Answer:
<h2>a. BbTt;</h2><h2>b. BbTt X BbTt
;</h2><h2>
Parent 1: BT bt Bt bT
</h2><h2>
Parent 2: BT bt Bt bT
;</h2><h2>c. BBTT
, BbTt
, BBTt
, BbTT
, BbTt
, BbTt
, BBTt
, BbTT
, BBTt
, Bbtt
, BBtt
, BbTt
, BbTT
, bbTt
, BbTt
, bbTT.</h2><h2>
d. 12:1:1:2</h2>
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
