Answer: 6/12 are white, 3/12 are colored and 3/12 are albino.
Step-by-step explanation: If the horses are white and their parents are ccww (albino) and CCWw (white horse), according to Mendel's premises, they both must be CcWw, since the crossing provides one C from one parent and other c from the other parent, one W and the other w. Using Mendel's chess and the principle of independent segregation, the crossing between CcWw results in the following fenotypical ratio:
1/16 CCWW (lethal)
2/16 CCWw (white)
2/16 CcWW (lethal)
4/16 CcWw (white)
1/16 CCww (normal)
2/16 Ccww (normal)
2/16 ccWw (albino)
1/16 ccWW (lethal)
1/16 ccww (albino)
Excluding the 4 individuals that have the lethal locus, we have 6/12 that are white (2/12 + 4/12) and 3/12 (1/12 + 2/12) that are colored. Also, there are 3/12 of albino individuals as well.
All my work is shown in the image.
A, B, and C. all of them are reasonable choices and wouldn’t give any conclusion, because not everyone has the flu
Answer:
(3/2, 6)
Step-by-step explanation:
y = 4x
8x + y = 18
this says y is 4x so you can replace y with 4x
8x + 4x = 18
12x = 18
/12 /12
x = 3/2
now sub x into y = 4(3/2)
y = 12/2
y = 6
Answer:
x > 4/33
Step-by-step explanation:
(-4 +8x) +3> x/-4
8x - 1 > x/-4
-32x + 4 < x
4 < 33x
4/33 < x
