Download Inheritance Problems

Inheritance Problems
Some aliens have the following traits:
Trait
Dominant Phenotype
Body color
(Y) Yellow
Number of Antennae (A) 2
Eye color
(P) Purple
Eyesight
(E) Glasses needed
Number of Body Rings (R) 3
Recessive Phenotype
(y) Orange
(a) 1
(p) White
(e) Glasses not worn
(r) 5
1.
A heterozygous male mates with a pure yellow female. What is the chance this couple will produce an orange baby?
Yy x YY; Offspring are 50% YY and 50%Yy; no orange babies.
2. Two purple-eyed aliens mate. Both aliens are heterozygous for the eye color trait. What is the chance this couple will produce a
baby with a homozygous recessive genotype?
Pp x Pp; 25% pp
3. A heterozygous female with 3 body rings mates with a 5-ringed male. What is the chance that this couple will have a baby that
looks like its mother?
Rr x rr; 50% Rr offpring has 3 rings like mom.
4. A pure male alien displaying the dominant body color mates with a female that is homozygous recessive for this characteristic.
What is the chance that this couple will have a baby with a heterozygous genotype?
Yy x yy; all offspring will be Yy heterozygotes.
5. A male alien with 1 antenna mates with a female alien who has 2 antennae. The female is heterozygous for the antenna trait. What
is the chance that this couple will produce a baby with the recessive phenotype?
aa x Aa; 50% Aa and 50% aa offspring – so 50% will have the recessive phenotype.
6. An alien couple, both of which wear glasses, are having a baby. The male’s genotype is heterozygous. The female is
phenotypically dominant but does carry the recessive allele. What is the chance that this couple’s baby will have to wear glasses?
Ee x Ee; 75% of offspring will have the dominant phenotype and will therefore need glasses.
7. A 3-ringed female mates with a homozygous male. The female has been genetically tested and is carrying both the dominant and
the recessive allele for this trait. The male displays the recessive phenotype. What is the chance that this couple will have a
homozygous baby?
Rr x rr; 50% homozygous baby
8. In peas red flowers are dominant over white flowers. A pea plant with white flowers is crossed with one that has red flowers; of
the progeny 147 have white flowers and 161 have red flowers. What are the genotypes of the two parent plants?
Ll (red) X ll (white) gives offspring 50% white : 50% red
9. In peas red flowers (R) are dominant over white flowers (r). Two plants that had red flowers. There were 76 plants with red
flowers and 24 plants with white flowers in the F1 generation. What was the genotype of the two parents.
Offspring ratio looks like 3:1, so parents need to be Rr X Rr
10. Assume that white is the dominant color over yellow squash. If pollen from the anthers of a heterozygous white-fruited plant is
placed on the pistil of a yellow-fruited plant, show, using ration, the genotypes and phenotypes you would expect the seeds from
the cross to produce.
Ww X ww
1:1 offspring ratio
50% Ww white : 50% ww yellow
11. Jim did not have bent little fingers, but his mother did. Rosie, his wife had bent little fingers. Rosie's mother did not have bent
little fingers. What are the chances that they will have kids with bent little fingers? A dominant allele (B) causes the last joint of
the little finger to bend inward towards the fourth (ring) finger. Individuals whose little fingers are straight possess the
homozygous recessive condition, bb. To determine, lay both hands flat on the table with your fingers slightly spread and note
whether the last joint of your little finger bends inward.
B-
bb
Bb
Bb
Bb
bb
Jim
Rosie
Bb X bb
Jim can put a B or a b in his sperm. Rosie can put a b in her eggs.
Offspring will be 50% Bb and 50% bb, so half of them will have bent fingers.
12. Both Count Dracula and Natasha are heterozygous for widow's peak. What are the chances that their son Igor will also have
widow's peak? WIDOW'S PEAK INFORMATION: The dominant allele (W) causes the hairline to form a V-shape point in the
middle of the forehead. A person who is homozygous recessive has a straight hair line.
Ww X Ww
Igor has 75% chance for Widows peak to 25% chance for no widows peak
13. George has attached ear lobes. Barbara has free earlobes, but she is heterozygous for this trait. What are the chances of their child
having free earlobes? ATTACHED EARLOBE INFORMATION: The inheritance of a dominant gene E results in the free or
unattached earlobe. If the lobe is attached directly to the head, the individual is homozygous recessive, and the ee genotype is
present. Ask someone to look at your earlobes if you are interested about yourself. Other genes can influence this trait, so there
will be a considerable amount of variation in the size and appearance of the earlobes in a group of people.
George is attached, ee
Barbara is free (heterozygous) Ee
Offspring will be either Ee (free) or ee (attached) ; 50% chance
14. The allele for Huntington's Chorea is dominant. Woody Guthrie was heterozygous for this trait and Mrs. Guthrie did not have
Huntington's Chorea. What are the chances that their son Arlo will develop this disease?
Woody is Hh (will give either a H or a h in sperm)
Mrs. Guthrie is hh (will put h in eggs)
Offspring will be 50% Hh (Huntingtons) and 50% hh (no Huntingtons)
15. When Rev folded his arms, he always positioned/placed his left arm on top of the right. His mother always positioned/placed her
right arm on top when she folded her arms. When Boopsie folded her arms, she always positioned her right arm on top of the left.
Rev and Boopsie are having a child....it is a girl. Which way will this girl position her arms when she folds them. Genes
determine how you place your arms when you fold them across your chest. Positioning/placing the left arm on top of the right is
due to a dominant allele (A). Placing the right arm on top is due to a recessive allele (aa). Test this.....cross your arms. Then
reverse the position. Which way feels more comfortable?
Rev is left over right, so has one dominant A, but his mother was aa, so Rev must be Aa.
Boopsie is aa.
Aa X aa will give offspring Aa or aa, so the daughter has a 50% chance of either phenotype.
16. The allele for Tay-Sachs is recessive. The Tay-Sachs genetic testing program has enabled Bernie to determine that he is a
carrier....he has a recessive allele for Tay-Sachs......... (and that his wife, Beverly, is not). What are the changes that they could
have a child with Tay-Sachs?
Bernie is Tt (a carrier of the recessive, but doesn’t have Tay-Sachs)
Beverly is TT
TT X Tt will give offspring with either TT or Tt – none of them will have Tay-Sachs, but 50% will be carriers.
17. Persons who are heterozygous for sickle cell anemia are said to have the sickle trait. What are the chances that Sam and his wife
Ethyl, who both have the heterozygous trait, will have a child with sickle cell anemia (homozygous recessive). Note: The
phenotype for individuals with sickle cell anemia and sickle cell trait are slightly different. This is a codominant disorder
Ss X Ss
Offspring will be 25% SS (normal); 50% Ss (sickle-cell trait); 25% ss (sickle cell anemia)
18. John can roll his tongue, but his mother cannot. He also has attached earlobes. He marries Marsha, who cannot roll her tongue but
has free earlobes. Marsha's father, however, has attached earlobes. What are the chances that their child will have free earlobes
and be able to roll its tongue? (Dihybrid cross)
John can roll his tongue, so he has one R. But his mother is rr, so she gave John a r – he must be Rr. John has attached earlobes,
so must be ee. His genotype for both genes is: Rree
Marsha is rr for rolling and has one dominant E. Because her father has attached earlobes, he must be ee. So, she is Ee. Her
genotype for both genes is: rrEe
Rree X rrEe
John can put the following combinations in his sperm: Re and re (remember, meiosis results in one allele of each gene in a sex
cell)
Marsha can put the following combinations in her eggs: rE and re
Possible offspring: RrEe (roller – free) Rree (roller – attached) rrEe (not roller – free) rree (not roller – attached) 25%
chance
19. Roger is homozygous dominant for Huntington's and has cystic fibrosis. Elvira is homozygous normal for Huntington's but a
carrier of cystic fibrosis. Cystic fibrosis is inherited via a recessive allele. What are the chances that their child has neither
Huntington's nor cystic fibrosis? (Dihybrid cross)
Roger is HH for huntingtons and cc for cystic fibrosis. Elvira is hh Cc
So, Roger can make Hc sperm and Elvira can make hC or hc eggs.
Their children will either be HhCc or Hhcc. All of them will have Huntingtons.
20. Both Willy and Ethyl are heterozygous for earlobe attachment and also for hitchhiker's thumb. What are the chance of a child
with ....attached earlobes and a hitchhiker's thumb? ......free earlobes and no hitchhiker's thumb?
AaBb x AaBb
Chance of attached earlobes is ¼ chance of hitchhiker’s thumb is ¼ so, ¼ X ¼ = 1/16
Chance of free earlobes is ¾
chance of no hitchhiker’s thumb is ¾
so, ¾ X ¾ = 9/16
Or, you can do the 16 square punnett and count.
21. In Guinea pigs rough coat (fur) texture and black coat (fur) are dominant. A rough-coated, black guinea pig, whose mother was
smooth and white is mated with a smooth, white animal. What kinds of offspring would they produce and in what relative
numbers?
RrBb (mom was rrbb, so he must be carrier) x rrbb Offspring: RrBb, rrBb, Rrbb, rrbb in equal number
22. A rough, black guinea pig bred with a rough, white one gives 28 rough, black; 31 rough, white; 11 smooth, black, and 9 smooth
white. Give the genotypes for the parents and each offspring.
RrBb (if there are smooth, white offspring, then he must be heterozygous for both genes) x Rrbb (if there are smooth offspring, then
she must be heterozygous for R)
Offspring: RRBb or RrBb, RRbb or Rrbb, rrBb, rrbb
23. Two rough black guinea pigs when bred together have two offspring, one of them rough white and the other smooth, black. If
these same parents were to be bred together further, what offspring would you expect from them.
RrBb x RrBb parents For Offspring, see #26, 3rd set of parents
24. A purple, smooth Jimson weed plant crossed with white, spiny one gives 320 purple, spiny and 312 purple smooth. If these two
types of offspring are bred together, what will their offspring be like, both as to appearance and as to genotype.
PPSs x ppss Offspring: PpSs and Ppss
F2 Offspring: PPSs, PpSs, ppSs, PPss, Ppss, ppss
Genotypes of F2: purple smooth, white smooth, purple spiny, white spiny
25. Color-blindness is an X-linked recessive. A girl of normal vision whose father was color-blind married a color blind boy. What
type of vision will be expected in their children?
Girl has one XA, but also one Xa because dad is colorblind. So, she’s X A Xa
Boy is colorblind, Xa Y
Eggs: XA or Xa
Sperm: Xa or Y
Offspring: XA Xa (normal daughter) Xa Xa (colorblind daughter) X A Y (normal son) Xa Y (colorblind son)
26. Two normal-visioned parents produce a color-blind son. What are the genotypes of the parents? What are the chances of their
next child being a color-blind daughter?
Parents are XA, but mother also has a Xa, because she must have given this to her colorblind son.
Parents are: XAXa X XAY
Offspring: XAXA (normal daughter), XAXa (normal carrier daughter), XAY (normal son), XaY (colorblind son) So, none of
them will be a colorblind daughter.
27. A woman of normal vision, whose father was color-blind, marries a man of normal vision whose maternal grandfather was colorblind. What type of vision will be expected in their children? Color-blindness is sex linked recessive.
XAXa x XAY
Offspring: XAXA XAXa XAY XaY
28. Yellow body (y)...the recessive allele in Drosophila is a sex-linked allele recessive to gray body (Y)...the dominant allele. A
certain gray female mated with an unknown male produced some yellow and some gray offspring of both sexes. What was the
genotype of the original female? What was the phenotype of the male to which she was mated? Drosophilia and human males are
hemizygous. That is, they carry some alleles on the X chromosome and none of the homologous alleles on the Y.
XYXy x XyY (yellow)
because they had XyXy females, the male parent must have the recessive allele
29. A female fruit fly with sable body (s) is mated with a male having gray body (S). Their daughters are gray, their sons sable. In
what chromosomes is the S gene?
X chromosome. Males give their X to their daughters, so the gray allele is passed to the daughters. The sons are sable because they
get their X from their mothers, who are sable.
30. Albino pigmentation in humans is due to autosomal linked homozygous recessive. Hemophilia is sex linked recessive. An albino,
non-hemophilic man marries a normally-pigmented, non-hemophilic woman whose father was hemophilic and whose mother was
an albino. What kinds of children can they have and in what proportions?
31. A woman whose maternal grandfather suffered from hemophilia has parents that are normal. The woman's husband is normal.
She has a hemophiliac son. What is the chance that the next son will be normal? Will any of the daughters be hemophiliac? Will
any be carriers? Hemophilia is sex linked recessive.
Grandfather: XhY (hemophilia)
Parents: XHXh (Mother must be a carrier to pass it to her daughter, who passes it to her son) XHY (normal father)
Woman: XHXh (she must be a carrier for her son to have inherited the trait)
Husband: XHY (normal)
Son: XhY (hemophilia)
So, a cross of XHXh and XHY
Offspring will be: XHXH (normal daughter), XHXh (carrier daughter), XHY (normal son), XhY (hemophilic son)
47. Fred is type A and his father is type O. Wilma is type B and her father was type O. The Rh blood group is not linked to the ABO
system. It is a two allele group, with positive dominant to negative. Fred is positive, but his mother is negative. Wilma is also
negative. What are the chances that "Pebbles," Fred and Wilma's daughter, will be AB negative
Fred: AO+sperm: A+, A-, O+, O-
Wilma: BO-egg: B-, OPebbles needs a B- egg (1/2 chance) and an A- sperm (1/4 chance) to be AB—
½ x ¼ = 1/8
48. In human beings, migraine (a type of headache) is due to a dominant allele on the X chromosome. A normal vision woman who
has never suffered from migraine takes her daughter to the doctor for an examination. In the course of the examination the doctor
discovers that the girl is color-blind (XaXa), and suffers from migraine (XMXm). What does the doctor automatically know about
the father?
Woman: XmXm
Daugher: XMXmXaXa Mom is homozygous recessive for m, so the daughter got her X m from mom. She must have gotten the
XM from dad. For her to be colorblind, she must have gotten one X a from mom (carrier) and one Xafrom dad. He must be
colorblind.
Father: XMY XaY So, he must have a “M” allele and an “a” allele on his X chromosome.
49. In humans, brown eyes (B) are dominant over blue (b)*. A brown-eyed man marries a blue-eyed woman and they have three
children, two of whom are brown-eyed and one of whom is blue-eyed. Draw the Punnett square that illustrates this marriage.
What is the man’s genotype? What are the genotypes of the children? (* Actually, the situation is complicated by the fact that
there is more than one gene involved in eye color, but for this example, we’ll consider only this one gene.)
Bb man x bb woman; Offspring have 50% Bb and 50% bb
50. In gerbils, the wild type body color is Agouti (brown). There are several genes which contribute to that color, and different alleles
for some of these genes can cause unusual fur colors. One of these genes has a rare mutant allele which causes the fur to be a
lighter, redder color, called copper. This gene has two alleles, its wild type (brown) and copper. Brown is completely dominant to
copper. You are going to mate two brown gerbils (call them Betty and Bert). Both Betty’s mother and Bert’s mother were copper
colored.
a. What color were Betty’s father and Bert’s father? Both were Brown
b. What is Betty’s genotype? What about Bert’s? Both are Bb
c. What genotypic ratio should this mating produce? 25% BB; 50% Bb; 25% bb
d. What phenotypic ratio should this mating produce? 75% Brown; 25% copper
e. Assume you now allow Betty and Bert to mate. Betty eventually has eight babies, five brown and three copper. Does this
agree with the prediction you made (your phenotypic ratio)? If not, what do you think went wrong? Doesn’t quite agree, but
the small number of offspring produced makes it difficult to get the actual to meet the theoretical calculations. Each
offspring has a 3 brown: 1 copper chance. If they had 300 offspring the number would probably be closer to this ratio.
51. This involves the same gerbil gene as the first scenario. You have another brown gerbil, Gertrude. You know nothing about
Gertrude’s family history. If you want to figure out whether Gertrude is homozygous or heterozygous for this gene, what kind of
gerbil should you choose to mate her to, and why? Mate her to a homozygous recessive. If she is heterozygous she should have some
copper babies. If she’s Homozygous dominant she will have only Brown babies.
51. Curled ears in cats is caused by a mutated gene and has been developed into a breed, known as the
American Curl. A cat with curled ears was mated with a cat with normal ears and all of the offspring
have curled ears. Is this trait dominant or recessive? What is the genotype of the curled eared cat?
Dominant trait. The genotype of the curled ear cat must be CC, otherwise she would have some offspring
without curled ears.
have curled ears?
If one of the kittens from the above cross was mated to a cat normal ears what percentage of the offspring will
The offspring will be Cc. The mating will then be Cc x cc (normal); Offspring will be 50% curled, 50% normal.
53. In cats B allele produces black fur and b produces orange fur. This gene is X linked. If both alleles are present calico cats
are produced. A calico cat has 2 calico kittens, 1 male black kitten, and 1 female orange kitten.
a. What is the genotype of the mother__XBXb____phenotype___calico____
b.
What is the genotype of the father___XbY_____phenotype__orange_____
54. Label all the genotypes for the following pedigree.
a.
Is this trait dominant or recessive?
Dominant
b.
How do you know?
No two unaffected parents having affected offspring.