism and 0 group of blood?

30. The distance between genes A and B within a chromosome is equal 10 M (Morgan), the distance between genes A and C is equal 3 M, the distance between genes B and C is equal 7 M, the distance between genes D and B is equal 2 M, and the distance between genes D and C is equal 5 M. Build the chromosome map.

31. The genes A and B are localized on distance of 20 M. Define genotypes of F1 and their probability, if parents are heterozygotes.

7. Genes of eliptocytosis and Rh factor are localized in the first chromosome of human being on distance of 3 M. What gametes are formed by woman, and what is probability of their forming, if genes of eliptocytosis and Rh + factor were inherited from her mother. Her father is healthy and has Rh – factor.

32. The genes of color blindness and hemophilia are X-linked. The distance between these genes is equal 9,8 M (Morgan). The mother of wife has the both of pathological traits. But the father of wife is healthy. The husband is healthy too. What is probability of birthing of healthy and sick children?

33. Genes of cataract and polydactylia are autosome and dominant. They are complete linked each other. Wife inherited gene of polydactylia from mother and gene of cataract – from father. Her husband is healthy. What is probability of birthing of healthy and sick children?

34. The concordance of monozygotic twins by parotitis is 82%, and the concordance of dizygotic twins is 74%. Determine the role of hereditary and environmental factors in the genesis of the disease.

35. The concordance of monozygotic twins by epilepsy is 67%, and the concordance of dizygotic twins is 3%. Determine the role of heredity and environmental in the genesis of the disease.

36. The concordance of monozygotic twins by pancreatic (insular) diabetes (diabetes mellitus) is 65%, and the concordance of dizygotic twins is 18%. Determine the role of heredity and environmental in the genesis of the disease.

37. Among 120 pairs of monozygotic twins of a population 48 pairs are concordant by congenital femoral dislocation and 72 pairs are discordant. Determine the role of heredity and environmental in the genesis of the disease.

38. The fourth nucleotide in normal hemoglobin (peptide) has the sequence of amino acids val-gys-ley-tre-pro-glu-lis. Define the character of mutation, if the abnormal peptide is val-gys-ley-tre-pro-lis-glu-lis.

39. Build the pedigree. Proband is woman having schizophrenia. Proband's father, brother and sister are healthy. Relatives of proband's father are following: brother has schizophrenia and two sisters are healthy. Father and mother of proband's father (proband's grandfather and grandmother) are healthy, but sister of grandmother is sick. Proband's mother of, grandfather and grandmother from mother's side are healthy. Analyze the family tree to define the type of inheritance, the genotypes of members of family tree, and prognosis of birthing of sick children.

40. Build the pedigree. Proband is man having cataracta. His mother and mother's mother (proband's grandfather) have the cataracta too. Uncle and aunt from of proband from mother's side are healthy. Proband's father, aunt from side of father, grandmother and grandfather from side are healthy. Wife of proband, her sisters, and two her brothers, and her parents are healthy. Proband has two children: son is sick and daughter is healthy. Analyze the family tree to define the type of inheritance, the genotypes of members of family tree, and prognosis of birthing of sick children.

41. Build the pedigree. Proband is man having Duchenne Muscular Dystrophy. His mother, mother's mother (proband's grandmother), and grandfather are healthy. Mother's sister is healthy too, but mother's brother is sick. Proband's father, aunt from side of father, grandmother and grandfather are healthy. Analyze the family tree to define the type of inheritance, the genotypes of members of family tree, and prognosis of birthing of sick children.

42. Build the pedigree. Proband is healthy man. His sister, father and grandmother have vitamin – D – firm rickets. Father's sister has rickets too. Proband's mother, two her sisters, mother (proband's grandmother), and father (proband's grandfather) are healthy. Analyze the family tree to define the type of inheritance, the genotypes of members of family tree, and prognosis of birthing of sick children.

43. Build the pedigree. Proband is man having hypertrichosis of auricle. His brother, father, uncle (father's brother) and grandfather have hypertrichosis too. Proband's sister, mother, aunt (mother's sister), grandmother (mother's mother) and grandfather (father's father) are healthy. Analyze the family tree to define the type of inheritance, the genotypes of members of family tree, and prognosis of birthing of sick children.

44. The frequency of albinos birth is 1:20 000. Define the frequency of albino recessive gene and frequency of heterozygotes within population.

2. A population has 84 000 persons, and 21 persons within the population have infantile type of cerebral sphingolipidosis (Tay-Sachs disease). Define the frequency of the cerebral sphingolipidosis recessive gene and frequency of heterozygotes within population.

45. A frequency of autosomal recessive gene of congenital amaurotic idiocy (Norman-Wood syndrome) is 0,009. Define the quantity of sick people within population, if general quantity of inhabitants is 100000 persons.

46. A frequency of autosomal dominant gene of tuberous sclerosis is 0,7%. Define the frequency of sick people within population.

47. The quantity of the sick people having Rieger Syndrome in the population is 5. The quantity of inhabitants in the population is 1000000 persons. Define the frequency of autosomal dominant gene.

1.5. SUMMARIZING of the STUDY by the teacher and valuation of performance of works by each student.

1.6. PLACE and TIME of STUDY: an educational room, 2 academic hours.

1.7. EQUIPMENT of STUDY: microscopes, microslides, tables, schemes.

1.8. Literary Sources:

1. Medical Biology: Textbook. - Simferopol: IAD CSMU, 2002. – P. 111 – 237.

The aim of phonological analysis is to establish distinctive difference between sounds, create the inventory of the phonemes and describe the phonemic system of a language. In other words, the aim of phonological analysis is the identification of the phonemes and their classification.

When I say that we must identify the sound, I mean, we must decide what phoneme it belongs to.

There are two main approaches to phonological analysis. Formally distributional approach practised by American structuralists is focused on the position of the sound in the word, or its distribution. Semantic method attaches special importance to meaning. It is widely used in this country. Let's focus on the semantic method.

The analysis is performed through the system of phonological oppositions. It is based on the following fundamental phonological rule: phonemes can distinguish the meaning when opposed to one another in the same phonetic context ( day – they, sheep – ship). So to establish the phonemic status of a sound it is necessary to oppose one sound to some other sound in the same phonetic context. This procedure is called commutation test.

To conduct this test we must find the so-called minimal pairs. A minimal pair is a pair of words which differ in one sound only. So we replace one sound by another sound and try to find out if the opposed sounds belong to the same or different phonemes.

Now, the commutation test may have three possible results:

Pin – sin | the meaning is different so the opposed sounds belong to different phonemes.

P(h)in – pin | the meaning is the same so the opposed sounds belong to the same phoneme.

Pin – hin | we have a meaningless word, so we can't make any conclusion about the phonemic status of the second sound, we can't identify it.

It should be noted that there are different types of oppositions.

1) If the members of the opposition differ in one articulatory feature the opposition is called single.

Pen – ben | [p] is fortis (voiceless), [b] is lenis (voiced).

2) If there are two distinctive features, the opposition is double.

Pen – den | [p] is labial, fortis, [b] is forelingual, lenis.

3) If three distinctive features are marked, the opposition is triple (multiple).

Pen –then | the differentiating features: occlusive – constrictive, labial – interdental, fortis – lenis.

To establish the system of phonemes of a language it is necessary to oppose sounds in all possible positions (initial, medial, final). But there are cases when the sounds can't be used in the same position and can't be opposed. For example [h] is never used in final position, [n ] is never used in the initial position. These sounds are treated as different phonemes on the basis of native speakers' knowledge and their phonetic dissimilarity which illustrates that they cannot be allophones of one phoneme.

There is another interesting case which is analysed and explained by different schools of classical phonology. In some cases different sounds occur in the same position and in the same phonetic context, but the meaning of the word remains unchanged (калоши –галоши, шкаф – шкап). They are called “free variants”.

Despite these difficulties the semantic method of phonological analysis is widely used and fulfills the task of systematizing the sounds of a language.

The application of this method shows that the English language has 24 consonant and 20 vowel phonemes. As it was mentioned sounds are grouped into classes according to the distinctive (or phonemic) features. In English the following features are distinctive for consonants:

- place of articulation (labial, lingual, glottal)

- type of obstruction (manner of articulation)

- force of articulation (fortis, lenis)

The phonemic features of English vowels are:

- quality (height and front-back position; stability of articulation).