# KEY-GENETICSBOOKPROBLEMS_1 by lsy121925

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```									                  GENETICS BOOK PROBLEMS #1 (pg. 272-273; #3, 8, 9, 15, 16)
3. A black guinea pig crossed with an albino guinea pig produces 12 black offspring. When the albino
is crossed with a second black one, 7 blacks and 5 albinos are obtained. What is the best
explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.

B=black           Black is dominant over albino
b=albino

Parents= BB x bb                                       bb x Bb
Offspring= Bb (F1 all)                                 Offspring = bb & Bb

8. What is the probability that each of the following pairs of parents will produce the indicated
offspring? (Assume independent assortment of all gene pairs.)
a. AABBCC X aabbcc  AaBbCc
1/1 x 1/1 x 1/1 = 1/1

b. AABbCc X AaBbCc  AabbCC
(1 x ½) x (½ x ½) x (½ x ½)
½ x ¼ x ¼ = 1/32

c. AaBbCc X AaBbCc  AaBbCc
[(½ x ½) + (½ x ½)] x [(½ x ½) + (½ x ½)] x [(½ x ½) + (½ x ½)]
½ x ½ x ½ = 1/8

d. aaBbCC X AABbcc  AaBbCc
(1 x 1) x [(½ x ½) + (½ x ½)] x (1 x 1)
1/1 x ½ x 1/1= ½

9. Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of
their parents have the disease, and none of them have been tested to reveal sickle-cell trait Based
on the incomplete information, calculate the probability that if this couple had a child, the child
will have sickle-cell disease.
Karen= S -
Steve= S -
(2/3 chance they will each be a carrier; we know they don’t have sickle-cell so only 3 possibilities
left & 2 of the 3 are carriers)

2/3 x 2/3 x ¼ = 4/36 = 1/9 (chance of 2 carriers having a child w/ sickle cell)
Karen’s chance of being a carrier x Steve’s chance of being a carrier x probability of 2 carriers having a child with sickle cell
15. A man has six fingers on each hand and six toes on each foot. His wife and daughter have the
normal number of digits. Extra digits are a dominant trait. What fraction of this couple’s children
would be expected to have extra digits?
B= extra fingers/toes (dominant)
Man= Bb              Wife= bb             Daughter= bb
Cross: Bb x bb

50% chance of having a child w/ extra fingers/toes

16. Imagine that you are a genetic counselor, and a couple planning to start a family come to you for
information. Charles was married once before, and he and his first wife had a child with cystic
fibrosis. The brother of his current wife Elaine died of cystic fibrosis. What is the probability
that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine have
cystic fibrosis.)

Charles= Ff, Elaine= F-
 probability of Charles being a carrier = 1/1
 probability of Elaine being a carrier = 2/3
 probability 2 carriers having a child with CF = ¼

1 x 2/3 x ¼ = 2/12 = 1/6

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