In horses, black color(B) dominates chestnut color(b). The trotting gait(T) dominates the pacing gait (t). A cross is made between a horse homozygous for black color and the pacing gait and a horse homozygous for chestnut color and the trotting gait.

a. List the probable genotype and phenotype of offspring resulting from such a cross.

b. a cross is made between two horses, both of who are heterozygous black trotter.

- Give a genotype of each parent horse.

- Determine all the possible kinds of gametes each parent can produce





c. Construct a Punnett square and show all the possible gene combinations that results.

d. What is the expected phenotype ration of the offspring?

Explanation:

As given Block color is dominant over chesnut color, so B is dominant over b.

Chestnut color=bb ;

Trotting gait is dominant over  pacing gait, so T is dominant over t .

Pacing gait=tt

Now;  cross between

(a) homozygous black, pacing gait is= BBtt; and horse homozygous for chestnut color, trotting gait is= bbTT

BBtt x bbTT

gametes are ; Bt and bT,

Progeny are BbTt, BbTt,

All progeny will be BbTt  heterozygous Black color and heterozygous trotting gait

b. a cross is made between two horses, both of who are heterozygous black trotter = BbTt X BbTt

 Possible gametes of each parent are;

Parent 1: BT bt Bt bT

Parent 2: BT bt Bt bT

c.  Punnett square and  all the possible gene combinations that results are;

Cross between these gametes

BT bt Bt bT

BT bt Bt bT

possible gene combinations ; BBTT , BbTt , BBTt , BbTT , BbTt , BbTt , BBTt , BbTT , BBTt , Bbtt , BBtt , BbTt , BbTT , bbTt , BbTt , bbTT.

d. expected phenotype ration of the offspring.

Black color and trotting gait  horse 12: chestnut and trotting gait horse 1: Black color pacing gait horse  1: chestnut and trotting gait  horse 2.

so 12:1:1:2

lets get this bread mom

its probably to later but whats the question