BRAIN TEASERS by VKQ2ko

VIEWS: 44 PAGES: 12

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                                         BRAIN TEASERS


1. Several black guinea pigs of the same genotype were mated and they produced 29 black and 9 white
   offsprings. What would be the genotypes of the parents?

2. If a black female guinea pig is testcrossed and produces at least one white offspring, determine the
   genotype and phenotype of the male and parent?

3. In foxes, silver black coat colour is governed by a recessive allele b and red colour by its dominant
   allele B. Determine the genotypic and phenotypic ratios expected from the following matings: (a)
   pure red and carrier red, (b) carrier red and silver black, (c) pure red and silver black.

4. The shape of radishes may be long (LL), round (RR) or oval (LR). If long radishes are crossed to oval
   radishes and the F1 then allowed to cross at random among themselves, what phenotypic ratio is
   expected in the F2 generation?

5. In Mirabilis ( four o’clock), a plant hybrid for red (R) and white flower (r) is pink (Rr). A plant with
   pink flowers is crossed with (i) one having red flowers and (ii) one having white flowers. Give the
   phenotypic ratios expected in its progenies.

6. In man, brown eyes (B) are dominant to blue (b) and dark hairs (R) to red hairs (r). A man with brown
   eyes and red hairs marries a woman with blue eyes and dark hair. They have two children, one with
   brown eyes and red hairs and the other with blue eyes and dark hairs. Give the genotypes of the
   parents and children.

7. In some plants, such as corn, there is a recessive gene on one chromosome which when homozygous,
   produces white plant colour, and also a dominant gene on another chromosome which produces white
   whether homozygous or heterozygous. What proportion of white plants will occur among the
   offspring of a self fertilized plant that is heterozygous for both of these genes?

8. In garden peas the tall allele (T) is dominant over dwarf allele (t), and round-seeded allele (S) is
   dominant over that of wrinkled (s). These two gene pairs are also known to assort independently of
   each other.
   (a) What proportions of phenotypes are expected among the progeny of tall, round seeded F1 plants
   crossed to each other if each such F1 parent is derived from a cross between a pure breeding tall,
   round seeded variety (TTSS) and a dwarf, wrinkled seeded variety (ttss)?
   (b) Would the proportions of phenotypes in the F2 generation be changed if the F1 plants were
   derived from a cross between a tall, wrinkled seeded variety (TTss) and a dwarf, round seeded variety
   (ttSS)?
   (c) What phenotypic results would you expect if the F1 plants in (a) were crossed to a dwarf, wrinkled
   seeded plant?

9. If two gene pairs A and a and B and b are assorting independently, the capital letters being dominant,
   what is the probability of obtaining:
   (a) An AB gamete from an AaBb individual?
   (b) An AB gamete from an AABB individual?
   (c) An AABB zygote from a cross AaBb × AaBb?
   (d) An AABB zygote from a cross aabb × AABB?
   (e) An AB phenotype from a cross AaBb × AaBb?
   (f) An aB phenotype from a cross AaBb × AaBB?

10. Two pair of alleles govern the colour of onion bulbs. A pure-red strain crossed to a pure white strain
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    produces an all-red F1. The F2 was found to consist of 47 white ,38 yellow, and 109 red bulbs.
    (a) what epistatic ratio is approximated by the data?
    (b) what is the name of this type gene interaction?
    (c) if another F2 is produced by the same kind of cross, and eight bulbs of the F2 are found to be of
    the double- recessive genotype, how many bulbs would be excepted in each phenotypic class ?

11. The colour of corn of aleurone is know to be controlled by several genes; A, C, and R are all
    necessary for colour to be produced. The locus of a dominant inhibitor of aleurone colour, I, is very
    closely linked to that of C. Thus, any one or more of the genotype I-, aa-, or rr- produces colorless
    aleurone.
    (a) What would be coloured : colourless ratio among F2 progeny from the cross AAIICCRR ×
    aaiiCCRR?
    (b) What proportion of the colourless F2 is expected to be homozygous ?

12. In tomato, genotypes aabbcc produces 100g tomatoes, and AABBCC produces 160g tomatoes, each
    dominant gene causing an increase of 10g. Give the weight of tomatoes in the parents and progenies
    in the following crosses: (i) AAbbcc × aaBBcc (ii) AaBbCc × aaBBCc.

13. In a given plant, long leaves (S) are dominant over short leaves (s) and green veins (V) are dominant
    over yellow veins (y). The cross SSYY × ssyy produces an F1 SsYy. Assuming that S and Y are
    linked, if 760 individuals are produced in F2 generation, work out their phenotypic ratio?

14. Two dominant mutants in the first linkage group of guinea pigs govern the traits pollex (Px), which is
    the atavistic return of thumb and little toe, and rough fur (R). When dihybrid pollex, rough pigs were
    crossed to normal wild type pigs, their progeny fall into four phenotypes: 79 rough, 103 normal, 95
    rough, pollex and 75 pollex. (a) Determine the genotypes of the parents. (b) Calculate the amount of
    recombination between Px and R.

15. A woman of blood group A marries a man of blood group O. There are three children in the family
    having blood group A, AB and O. Which child was definitely adopted?

16. A woman whose father was of blood group AB and whose mother was A, marries a man of blood
    group B. Of their two children one has blood group O and the other has A. (a) What is the genotype
    of the woman? (b) What is the genotype of her husband? (c) What is the genotype of her mother?

17. What phenotypes and ratios would you expect among the offspring of the following crosses: (a) IAIA
    × ii (b) IAIA × IAIB (c) IAIA × IBi (d) IAIA × IA i ?

18. In case of a disputed parentage the blood group of mother is A and her child’s is B. List all the
    possible blood groups which the father may possess.

19. A woman of A blood group and normal vision has five children as follows: (a) male, A blood group,
    colourblind. (b) male, O blood group colourblind. (c) female, A blood group colourblind. (d) female,
    B blood group, normal colour vision (e) female, A blood group, normal colour vision. Of the two men
    that have mated with this women at different times, no.1 had AB blood group and was colourblind,
    and no.2 had A blood group with normal colour vision. Which of these men is the most probable
    father in each case?

20. A sex linked recessive gene c produces red-green colour blindness in humans. A normal woman
    whose father was colour blind marries a colour blind man.
    (a) What genotypes are possible for the mother of the colour blind man?
    (b) What are the chances that the first child from this marriage will be a colour blind boy?
    (c) Of all the girls produced by these parents, what percentage is expected to be colour blind?
    (d) Of all the children from these parents, what proportion is expected to be normal ?
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21. Sex determination in the grasshopper is by XO method. The somatic cells of a grasshopper are
    analyzed and found to contain 23 chromosomes. (a) What sex is this individual? (b) Determine the
    frequency with which different types of gametes can be formed in this individual. (c) What is the
    diploid number of the opposite sex?

22. In Drosophila, the ratio between the number of X-chromosomes and the number of sets of autosomes
    is called the sex index. Calculate the sex index and the sex phenotype in the following individuals: (a)
    AAX (b) AAXXY (c) AAAXX (d) AAXXX (e) AAAXXX (f) AAY.

23. If the diploid number of honey bee is 16. (a) How many chromosomes will be found in the somatic
    cells of the drone? (b) How many bivalents will be seen during the process of gametogenesis in the
    male? (c) How many bivalents will be seen during the process of gametogenesis in the female ?

24. A dominant sex limited gene is known to affect premature baldness in men but is without effect in
    women. (a) What proportion of the male offspring from both heterozygous parents, is expected to be
    bald prematurely? (b) What proportion of all their children is expected to be prematurely bald ?

25. A holandric gene is known in humans that causes long hair to grow on the external ears. When men
    with hairy marry normal women, (a) what percentage of their sons would be expected to have hairy
    ears, (b) what proportion of the daughters is expected to show the trait? (c) what ratio of hairy eared :
    normal children is expected?

26. In fowl, barring (B) is sex linked and dominant, the recessive allele (b) producing solid black colour
    when homozygous. Silky feathers (s) is a recessive autosomal allele, as opposed to non-silky (S). If
    black cocks, heterozygous for silky, are crossed to barred, silky hens, what genotypes and phenotypes
    will be produced, including their ratio? Hint: In birds females are XY and males are XX.

27. A married couple, both of whom had normal vision, produced a colourblind son. Examination of cell
    samples from the son showed the presence of Barr bodies. (a) What is the genotype of the son? (b)
    What is the cause for this genotype?

28. Suppose that a female undergoes sex reversal to become a functional male and is then mated to a
    normal female. Determine the expected F1 sex ratios from such matings in species with (a) ZW
    method of sex determination (b) XY method of sex determination.

29. Normal women possess two sex chromosomes (XX) and normal men have a single X chromosome
    plus a Y chromosome that carries male determiners. Rarely, a woman is found with marked
    abnormalities of primary and secondary sexual characteristics, and she has only one X chromosome
    (XO), Turner’s syndrome. Likewise men are occasionally discovered with an XXY constitution
    called as Klinefelter’s syndrome. Colour blindness is a sex linked recessive trait.
    (a) A husband and wife both have normal vision, but one of their children is a colour blind Turner
    girl. Diagram this cross, including the gametes that produced this child.
    (b) In another family the mother is colour blind and the father has normal vision. Their child is a
    Klinefelter with normal vision. What gametes produced this child?
    (c) Suppose the same parents in part (b) produced a colour blind Klinefelter, what gametes produced
    this child?
    (d) The normal diploid number for humans is 46. A trisomic condition for 21 results in Down’s
    syndrome. At least one case of Down-Klinefelter has been recorded. How many chromosomes would
    this individual be expected to possess?

30. Assume the eye colour in humans is controlled by a single pair of genes of which the effect of that for
    brown (B) is dominant over the effect of that for blue (bb). (a) What is the genotype of a brown-eyed
    individual who marries a blue eyed individual and produces a first offspring that is blue-eyed? (b) For
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    the same mating as in (a), what proportions of the two eye colours are expected among further
    offsprings? (c) What are the expected proportions of eye colours among the offspring of a mating
    between the two brown-eyed individuals who each had one parent that was blue-eyed?

31. Colourblindness is a recessive sex-linked disorder, mostly found in humans. (a) can two colourblind
    parents produce a normal son? (b) can two colourblind parents produce a normal daughter? (c) can
    two normal parents produce a colourblind son? (d) can two normal parents produce a colourblind
    daughter? (e) can a normal daughter have a colorblind father? (f) can a colourblind daughter have a
    normal father?

32. A colourblind woman marries a normal visioned man. They have two children, a boy and a girl. (a)
    What will be the genotype and phenotype of the boy? (b) What will be the genotype and phenotype of
    the girl ?

33. A normal woman whose father was colourblind marries a colourblind man. They produce a son and a
    daughter. (a) What is the probability that the son is colourblind? (b) What is the probability that the
    daughter is colourblind?

34. A normal woman whose mother was colourblind has a son. Nothing is known of the colour-vision
    phenotype of the father. What is the probability that the son will be colourblind?

35. If an individual with Klinefelter’s syndrome (XXY) was found to be fertile in a mating with a normal
    female, what proportions of his sons would be expected to have this syndrome?

36. According to the sex-determination mechanism proposed by Bridges, what are the sexual phenotypes
    of the following tetraploids Drosophila: (a) XXYY (b) XXXY (c) XXY (d) XXXXY ?

37. The diploid number of an organism is 12. How many chromosomes would be expected in the
    following: (a) a monosomic (b) a trisomic (c) a tetrasomic (d) a double trisomic (e) a nullisomic (f) a
    monoploid (g) a triploid (h) an autotetraploid ?

38. The European raspberry (Rubus idaeus) has 14 chromosomes. The dewberry (R. caesius) is a
    tetraploid with 28 chromosomes. Hybrids between these two species are sterile F1 individuals. Some
    unreduced gametes of the F1 are functional in back crosses. Determine the chromosome number and
    level of ploidy for each of the following: (a) F1 (b) F1 backcrossed to R. idaeus. (c) F1 backcrossed to
    R. caesius. (d) chromosome doubling of F1 ?

39. Assume a chromosome with the following gene sequence, the dot (●) represents centromere.
    ABCD●EFGH. Identify the aberrations for the following sequence. (a) ABCD●EFH (b)
    ADCB●EFGH (c) ABCDCD●EFGH.

40. If the mutation rate of a certain gene is directly proportional to the radiation dosage and the mutation
    rate of Drosophila is observed to increase from 3% at 1000R to 6% at 2000R, what percentage of
    mutations would be expected at 3500R?

41. Down’s syndrome occurs in humans when a particular chromosome is present in triplicate instead of
    in the usual diploid state. Such individuals have 47 chromosomes instead of the normal 46. What
    proportion of offspring produced by an affected mother with 47 chromosomes mated to a normal man
    would be similarly affected?

42. A single stranded DNA phage has an A/T base ratio of 0.33, a G/C ratio of 2.0 and an A+T/G+C ratio
    of 1.33. (a) What is the A+G/T+C ratio in this molecule? (b) If this single stranded molecule forms a
    complementary strand, what are these four ratios in the complementary strand? (c) What are these
    four ratios in both the original and complementary strands together?
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43. Given a single strand of DNA- 3' TACCGAGTACTGACT 5' . Construct (a) the complementary
    DNA chain, (b) the mRNA chain that would be made from the given strand, (c) Which strand is the
    sense strand?

44. Meiotic non-disjunction of the sex chromosomes in either parent can produce a child with Klinefelter
    syndrome (XXY) or Turner syndrome (XO). (a) If a colourblind woman and man with normal vision
    produce a colourblind Klinefelter child, in which parent did the non-disjunction even occur?

45. Given the frequency of gene A as 0.2, gene B as 0.6, find the equilibrium frequencies of the gametes
    AB, Ab, aB and ab.

46. The gene frequency of a sex-linked allele in human population is 0.4 in females and 0.8 in males.
    Find the frequency in the entire population.

47. Baldness is a sex-influenced trait that is dominant in males and recessive in females. In a sample of
    10,000 men, 7225 were found to be normal. In a sample of same size, how many females are
    expected to be normal?

48. Haemophilia is a sex-linked recessive disease in humans. In a local population, survey was done on
    500 men and 20 were found to be haemophilic. What is the gene frequency of normal allele in the
    population?

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                                                   ANSWERS

1. Bb × Bb                                                17. (a) All IAi; (b) 50% IAIA, 50% IAIB; (c) 50%
2. bb and Bb                                                  IAIB, 50% IAi; (d) 50% IAIA, 50% IAi
3. (a). ½ BB, ½ Bb, all red; (b). ½ Bb, red: ½            18. AB or B
    bb, silver black; (c). all Bb, red.                   19. man of AB blood group is the father of all
4. 9 long, 6 oval, 1 round.                                   three females; man of blood group A is the
5. (i) 1 red : 1 pink (ii) 1 pink : 1 white                   father of both the males.
6. Father. Bbrr; Mother. bbRr; Children. Bbrr             20. (a) Cc or cc (b) 1/4 (c) 50% (d) 1/2
    and bbRr                                              21. (a) Male (b) 1/2 (11A + 1X) : 1/2 (11A) (c)
7. All white                                                  24
8. (a) 9 : 3 : 3 : 3 : 1; (b) no change; (c) 1 : 1 : 1    22. (a) 0.5 male (b) 1.0 female (c) 0.67 intersex
    : 1, test cross                                           (d) 1.5 superfemale (e) 1.0 female (f) lethal
9. (a) 25%; (b) 100%; (c) 1/16; (d) 0%; (e)               23. (a) 8 (b) None, meiosis cannot occur in
    9/16; (f) 4/16                                            haploid males (c) 8
10. (a) 9:3:4 (b) Recessive epistasis (c) 32 white        24. (a) 3/4 (b) 3/8
    : 24 yellow : 72 red                                  25. (a) 100% (b) None (c) 1 hairy : 1 normal
11. (a) 13 colourless : 3 coloured (b) 3/13               26. Males. 50% BbSs, barred non-silky, 50%
12. (i) Parents 120g each, progenies 120g. (ii)               Bbss, barred silky; Females. 50% Ssb, black
    Parents 130g each, progenies 130g                         nonsilky, 50% ssb, black silky.
13. 570 long, green and 190 short, yellow                 27. (a) XCXCY; (b) Mother heterozygous for
14. (a) PxR, pollex, rough × pxr normal; (b) 79               colourblindness; non-disjunction of X-
    rough and 75 pollex types are recombinants,               chromosomes during meiosis and its
    constituting 154 out of 352 individuals, i.e.             fertilization with a Y bearing sperm resulted
    43.8% recombination                                       into this type of son.
15. Child AB                                              28. (a) 2 females : 1 male; (b) All females
16. (a) IAi; (b) IBi; (c) IAi
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29. (a) P: XCXc × XCY; F1: XCO (b) XC and                   40. 10.5%
    XCY (c) Xc Xc and Y (d) 48                              41. 50% sons and 50% daughters will be
30. (a) XbY and XBXb; (b) 1 : 1; (c) 3 brown : 1                affected
    blue                                                    42. (a) 0.75; (b) 3.03, 0.5, 1.33, 1.33; (c) 1.0,
31. (a) No; (b) No; (c) Yes, if the mother is a                 1.0, 1.33, 1.0
    carrier; (d) No; (e) Yes; (f) No                        43. (a) 5' ATGGCTCATGACTGA 3' ; (b) 5'
32. (a) XCY, colourblind; (b) XCX, normal but                   AUGGCUCAUGACUGA 3' ; (c) the strand
    carrier                                                     given in question will be the sense strand as
33. (a) 50%; (b) 50%                                            it can code for AUG. The complementary
34. 50%                                                         strand cannot code for it and therefore it will
35. 0%                                                          be the antisense strand.
36. (a) male; (b) intersex; (c) male; (d) female            44. Mother.
37. (a) 11 (b) 13 (c) 14 (d) 14 (e) 10 (f) 6 (g) 18         45. AB = 0.12, Ab = 0.16, aB = 0.48, ab = 0.32
    (h) 24                                                  46. 0.6
38. (a) 21, triploid (b) 28, tetraploid (c) 35,             47. 9075
    pentaploid (d) 42, hexaploid                            48. 0.96
39. (a) Deletion; (b) Inversion; (c) Duplication




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                                              EXPLANATION

1. 29 black and 9 white means 3 : 1 ratio (genotypic) approximately. The phenotypic ratio must be 1
(homozygous black) : 2 (heterozygous black) : 1 (homozygous white).
                                          B       b
                                      B BB Bb
                                      b Bb bb

2. The black female is test crossed; i.e. it is crossed with bb. There are two possibilities of the genotype of
female BB or Bb.
                        B       B                                       B       b
                    b   Bb      Bb                                  b   Bb      bb
                    b   Bb      Bb                                  b   Bb      bb
                          (i)                                            (ii)
In cross (i) above, no white offspring is produced, therefore the genotype of the female is Bb, i.e. the (ii)
cross; and bb that of male.

3.           B          B                             B    b                             B    B
           B BB         BB                      b     Bb   bb                        b   Bb   Bb
           b Bb         Bb                      b     Bb   bb                        b   Bb   Bb

(a) ½ BB, ½ Bb, all red         (b) ½ Bb, red: ½ bb, silver black         (c) all Bb, red

4. It is a clear cut case of duplicate genes with cumulative effect in which both the dominant non allelic
alleles, when present together, give a new phenotype, but when allowed to express independently, they
give their own phenotypic expression separately, giving the ratio 9 : 6 : 1.
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5. Mirabilis shows incomplete dominance.
                          R         r                                    R      r
                        R RR        Rr                              r    Rr     rr
                        R RR        Rr                              r    Rr     rr
                       (i) 1 red : 1 pink                         (ii) 1 pink : 1 white


6.                                               Br                 br

                                                BbRr              bbRr
                                  bR
                                            (brown, dark)     (blue, dark)

                                                Bbrr             Bbrr
                                  br
                                            (brown, red)     (brown, dark)


7. Recessive gene on one chromosome produces white (ww; homozygous).
Dominant gene on another chromosome produces white (AA; homozygous or Aa; heterozygous).
Heterozygous F1 for both the alleles will be Aw (white).
                                             A     w
                                         A AA Aw
                                         w Aw ww
The offspring will be 1 AA (white) : 2 Aw (white) : 1 ww (white)

8. (a) The F1 is TtSs obtained by crossing TTSS and ttss. When these F1 are selfed, they will show the
normal Mendelian ratio as 9 : 3 : 3 : 1.
(b) If the P generation is changed then also the F1 obtained will be TtSs. So the F2 ratio will not change.
(c) If F1 is crossed to recessive parents (dwarf, wrinkled), i.e. test cross, the ratio will be 1 : 1 : 1 : 1.

9. (a) Gametes of AaBb are AB, Ab, aB and ab. AB is 1 out of 4, i.e. 25%.
(b) Gametes of AABB are all AB, i.e. 100%.
(c) AABB zygote is homozygous dominant and we know that in a F2 ratio, pure dominant and pure
recessive are formed singly, i.e. 1/16.
(d) AABB zygote canot be formed because the other parent (aabb) does not have any dominant allele, so
the answer is 0%.
(e) AB phenotype means dominant phenotype. In the ratio 9 : 3 : 3 : 1, the number 9 shows the dominant
phenotype, i.e. 9/16.
(f) In the ratio 9 : 3 : 3 : 1, the last numbers 3 and 1 shows the aB phenotype, i.e. 4/16.

10. (a) 109 : 38 : 47 is approximately 9 : 3 : 4. (b) This ratio depicts recessive epistasis. (c) Eight bulbs are
double recessive, so by multiplying the 9 : 3 : 4 by eight we get the answer as 72 (red) : 32 (white) : 24
(yellow).

11. Here the dominant allele, either in homozygous or heterozygous condition, of one gene and the
homozygous recessive allele of other gene produces the same phenotype. This is the case of dominant
recessive interaction and the ratio comes to be 13 : 3, out of which only 3 will be homozygous colourless.

12. (i) Weight of AAbbcc is 100 + (10A +10A) = 120g. Weight of aaBBcc is 100 + (10B +10B) = 120g.
The genotype of the progeny will be AaBbcc; its weight is 100 + (10A +10B) = 120g.
(ii) Weight of AaBbCc is 100 + (10A +10B + 10C) = 130g. Weight of aaBBCc is 100 + (20B +10C) =
130g. The genotype of the progeny will be AaBbCc; its weight is 100 + (10A +10B + 10C) = 130g.
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13. S and Y are linked so they will probably give the ratio of 3 : 1 in F2 generation. Therefore out of 760,
570 will be long, green and 190 will be short, yellow.

14. (a) One parent is dihybrid pollex (Px), rough (R), means its genotype is PxR. The other is wild i.e. px
and normal i.e. r, so its genotype is pxr. (b) 79 rough (R) and 75 pollex types (Px) are recombinants,
constituting 154 (79 + 75) out of 352 (79 + 75 + 103 + 75) individuals, i.e. 154/352 x 100 = 43.8%
recombination.

15.
                                IA     IO                                          IA    IA
                         I O IA IO    IA IO                                 IO IA IO    IA IA
                           O    A O                                           O
                         I     I I    IA IO                                 I      A O
                                                                                  I I   IA IA
                        (i) Hetreozygous A                                 (ii) Homozygous A

Woman of blood group A may be homozygous or heterozygous. Both the possibilities have been shown
here. But none of the case is found to be of blood group AB of the child. So, AB child is the adopted one.

16. (a)
                              IA      IB                                           IA      IB
                        A          A AA B                                   A         A A  A B
                       I     I I     I I                                    I     I I     I I
                         O    A O                                             A
                       I     I I     IB IO                                  I      A A
                                                                                  I I     IA IB
                       (i) Heterozygous A                                  (ii) Heterozygous A

The mother of the woman may be homozygous or heterozygous. Both the cases are solved above. The
woman may be of blood group AB or A. When she marries a man with B blood group and children have
blood group O and A, then it is sure that woman is of A blood group. Now the question arises
homozygous A or heterozygous A. This is solved below.

                                    IA           IO                                    IA        IA
                        B          A B          B O                         B         A B       A B
                       I           I I          I I                        I          I I       I I
                       IO          IA IO        IO IO                      IO         IA IO     IA IO

                       (iii) Heterozygous A                                (iv) Homozygous A

In the second case there is no possibility of the O group child. So the woman is heterozygous A, i.e. IA IO .
(b) The genotype of her husband is IB IO as shown in (iii) above. (c) Since it is confirmed that woman is
IAIO , the genotype of her mother is also IAIO as shown in (i) above.

17.
                 IA          IA                    IA          IA            IA           IA              IA       IA
                                           A                           B                           A
          i     IA i        IA i           I       A A
                                                  I I          A A
                                                              I I     I     A B
                                                                           I I           A B
                                                                                        I I       I      A A
                                                                                                        I I       A A
                                                                                                                 I I
          i     IA i        IA i           IB     IA IB       IA IB    i    IA i         IA i      i     IA i     IO i
                 (a)                                    (b)                     (c)                       (d)

18. Four possibilities arises, either the father is of A, B, AB or O blood group. All of these are illustrated
below.
                IA         IO                      IA          IO           IA           IO               IA      IO
          IA   IA IA     I A IO            IB     IA IB       IB IO   IA   IA IA        IB IO     IO    I A IO   IO IO
          IO   IA IO     I O IO            IO     IA IO       IO IO   IB   IA IB        IB IO     IO    I A IO   IO IO
                (a)                      (b)                 (c)                   (d)
In only (b) and (c), the child is having B blood group, i.e. the father belongs to either B or AB blood
group.
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19. All the sons are colourblind, this suggests that mother is carrier; father never transmits sex-linked
disease to his son. Sons are of blood group A and O, therefore their father must be of A blood group. The
father of all the daughters is of AB blood group.
The colourblind female is produced due to one allele passed on to her from father and one from the
mother. Rest of the females are normal as they received only one allele of colourblindness from their
father.

20. (a) If the woman is normal and her father is colourblind, then her mother could either be normal (cc)
or a carrier (Cc), but not colourblind (CC).
 (b) If the woman father is colourblind, the woman must be recessive to colourblindness.
                                                   XC          X
                                            C
                                           X      XCXC        XC X
                                           Y      XC Y        XY
We know that the chance of boy or a girl is 50%, and as shown above, the chances of a colourblind child
is 50%, so the chances of a colourblind boy is 25% or 1/4 .
(c) 50% girls are normal but carrier and 50% girls are colourblind.
(d) Out of the four children shown above, 50% are normal and 50% are colourblind.

21. (a) In XO method of sex determination, the number of chromosomes in males (XO) will be always in
odd number and even number in females (XX). (b) Out of 23 chromosomes, 22 are autosomes and 1 X-
chromosome. The gametes will be 11A + X and 11A. (c) The chromosome number of female will be 24
(22 autosomes + 2 X chromosomes).

22. (a) X/A = 1/2 = 0.5, i.e. diploid male; (b) X/A = 2/2 = 1.0, i.e. diploid female; (c) X/A = 2/3 = 0.67,
i.e. intersex; (d) X/A = 3/2 = 1.5, i.e. superfemale; (e) X/A = 3/3 = 1.0, i.e. triploid female; (f) X
chromosome is absent, lethal effect; this Drosophila will not survive.

23. (a) Drones are formed without fertilization and are thus always haploid, their chromosome number
will be 8. (b) Gametogenesis means meiosis. Since the males are haploid, meiosis does not occurs in
haploid cells. (c) In females the bivalents will be half the chromosome number, i.e. 8.

24. (a)
                                       ♂         B       b
                                       B         BB      Bb
                                       b         Bb      bb
BB and Bb phenotype in males causes premature baldness, therefore 3/4 males will be prematurely bald.
                                       ♀         B       b
                                       B         BB      Bb
                                       b         Bb      bb
Only BB female will be bald. Therefore out of total 8 individuals, 3 males and one female will be
prematurely bald, i.e. 4/8.

25. (a) Hypertrichosis is transferred by the Y-chromosome, so all the sons will inherit the disease. (b)
None of the daughters will be affected because Y-chromosome is absent in females. (c) All females
normal : all males hairy, i.e. 1 : 1.

26.                                                X bS   X bs
                                            Bs     Bs bS
                                           X      X X    XBsXbs
                                            Y      XbSY   XbsY
27. Presence of Barr body in son clearly indicates XXY, i.e. Klinefelter’s syndrome caused by non-
disjunction at the time of meiosis. Refer to the chapter of cell cycle.
                                                                                                      10



28. (a) The functional male derived through sex reversal will still remain genetically female (ZW); it is
crossed to a normal female (ZW), as follows. Furthermore, at least one sex chromosome (Z) is essential
for life.
                                                 Z        W
                                          Z     ZZ       ZW
                                          W    ZW       WW
The ratio obtained is 2 females (ZW) and one male (ZW). WW will not survive.
(b) The functional male derived through sex reversal will still remain genetically female (XX); it is
crossed to a normal female (XX), as follows.


                                                    X      X
                                          X        XX      XX
                                          X        XX      XX
The ratio obtained is all females (XX).

29. (a)          XCX             x                 XY



                 XC O               X (The chromosome from the father is unable to combine with XC of
mother, i.e. non-disjunction, producing a Turner colourblind girl).

(b)                            XCXC            x
                                                                XY


                                 XCXY (an example of non-disjunction)

(c)                              XY                x
                                                             XCXC


                                 XCXCY (an example of non-disjunction)

(d) 46 + 1 (due to extra 21st chromosome) + 1 (due to extra X-chromosome) = 48.

30. (a) Mother is heterozygous brown eyed and father is blue eyed.
                                                    XB     Xb
                                          Xb       XBXb   Xb Xb
                                          Y        XB Y   Xb Y
(b) For the same mating, the ratio will be 1 (brown) : 1 (blue).
(c) Woman have blue eyed parent, so she will be heterozygous brown (Bb). The eye colour proportion
will be 3 (brown) : 1 (blue).
                                                    XB     Xb
                                          XB       XBXB   XB Xb
                                          Y        XB Y   Xb Y
                                                                                                        11


31.             XC         XC                     XC       XC                      X       XC
           C                                 C
          X    XCXC       XC XC            X     XCXC     XC XC             X     XX      XC X
          Y    XC Y       XC Y             Y     XC Y     XC Y              Y     XY      XC Y

                (A)                                (B)                            (C)
                 X         XC                      X        X                     XC       X
          X     XX        XC X             XC     XCX     X XC
                                                                            XC   XCXC     XC X
          Y     XY        XC Y             Y      XY      XY                Y    XC Y     XY

                (D)                               (E)                             (F)

32. All the sons are colourblind and all the daughters are normal but carrier.
                                                  XC        XC
                                            X    XCX       XC X
                                            Y    XC Y      XC Y

33. 50% of the sons and daughters will be colourblind .
                                                  XC        X
                                             C
                                           X     XCXC      XC X
                                           Y     XC Y      XY

34. There are two possibilities, either the father is normal (a) or colourblind (b). But as we know father
does not inherit his disorder to his sons, so his genotype will not effect the son. 50% 0f the sons will be
colourblind in both the cases as shown below.
                            XC      X                                   XC        X
                                                                   C
                      X    XCX      XX                            X    XCXC      XC X
                      Y    XC Y     XY                            Y    XC Y      XY
                              (a)                                          (b)

35. Klinefelter’s syndrome is not a sex-linked disorder; it is a result of non-disjunction at the time of
meiosis.

36. Tetraploid means the number of autosomes is four. (a) X/A = 2/4 = 0.5, i.e. male; (b) X/A = 3/4 =
0.67, i.e. intersex; (c) X/A = 2/4 = 0.5, i.e. male; (d) X/A = 4/4 = 1.0, i.e. female.

37. (a) Monosomy is 2n – 1 = 12 – 1 = 11 (b) Trisomy is 2n + 1 = 12 + 1 = 13 (c) Tetrasomy is 2n + 2 =
12 + 2 = 14 (d) Double trisomy is 2n + 1 + 1 = 12 + 1 + 1 = 14 (e) Nuliisomy is 2n – 2 = 12 – 2 = 10 (f)
Monoploid is n = 6 (g) Triploid is 3n = 6 x 3 = 18 (h) Autotetraploid is 4n = 6 x 4 = 24.

38. (a)         Rubus idaeus (2n = 14)                    R. caesius (2n = 28)


                   Gametes(n = 7)                         Gametes (n = 14)



                                         21 (triploid)

(b) First the chromosome doubling will take place of the F1, so it will become 42, hexaploid. Gametes of
F1 = 21, gametes of Rubus idaeus = 7. Offspring will be 21 + 7 = 28, tetraploid.
(c) Gametes of F1 = 21, gametes of Rubus caesius = 14. Offspring will be 21 + 14 = 35, pentaploid.
                                                                                                          12


(d) Chromosome doubling = 21 x 2 = 42, hexaploid.

39. (a) ABCD●EFH shows deletion of ‘G’. (b) ADCB●EFGH shows inversion of BCD to DCB. (c)
ABCDCD●EFGH shows duplication of CD.

40. Mutation rate is increasing at the rate of 3% per 1000. So at 3500R, the mutation rate will be 6%
(2000R) + 3% (1000R) + 1.5% (500R) = 10.5%.

41. The chromosome complement of the mother will be 45 + XX = 47. The gametes will be 22 + X and
23 + X. As shown below, if such a woman is crossed to a normal man, 50% of the sons and 50% of the
daughters will be affected.
                                        22 + X        23 + X
                              22 + X   44 + XX       45 + XX
                              22 + Y   44 + XY       45 + XY

42. A/T base ratio is 0.33, i.e. 1/3. Therefore A = 1 and T = 3. G/C base ratio is 2.0, i.e. 4/2. Therefore G
= 4 and C = 2. A+T/G+C ratio is 1.33, A+T = 133 and G+C = 100, i.e. A = 33, T = 100, G = 67, C = 33.
(a) A+G/T+C ratio is 33 + 67 / 100 + 33 = 100/133 = 0.75.
(b) In complementary strand, A = 100, T = 33, G = 33, C = 67.
A/T base ratio = 100/33 = 3.03; G/C base ratio = 33/67 = 0.5; A+T/G+C ratio = 100 + 33 / 33 + 67 =
133/100 = 1.33; A+G/T+C ratio = 100 + 33 / 33 + 67 = 133/100 = 1.33
(c) In both the strands A = 33 + 100 = 133; T = 100 + 33 = 133; G = 67 + 33 = 100; C = 33 + 67 = 100.
A/T base ratio = 133/133 = 1.0; G/C base ratio = 100/100 = 1.0; A+T/G+C ratio = 133 + 133 / 100 + 100
= 266/200 = 1.33; A+G/T+C ratio = 133 + 100 / 133 + 100 = 233/233 = 1.0

43. Explanation already given with the answer.

44. Since the child is colourblind, that means the extra X-chromosome comes from the mother, in which
non-disjunction had occur.

45. Equilibrium frequency of AB = 0.2 x 0.6 = 0.12; Equilibrium frequency of ab = (B)2 – (A)2 = 0.36 –
0.04 = 0.32; NOTE: Frequency of homozygous recessive is twice the frequency of heterozygotes (2pq).
q2 = 2pq = 2 x 0.32 = 0.64; q = 0.8. Equilibrium frequency of Ab = 0.2 x 0.8 = 0.16; Equilibrium
frequency of aB = 0.8 x 0.6 = 0.48

46. Suppose the population consists of 100 males and 100 females. Percentage of males having sex-linked
allele is 80% and percentage of females is 40%. Combining these two the resultant frequency comes out
to be (120/200) x 100 = 60%. Thus the gene frequency of the entire population is 0.6

47.
         ♀      B        b                                          ♂      B        b
         B      BB       Bb                                         B      BB       Bb
         b      Bb       bb                                         b      Bb       bb

Out of 10,000 males 7225 are normal, that means 2775 are bald. In the above cross, it has been shown
that 3/4 males are generally bald. Here the 3 shaded blocks represents the figure 2775, i.e. one block =
925 individuals. In females only 1/4 are found to be bald, i.e. represented by one block only (shaded),
which is equal to 925. Therefore normal females will be 10,000 – 925 = 9075.

48. (a) 20 men are haemophilic, i.e. 480 are normal; gene frequency of normal allele = 480/500 = 0.96.

								
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