Heredity

Sex is determined by the sex chromosomes at fertilization.

              Father (44 + XY)          Mother (44 + XX)
                    |                          |
              Meiosis (sperms)           Meiosis (eggs)
                  /     \                    /    \
Gametes:    22 + X   22 + Y            22 + X    22 + X
                  \       \               /        /
                   Fertilization (Punnett square)

             |  22 + X (egg)     | 22 + X (egg)
   ----------+-------------------+------------------
   22 + X    |  44 + XX (girl)   | 44 + XX (girl)
   22 + Y    |  44 + XY (boy)    | 44 + XY (boy)

If a sperm carrying X fertilizes the egg, the child is a girl (XX). If a sperm carrying Y fertilizes the egg, the child is a boy (XY). The chance of a boy or a girl is 50% each.

Figure A is RNA and Figure B is DNA.

1. Bottle gourd is a large climber that needs a lot of space and support, so raising many plants over several generations would be difficult.
2. Its male and female flowers are separate on the same plant, so controlling cross-pollination by hand is difficult.

1. Pea plants have many clear contrasting characters (7 pairs, e.g. tall/dwarf, red/white).
2. They are self-pollinating, annual, easy to grow, with a short life cycle and many seeds, and artificial cross-pollination is easy.

Parents (P):      BB (black)    x    bb (white)
                    |                  |
Gametes:          B  B              b  b
                      \              /
F1 generation:      Bb x4  (all black, hybrid)

F1 x F1:          Bb (black)   x   Bb (black)
Gametes:          B  b              B  b

F2 (Punnett square):
            | B    | b
        ----+-------+------
         B  | BB    | Bb
         b  | Bb    | bb

F2: BB (black, pure) : Bb (black, hybrid) : bb (white, pure)

• Phenotypic ratio in F$_2$ = black : white = 3 : 1
• Genotypic ratio in F$_2$ = BB : Bb : bb = 1 : 2 : 1

The F$_1$ dogs are all black (Bb), showing that black is dominant over white.

(a) Mendel's law of dominance states that when two pure individuals having contrasting characters are crossed, only one character (the dominant one) appears in the F$_1$ generation, and the other (recessive) character remains hidden.

(b)

Parents (P):      BB (black)    x    bb (white)
                    |                  |
Gametes:          B  B              b  b
                      \              /
F1 generation:      Bb x4  (all black, hybrid)

F1 x F1:          Bb (black)   x   Bb (black)
Gametes:          B  b              B  b

F2 (Punnett square):
            | B    | b
        ----+-------+------
         B  | BB    | Bb
         b  | Bb    | bb

F2: BB (black, pure) : Bb (black, hybrid) : bb (white, pure)

• Phenotypic ratio in F$_2$ = black : white = 3 : 1
• Genotypic ratio in F$_2$ = BB : Bb : bb = 1 : 2 : 1

No. The husband produces two kinds of sperms, one carrying X and the other carrying Y, in equal numbers. A daughter is born only when an X-bearing sperm happens to fertilize the egg. Having two daughters is a matter of chance, so it does not mean that he lacks Y-bearing sperms.

              Father (44 + XY)          Mother (44 + XX)
                    |                          |
              Meiosis (sperms)           Meiosis (eggs)
                  /     \                    /    \
Gametes:    22 + X   22 + Y            22 + X    22 + X
                  \       \               /        /
                   Fertilization (Punnett square)

             |  22 + X (egg)     | 22 + X (egg)
   ----------+-------------------+------------------
   22 + X    |  44 + XX (girl)   | 44 + XX (girl)
   22 + Y    |  44 + XY (boy)    | 44 + XY (boy)

Each fertilization has a 50% chance of giving a girl and 50% of giving a boy, so a couple may have only daughters by chance. The sex of the child depends on the father's sperm, not on the mother, because the mother produces only X-bearing eggs.

Mendel crossed pure red-flowered (RR) and pure white-flowered (rr) pea plants.

Parents (P):      RR (red)    x    rr (white)
                    |                  |
Gametes:          R  R              r  r
                      \              /
F1 generation:      Rr x4  (all red, hybrid)

F1 x F1:          Rr (red)   x   Rr (red)
Gametes:          R  r              R  r

F2 (Punnett square):
            | R    | r
        ----+-------+------
         R  | RR    | Rr
         r  | Rr    | rr

F2: RR (red, pure) : Rr (red, hybrid) : rr (white, pure)

• Phenotypic ratio in F$_2$ = red : white = 3 : 1
• Genotypic ratio in F$_2$ = RR : Rr : rr = 1 : 2 : 1

Parents (P):      BB (black)    x    bb (white)
                    |                  |
Gametes:          B  B              b  b
                      \              /
F1 generation:      Bb x4  (all black, hybrid)

F1 x F1:          Bb (black)   x   Bb (black)
Gametes:          B  b              B  b

F2 (Punnett square):
            | B    | b
        ----+-------+------
         B  | BB    | Bb
         b  | Bb    | bb

F2: BB (black, pure) : Bb (black, hybrid) : bb (white, pure)

F$_1$: all black (Bb). F$_2$: black : white = 3 : 1; genotypes BB : Bb : bb = 1 : 2 : 1.

Parents (P):      LL (long wing)    x    ll (short wing)
                    |                  |
Gametes:          L  L              l  l
                      \              /
F1 generation:      Ll x4  (all long wing, hybrid)

F1 x F1:          Ll (long wing)   x   Ll (long wing)
Gametes:          L  l              L  l

F2 (Punnett square):
            | L    | l
        ----+-------+------
         L  | LL    | Ll
         l  | Ll    | ll

F2: LL (long wing, pure) : Ll (long wing, hybrid) : ll (short wing, pure)

F$_1$: all long-winged (Ll). F$_2$: long : short = 3 : 1; genotypes LL : Ll : ll = 1 : 2 : 1.

(a) Sex determination chart

              Father (44 + XY)          Mother (44 + XX)
                    |                          |
              Meiosis (sperms)           Meiosis (eggs)
                  /     \                    /    \
Gametes:    22 + X   22 + Y            22 + X    22 + X
                  \       \               /        /
                   Fertilization (Punnett square)

             |  22 + X (egg)     | 22 + X (egg)
   ----------+-------------------+------------------
   22 + X    |  44 + XX (girl)   | 44 + XX (girl)
   22 + Y    |  44 + XY (boy)    | 44 + XY (boy)

(b) Cross between black (BB) and white (bb) rabbits

Parents (P):      BB (black)    x    bb (white)
                    |                  |
Gametes:          B  B              b  b
                      \              /
F1 generation:      Bb x4  (all black, hybrid)

F1 x F1:          Bb (black)   x   Bb (black)
Gametes:          B  b              B  b

F2 (Punnett square):
            | B    | b
        ----+-------+------
         B  | BB    | Bb
         b  | Bb    | bb

F2: BB (black, pure) : Bb (black, hybrid) : bb (white, pure)

Genotypic ratio in F$_2$ = BB : Bb : bb = 1 : 2 : 1 (phenotypic ratio black : white = 3 : 1).

(a) The second filial generation (F$_2$), because Rr x Rr gives RR, Rr, Rr, rr.

(b) RR (pure red) and rr (pure white) show pure characters.

(c) Genotypic ratio = RR : Rr : rr = 1 : 2 : 1.

(d) Mendel's law of segregation (purity of gametes): the two alleles of a pair separate during gamete formation, so each gamete gets only one allele, and the recessive character reappears unchanged in F$_2$ (the rr plants).

(a) F$_2$ generation (cross of two F$_1$ hybrids, Rr x Rr).

(b) Phenotypic ratio: red : white = 3 : 1. Genotypic ratio: RR : Rr : rr = 1 : 2 : 1.

(c) RR represents the pure red-flowered pea plant.

(d) It proves the law of segregation (purity of gametes): the hidden recessive character (white, rr) reappears in F$_2$ and the pure characters RR and rr are unchanged.

(iii) 7. Mendel studied seven contrasting pairs of characters in the pea plant, such as height, flower colour and seed shape.

Father (44 + XY) forms sperms (22 + X) and (22 + Y). Mother (44 + XX) forms eggs (22 + X) and (22 + X).

Parents:   Father (44 + XY)    x    Mother (44 + XX)
Meiosis:   Spermatogenesis           Oogenesis
Gametes:   (22 + X) (22 + Y)         (22 + X) (22 + X)

(c) pure tall : hybrid tall : pure dwarf = 1 : 2 : 1. In F$_2$ the genotypes are TT : Tt : tt = 1 : 2 : 1.

Law of purity of gametes: the two alleles of a pair separate (segregate) during gamete formation, so each gamete carries only one allele and remains pure.

Parents:    Bb  x  Bb
Gametes:    B  b    B  b

         |  B    | b
     ----+-------+------
      B  |  BB   | Bb
      b  |  Bb   | bb

The hybrids (Bb) form two types of pure gametes, B and b, never a mixed 'Bb' gamete. When they combine randomly, the hidden recessive character (bb, white) reappears in 1/4 of the offspring unchanged. Ratio: BB : Bb : bb = 1 : 2 : 1 and black : white = 3 : 1. This justifies the law of purity of gametes.

(b) 2 : 2 hybrid tall and pure dwarf. Crossing T t (parent) with t t gives Tt, Tt, tt, tt, i.e. two hybrid tall (Tt) and two pure dwarf (tt).

• DNA = Deoxyribonucleic acid. Its major function is to store genetic information and pass hereditary characters from parents to offspring.
• RNA = Ribonucleic acid. Its major function is to help in protein synthesis by carrying the genetic message from DNA to the ribosomes.

(a) 46. Human body cells are diploid, having 23 pairs = 46 chromosomes.

Figure A shows mitosis (2n to two 2n cells) and Figure B shows meiosis (2n to four n cells).

(i) Figure B (meiosis) forms gametes. The mother cell has 4 chromosomes, so each gamete has 2.

(ii) Figure A (mitosis) forms daughter cells identical to the mother cell, each with 4 chromosomes.

(iii) Importance of A (mitosis): it helps in growth of the body, repair of damaged tissues, replacement of dead cells and asexual reproduction, and it keeps the chromosome number constant.

(iii) The fact is correct: in IVF, egg and sperm are united outside the body in a glass vessel and the embryo is placed in the uterus. Statement 1 is incorrect because sperm is essential for fertilization. Statement 2 is correct because a surrogate mother can carry the embryo in her uterus.

RNA = Ribonucleic acid.

• Monohybrid cross: a cross between two individuals (parents) differing in a single pair of contrasting characters, e.g. tall (TT) x dwarf (tt).
• Phenotype: the outwardly visible or observable character of an organism, e.g. tall, dwarf, red flower.

(b) 23. Reproductive cells (gametes) are haploid, formed by meiosis, so they contain half of 46 chromosomes.

Half of the offspring are black (Bb) and half are brown (bb), i.e. Bb : bb = 1 : 1.

Parents:    Bb (black)  x  bb (brown)
Gametes:    B  b           b  b

         |  b    | b
     ----+-------+------
      B  |  Bb   | Bb
      b  |  bb   | bb

(a) Dominant character. Red appeared in all F$_1$ plants even though the plant also received the white allele, so red is dominant over white.

DNA = Deoxyribonucleic acid.

No type of blood cell is imbalanced in haemophilia. Haemophilia is caused by a deficiency of clotting factors (usually factor VIII or factor IX) in the blood plasma because of a defective gene on the X chromosome, so the blood does not clot easily. Platelets (thrombocytes) remain normal in number and function.

1. It helps to treat and prevent genetic diseases, e.g. gene therapy, and to identify criminals and family relations by DNA test.
2. It improves crops and livestock by producing high-yielding, disease-resistant varieties and gives medicines such as insulin through genetically modified organisms; IVF helps childless couples.

(i) The parents of the offspring shown in the table were both hybrid long-winged: genotype $Ll$, phenotype long-winged. (The purebred long-winged $LL$ and short-winged $ll$ were the P generation; the table shows the F$_2$ generation obtained from $Ll \times Ll$.)

(ii) Genotypic ratio = $LL : Ll : ll = 1 : 2 : 1$.

(iii) In $Ll$, the long-wing allele $L$ is dominant and the short-wing allele $l$ is recessive. The dominant allele expresses itself and masks the recessive allele, so only the long-wing character is expressed while the short-wing factor remains hidden (law of dominance).

(iv) Test cross of hybrid ($Ll$) with recessive ($ll$):

Parents:    Ll (long)  x  ll (short)
Gametes:    L  l          l  l

         |  l    | l
     ----+-------+------
      L  |  Ll   | Ll
      l  |  ll   | ll

Offspring: $Ll$ (long-winged) : $ll$ (short-winged) = $1 : 1$.

(c) 23 pairs. A human somatic cell has 46 chromosomes, which is 23 pairs (22 pairs of autosomes and 1 pair of sex chromosomes).

1. Mustard plants are very small with tiny flowers, so artificial cross-pollination (removing anthers by hand) is very difficult.
2. It has few clear, easily observable contrasting characters, and it is hard to keep its lines pure.

1. Reduce the use of chemical fertilizers, pesticides, detergents and plastic, and use organic manure and biological pest control instead.
2. Dispose of and recycle wastes properly (do not burn or dump them), and avoid direct contact with such chemicals, using protective equipment.

(d) 46. A skin cell is a somatic (diploid) cell, containing 46 chromosomes (23 pairs).

Black (Bb) x brown (bb) gives F$_1$ with Bb (black) and bb (brown) in the ratio 1 : 1. When the black F$_1$ dogs (Bb) are crossed together, the F$_2$ genotypes are BB : Bb : bb = 1 : 2 : 1.

1. Apple is a tree with a very long life cycle (several years to bear fruit), so it would take a very long time to obtain generations.
2. It is a large tree that needs a lot of space, and controlling its crosses and observing the offspring is hard, so results could not be obtained quickly.

(ii) 44 + xy. A human male body (somatic) cell has 44 autosomes and the sex chromosomes X and Y. 22 + y and 22 + x are gametes.

(b) 3 : 1. In F$_2$ of a monohybrid cross, dominant : recessive = 3 : 1.

Advantages

1. The offspring produce more milk than local buffaloes, so income rises.
2. The offspring are larger and grow faster, giving more meat and better draught power.

Disadvantages

1. The crossbreeds need more and better feed, care and veterinary services, which is costly.
2. They are less resistant to local diseases and climate, and the pure local breed may be lost in the long run.

(d) Thymine. The figure is DNA, and thymine is found only in DNA (RNA has uracil instead). Adenine, guanine and cytosine occur in both.

Zorse (also called zebroid), a hybrid of a zebra and a horse.

Parents:    Tt (hybrid tall)  x  TT (pure tall)
Gametes:    T  t                 T  T

         |  T    | T
     ----+-------+------
      T  |  TT   | TT
      t  |  Tt   | Tt

Offspring: TT : Tt = 1 : 1 (genotypic ratio). All offspring are tall (phenotype), half pure tall and half hybrid tall.

No. The semen of the father contains two kinds of sperms, 22 + X and 22 + Y, in equal numbers, so X is not absent. A couple having four sons is a matter of chance: each time a Y-bearing sperm happened to fertilize the egg.

              Father (44 + XY)          Mother (44 + XX)
                    |                          |
              Meiosis (sperms)           Meiosis (eggs)
                  /     \                    /    \
Gametes:    22 + X   22 + Y            22 + X    22 + X
                  \       \               /        /
                   Fertilization (Punnett square)

             |  22 + X (egg)     | 22 + X (egg)
   ----------+-------------------+------------------
   22 + X    |  44 + XX (girl)   | 44 + XX (girl)
   22 + Y    |  44 + XY (boy)    | 44 + XY (boy)

If an X sperm fertilizes the egg the child is a daughter (XX); if a Y sperm fertilizes it the child is a son (XY). The father's sperm decides the sex, and each child has a 50% chance of being a boy or a girl, independently of earlier children.

(iv) jellyfish, earthworm, fish, frog. This is the order of appearance in evolution: invertebrates (jellyfish, then earthworm) before vertebrates, and among them fish before amphibians (frog).

Yes. If both black parents are hybrid (Bb), they carry the hidden recessive allele for white (b). Black (B) is dominant over white (b).

Parents:    Bb (black)  x  Bb (black)
Gametes:    B  b           B  b

         |  B    | b
     ----+-------+------
      B  |  BB   | Bb
      b  |  Bb   | bb

Offspring: BB : Bb : bb = 1 : 2 : 1, i.e. black : white = 3 : 1. The child bb is white because it received the recessive allele b from both parents. The recessive character is masked, not lost (law of segregation), so each child has a 25% chance of being white.

In-vitro fertilization (IVF) is the process in which an egg and a sperm are united outside the body in a glass dish (test tube) and the embryo is then placed in the uterus.

Mature eggs taken from ovary  +  sperms from father
              \                 /
          mixed in a glass dish (in vitro)
                      |
             fertilization -> zygote
                      |
          grows into embryo (2-5 days)
                      |
   embryo placed in the uterus of mother / surrogate mother
                      |
               pregnancy and baby

Steps: (1) the mother is given hormones so that many eggs mature; (2) eggs are collected and sperms are taken from the father; (3) they are mixed in the dish where fertilization occurs; (4) the embryo is placed in the uterus, where it implants and grows.

Differences between inbreeding and line breeding

Shark (A) is a fish that breathes through gills and lays eggs, and it appeared earlier in evolution. Whale (B) is an aquatic mammal that breathes by lungs and gives birth to young, and it evolved later from land mammals.

Black eyes are dominant over brown eyes, which are recessive. The father is pure black-eyed (BB) and the mother is brown-eyed (bb). All children get B from the father and b from the mother, so they are Bb. As B is dominant, it masks b, so the children have black eyes and not brown.

Parents (P):      BB (black-eyed)    x    bb (brown-eyed)
                    |                  |
Gametes:          B  B              b  b
                      \              /
F1 generation:      Bb x4  (all black-eyed, hybrid)

F1 x F1:          Bb (black-eyed)   x   Bb (black-eyed)
Gametes:          B  b              B  b

F2 (Punnett square):
            | B    | b
        ----+-------+------
         B  | BB    | Bb
         b  | Bb    | bb

F2: BB (black-eyed, pure) : Bb (black-eyed, hybrid) : bb (brown-eyed, pure)

• Phenotypic ratio in F$_2$ = black-eyed : brown-eyed = 3 : 1
• Genotypic ratio in F$_2$ = BB : Bb : bb = 1 : 2 : 1

(b) The embryo develops into a male because of the Y chromosome. An XY embryo is male; the Y chromosome carries the gene that makes the embryo male, and XX develops into a female.

A male produces two types of sperms in equal numbers: half carry X and half carry Y. The egg always carries X. If an X sperm fertilizes the egg the child is a girl (XX), and if a Y sperm fertilizes it the child is a boy (XY). Which sperm reaches the egg is a matter of chance, so some men by chance have only sons and others only daughters, though each has both X and Y.

(a) Labelled diagram of DNA strands carrying hereditary characters with nitrogen bases:

        5' end                                     5' end
          |                                           |
   S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S
          |     |     |     |     |     |     |     |
          A     T     G     C     A     T     G     C
          |     |     |     |     |     |     |     |
          T     A     C     G     T     A     C     G
          |     |     |     |     |     |     |     |
   S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S—P—S
          |                                           |
        3' end                                     3' end

Labels:
S = deoxyribose sugar; P = phosphate; A = adenine; T = thymine; G = guanine; C = cytosine; vertical lines = hydrogen bonds.
A pairs with T by 2 hydrogen bonds; G pairs with C by 3 hydrogen bonds. The two strands are antiparallel and twist into a double helix. Genes on DNA carry hereditary characters.

(b) In-vitro fertilization (IVF): It is the process of fertilizing an egg by a sperm outside the mother’s body in a glass dish, and then transferring the resulting embryo into the uterus. It is also called the test-tube baby technique.

   Ovary
     |
  Egg cell  +  Sperm cell
      \        /
       \      /
    +----------------+
    |  Glass dish    |
    |  (in vitro)    |
    +----------------+
            |
       Fertilization
            |
         Zygote
            |
         Embryo
            |
   +----------------+
   |     Uterus     |
   | embryo transfer|
   +----------------+
            |
        Pregnancy
            |
          Baby

It helps couples who cannot have children naturally due to blocked fallopian tubes, low sperm count, etc. The embryo may also be carried by a surrogate mother if needed.

(a) The ratio of black and white is equal (1 : 1) when a hybrid black rabbit (Bb) is crossed with a white rabbit (bb), which is a test cross.

Parents:    Bb (black)  x  bb (white)
Gametes:    B  b           b  b

         |  b    | b
     ----+-------+------
      B  |  Bb   | Bb
      b  |  bb   | bb

Offspring: Bb (black) : bb (white) = 1 : 1.

(b) The parents are pure black (BB) and white (bb). Every offspring is Bb and black because B is dominant.

(c) Law of purity of gametes (segregation): the two alleles of a gene separate during gamete formation, so each gamete carries only one allele (B or b), and gametes are always pure.

The chart represents sex determination (fertilization of gametes to determine the sex of the child).

The organism with (44, xx) is a female, having two X sex chromosomes, while the organism with (44, xy) is a male, having one X and one Y sex chromosome. The Y chromosome makes the child male.

It is called a monohybrid cross.

A housefly cell has 6 pairs of chromosomes (2n = 12).

The karyotype of Klinefelter syndrome is 47, XXY (44 autosomes + XXY).

There are 23 pairs of chromosomes in human beings (2n = 46).

A chromosome is a thread-like structure in the nucleus, made of DNA and protein, that carries genes and passes characters from parents to offspring.

The ratio of dominant to recessive characters in the $F_2$ generation is 3 : 1.

Mendel's law of purity of gametes states that the two alleles (factors) of a character separate during gamete formation, so each gamete receives only one of them and is pure for that character.

A hybrid is an organism produced by crossing parents with contrasting characters; it carries two different alleles for the character (heterozygous), e.g. $Tt$.

(a) Sex in gorilla is determined by the sex chromosomes: female XX, male XY.

Parents:      Female (XX)      x      Male (XY)
Gametes:        X   X                  X    Y
                    
                 X        Y
        X       XX       XY
        X       XX       XY

Offspring: 50% XX (female) : 50% XY (male)

(b) The male is mainly responsible. All eggs of the female carry X, but the male produces two kinds of sperm: X-bearing (giving a daughter) and Y-bearing (giving a son). A son is born only when a Y-bearing sperm fertilises the egg.

Suggestion: the females should not be removed because they are not responsible. The chance of a male gorilla at each birth is 50% by chance, so the centre should continue breeding the females with a healthy male and wait for more births.

1. Non-disjunction of chromosome 21 during gamete formation, so a gamete carries two copies and the zygote gets three (trisomy 21, total 47 chromosomes).
2. Late motherhood (mother older than about 35 years), which increases the chance of non-disjunction.

It occurs in both boys and girls because chromosome 21 is an autosome, not a sex chromosome.

(a) The person has Down's syndrome, caused by an extra chromosome 21 (trisomy 21), which is an autosome.

(b) It occurs in both males and females.

(a) Chromosome B, which has the shorter arm, is the Y chromosome (chromosome A is the X chromosome).

(b) The structure responsible for sex determination is the sex chromosome (X and Y).

OR

The father's chromosome is responsible. The mother (XX) produces only X-bearing eggs, but the father (XY) produces X-bearing and Y-bearing sperms. If an X sperm fertilises the egg, the child is a girl (XX); if a Y sperm does, the child is a boy (XY).

(a) This result is seen in the $F_1$ generation (first filial generation).

(b) It proves Mendel's law of dominance: when pure parents with contrasting characters are crossed, only the dominant character (black) appears in $F_1$.

Let R = red (dominant) and r = white (recessive).

Parents (P):      Pure red (RR)    x    Pure white (rr)
Gametes:             R   R              r   r
                           |
F1:                  Rr (all hybrid red)
F1 x F1:             Rr   x   Rr
Gametes:           R, r        R, r

F2:          R      r
      R     RR     Rr
      r     Rr     rr

Genotypic ratio  RR : Rr : rr = 1 : 2 : 1
Phenotypic ratio red : white = 3 : 1

OR: differences between pure and hybrid characters

(a) Dominant : recessive = 3 : 1 (AA, Aa, Aa show the axial flower and aa the terminal flower).

(b) The plant with genotype aa has pure terminal flowers.

(a) The result appears in the $F_2$ generation, because it comes from crossing two hybrid ($Rr$) plants of $F_1$.

(b) RR is a pure (homozygous) round-seeded plant with two identical alleles, and it produces only R gametes. Rr is a hybrid (heterozygous) round-seeded plant, which carries the recessive allele hidden and produces both R and r gametes. Both look round, but only RR breeds true.

Pure round (RR) $\times$ pure wrinkled (rr) gives all hybrid round (Rr) in $F_1$. Selfing $F_1$ gives the Mendelian genotypic ratio $1\,RR : 2\,Rr : 1\,rr$ in $F_2$ (phenotypic ratio round : wrinkled $= 3:1$).

Total seeds $= 7524$.

$\text{Pure wrinkled (rr)} = \frac{1}{4}\times 7524 = 1881$

$\text{Hybrid round (Rr)} = \frac{2}{4}\times 7524 = 3762$

So there are 1881 pure wrinkled and 3762 hybrid round seeds (and 1881 pure round).

Let B (black) be dominant over b (brown).

(a) The pure brown dog, genotype bb, cannot produce pure black (BB) offspring, since it has no B allele to give.

(b) Pure brown (bb) $\times$ pure black (BB): all gametes of one parent are b and of the other are B, so every offspring is hybrid black (Bb).

(c) Black is dominant and brown is recessive. Every offspring receives the dominant B from the pure black parent, so it masks b and no brown dog (bb) can form in this cross.

(a) The disorders are caused by abnormal sex chromosomes (X and Y), not by autosomes.

(b) The man has Klinefelter's syndrome.

(c) The woman has Turner's syndrome, with the chromosome pattern 44 + XO (45 chromosomes in all, only one X).

(a) He suffers from Klinefelter's syndrome, a sex chromosomal disorder of males.

(b) It is caused by non-disjunction of sex chromosomes during gamete formation, so an egg carrying XX is fertilised by a Y sperm (or an X egg by an XY sperm), giving an extra X chromosome.

(c) Karyotype: 44 + XXY (47 chromosomes in total).

(i) No. The number of chromosomes is not related to the size or complexity of an organism. Human has 23 in the gamete (46 in the body) and the small potato plant has 24 (48 in the body), while the pine, a huge tree, has only 12. The number is characteristic of the species and depends on its genetic make-up, not its size.

(ii) The frog's gametes are haploid ($n = 13$), formed by meiosis, which halves the chromosome number. At fertilisation the sperm and the egg fuse and restore the diploid number ($2n = 26$) in the zygote. Mitosis then passes this same number to every body cell.

Colour blindness is caused by a defective recessive gene carried on the X chromosome (a sex-linked genetic disorder).

Parents:      BB (black)      x      Bb (hybrid black)
Gametes:      B   B                  B   b

             |  B    | b
         ----+-------+------
          B  |  BB   | Bb
          B  |  BB   | Bb

Offspring: BB : Bb = 1 : 1 (genotypic ratio); all are black (phenotype).

i. Dolly is identical to sheep 1, the sheep from whose udder the nucleus was taken.

ii. It means a single cell (only its nucleus was used), because the nucleus was removed from the piece and transferred into the egg cell.

iii.

The first is a biological risk that can be tested, while the second is a religious or moral opinion.

1. The offspring give much more milk than local buffaloes, which increases the farmers' income.
2. The offspring grow faster, are stronger and healthier (better meat and disease resistance), so they fetch higher prices.