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Basic Heredity Worksheet

 

DETERMINING SYMBOLS FOR GENES

For each of the following, give the symbols to all the alleles that are consistent with the methods you learned in lecture.

  1. The mother in a family has dimples. The father does not, but all of their 8 children have dimples. A single gene controls dimples.
  2. D=dimples; d=no dimples

  3. There are two different pairs of genes in peaches. One pair determines whether there will be glands at the base of the leaves (presence of glands is dominant to absence), the other determines skin type (fuzzy is dominant to smooth skin).
  4. G=glands; g=no glands

    F=fuzzy; f=smooth

  5. Sickle-cell anemia is related to one autosomal gene in humans. People homozygous for the sickling allele have mostly sickled red blood cells and sever anemia. People who are heterozygous for this gene have a few sickled cells and mild anemia.
  6. Bs=sickle cell blood cells; Bn=normal blood cells

  7. In cattle, when purebred red-colored bulls are mated with white cows, the offspring are roan with both red and white hairs in the coat.
  8. Cr=red color; Cw=white color

  9. Handedness in humans is hereditary with right-handedness dominant to left handedness.
  10. R=right handed; r = left handed

  11. Cystic fibrosis is due to the expression of a recessive autosomal allele.

C=normal; c=cystic fibrosis

 

PREDICTING GENOTYPES

Using the traits and symbols that you developed in the previous section, predict the genotypes of the individuals in the following.

  1. Two normal parents have a child with cystic fibrosis. What are the genotypes of the parents and the child?
  2. parents: Cc x Cc child: cc

  3. A right-handed man has a left-handed mother. What is his genotype?
  4. mother = rr; man = Rr

  5. Two left-handed parents have children. What are the possible genotypes of the children?
  6. parents: rr x rr kids: rr

  7. A peach tree has glands and produces fuzzy fruit. What are the possible genotypes of this tree?
  8. tree: F_G_ possibilities: FFGG, FFGg, FfGG, FfGg

  9. The tree in #4 was cross-pollinated with a tree lacking glands and producing smooth fruit. The resulting trees were of four different phenotypes
    1. glands and fuzzy fruit F_G_
    2. glands and smooth fruit ffG_
    3. no glands and fuzzy fruit F_gg
    4. no glands and smooth fruit ffgg

What are the genotypes of each of these offspring?

What is the genotype of each of the parents? FfGf x ffgg

 

STEPS IN PROBLEM SOLVING

The following are some steps in solving genetics problems:

Steps:

3 Determine the possible gametes that may be formed by each parent.

2 Write out the cross using genotypes of the parents

6 Determine the genotypic ratio of the offspring

4 Set up a Punnett Square with the gametes entered along the left side and across the top of the matrix.

7 Determine the phenotypic ratio of the offspring.

5 Fill in the cells of the Punnett Square to determine what offspring genotypes should be expected from the cross.

1 Determine the correct symbols for the alleles.

 

GAMETE DETERMINATION

For each of the crosses below, determine the gametes that would be formed by the parents and set up a Punnett Square. You do not need to fill in the square now, but save them for the next section.

  1. A mother that is heterozygous for the gene for dimples with her husband who has no dimples.
  2. parents:

    Dd

    dd

    gametes:

    D, d

    d

     

    d

    D

     

    d

     

     

  3. A man with no sickled cells in his blood with a woman having a few sickled cells and mild anemia.
  4. parents:

    BnBn

    BnBs

    gametes:

    Bn

    Bn, Bs

     

    Bn

    Bn

     

    Bs

     

     

  5. A peach tree that is heterozygous for both the gland gene and the skin-type gene, with a tree that is homozygous recessive for both genes.
  6. parents:

    FfGg

    ffgg

    gametes:

    FG, fG, Fg, fg

    fg

     

    FG

    fG

    Fg

    fg

    fg

     

     

     

     

  7. A roan cow with a white bull.
  8. parents:

    CRCW

    CWCW

    gametes:

    CR,CW

    CW

     

    CW

    CR

     

    CW

     

  9. A homozygous right-handed person and a left-handed person.
  10.  

    parents:

    RR

    rr

    gametes:

    R

    r

     

    r

    R

     

  11. A woman who is heterozygous with blood type A and a man who is AB.

parents:

IAi

IAIB

gametes:

IA, i

IA, IB

 

IA

i

IB

 

 

IA

 

 

 

GENETICS PROBLEMS

For each of the problems below:

      1. State the category of which it is an example (monohybrid cross with dominance, codominance, dihybrid cross, multiple alleles, sex-linked, other.)
      2. State what clues led to your decision.
      3. Solve the problem showing your work, step by step.
  1. Feathering on the legs of chickens is due to an allele, F. Absence of feathers is due to f. A heterozygous hen with feathers on her legs is mated with a rooster that has bare legs. Determine the expected genotypic and phenotypic ratios of the chicks produced by the mating.
  2. monohybrid cross

    parents:

    Ff

    ff

    gametes:

    F, f

    f

     

    f

    F

    Ff

    f

    ff

    genotypic ratio: 1 Ff : 1ff

    phenotypic ratio: 1 feathered : 1 bare

  3. One human hereditary condition is cystic fibrosis in which there is abnormally thick mucus produced with blocks the lung passageways. This leads to impaired breathing and infections. It is usually fatal at an early age. Cystic fibrosis (CF) is due to the expression of a recessive autosomal gene c. A man and woman planning to marry are concerned about their chances of having a CF child because both the man’s sister and the woman’s brother had CF. Both the man and woman have normal phenotypes.

monohybrid cross

  1. What are the genotypes of the man’s parents? Cc, Cc
  2. What are the genotypes of the woman’s parents? Cc, Cc
  3. What is the man’s genotype? C_ (either CC or Cc)
  4. What is the woman’s genotype? C_ (either CC or Cc)
  5. Determine the chances that the couple would have a CF child.

man has 50% chance of being Cc; woman has 50% chance of being Cc; probability that both man and woman are Cc = 0.5 x 0.5 = 0.25 (25%)

if both parents are Cc,

 

C

c

C

CC

Cc

c

Cc

cc

probability of having a cc child is 25%; therefore total probability is 0.25 x 0.25 = 0.625 or 6.25%

  1. Radishes are long, round or oval. When a plant producing long radishes is crossed with one producing round radishes, what will be the genotypes and phenotypes of the offspring? If two of these F1 offspring were crossed, what will be the outcome? Show genotypic and phenotypic ratios for both crosses.
  2. codominance; SL = long; SR = round

    parents:

    SL SL

    SR SR

    gametes:

    SL

    SR

     

    SL

    SR

    SLSR

    all offspring are SLSR; all are oval

    F1:

    SLSR

    SLSR

    gametes:

    SL, SR

    SL, SR

     

    SL

    SR

    SL

    SLSL

    SLSR

    SR

    SLSR

    SRSR

    F2 genotypic ratio = 1 SLSL : 2 SLSR : 1 SRSR

    F2 phenotypic ratio = 1long : 2 oval : 1 round

  3. Two roan cattle are mated repeatedly and the calves in the F1 are three different colors, red, white and roan. What are the genotypes of the parents? What is the probability of getting each of these colors (the expected phenotypic ratio)?
  4. codominance

    parents:

    CRCW

    CRCW

    gametes:

    CR,CW

    CR,CW

     

    CR

    CW

    CR

    CRCR

    CRCW

    CW

    CRCW

    CWCW

    genotypic ratio = 1 CRCR : 2 CRCW : 1 CWCW

    phenotypic ratio = 1 red : 2 roan : 1 white

  5. Feathering on the legs of chickens is dominant to absence of feathering (bare legs). Under the control of a second gene, a cap-shaped comb, called pea comb, is dominant to the familiar single erect comb. Give abbreviations to the alleles that effect these four phenotypes. When hens that are heterozygous for both leg feathering and pea combs are crossed with roosters that have no leg feathers and single erect combs, what are the expected phenotypic and genotypic ratios among the chicks?
  6. dihybrid cross; F = feathers; f = bare; C = pea comb; c = erect comb

    parents:

    CcFf

    ccff

    gametes:

    CF, cF, Cf, cf

    cf

     

    CF

    cF

    Cf

    cf

    cf

    CcFf

    ccFf

    Ccff

    ccff

    genotypic ratio: 1 CcFf : 1 ccFf : 1 Ccff : 1 ccff

    phenotypic ratio: 1 with feathers and pea comb : 1 with feathers and erect comb : 1 with bare legs and pea comb : 1 with bare legs and erect comb

  7. The gene locus for color vision/color-blindness is on the X chromosome, but is absent on the Y. A color-blind man marries a woman with normal vision whose father was color-blind. What are the chances that their male children will be color-blind? Their female children? Give the genotypic ratios and phenotypic ratios by sex as well as color-blindness.
  8. sex-linked; Xc = color blind; XC = normal

    parents:

    XcY

    XCXc

    gametes:

    Xc, Y

    XC, Xc

     

    Xc

    Y

    XC

    XCXc

    XCY

    Xc

    XcXc

    XcY

    genotypic ratio = 1 XCXc : 1 XcXc : 1 XCY : 1 XcY

    phenotypic ratio = 1 normal female : 1 colorblind female : 1 normal male : 1 colorblind male

    chances of male kids being colorblind = 50%

    chances of female kids being colorblind = 50%

  9. Singing voice is influenced by a gene with two autosomal alleles (va and vb). Men and women who are homozygous for va tend to be tenors and altos, respectively. Those homozygous for vb tend to be basses and sopranos, respectively. Those who are heterozygous tend to be baritones and mezzo-sopranos.
  1. Two professional singers, a baritone and a soprano, marry and have eight children, half boys and half girls. How many of each voice type would you expect among the children?
  2. If a baritone and mezzo-soprano marry and have four boys, how many of each voice type would be expected?

codominance (and sex influenced)

a:

parents:

vavb

vbvb

gametes:

va,vb

vb

 

va

vb

vb

vavb

vbvb

genotypic ratio = 1 vavb : 1 vbvb

phenotypic ratio = 1 baritone or mezzo-soprano : 1 bass or soprano

in eight kids: expect 2 baritones, 2 mezzo-sopranos, 2 basses and 2 sopranos

b.

parents:

vavb

vavb

gametes:

va,vb

va,vb

 

va

vb

va

vava

vavb

vb

vavb

vbvb

genotypic ratio = 1 vava : 2 vavb : 1 vbvb

phenotypic ratio (for males only)= 1 tenor : 2 baritones : 1 bass

  1. One gene with two alleles controls the expression of red blood cell shape in humans. There are three different phenotypes: 1. Normal blood cells only resulting in no anemia, 2. A few sickled cells resulting in mild anemia, and 3. A usually fatal condition with mostly sickled cells and severe anemia.
  2. Three couples go to get genetic counseling before marriage. In the first couple, both people are homozygous for the normal allele. In the second, the man is heterozygous, while the woman is homozygous for the normal allele. In the third couple, both parents are heterozygous. Show the expected genotypic and phenotypic ratios for the children of each couple. What is the probability that each couple will have a child with fatal sickle cell anemia?

    these are all monohybrid crosses

    parents:

    BnBn

    BnBn

    gametes:

    Bn

    Bn

     

    Bn

    Bn

    BnBn

    genotypic ratio = all BnBn

    phenotypic ratio = all normal

    parents:

    BnBn

    BnBs

    gametes:

    Bn

    Bn, Bs

     

    Bn

    Bn

    BnBn

    Bs

    BnBs

    genotypic ratio = 1 BnBn: 1 BnBs

    phenotypic ratio = 1 normal : 1 mild anemia

     

    parents:

    BnBs

    BnBs

    gametes:

    Bn, Bs

    Bn, Bs

     

    Bn

    Bs

    Bn

    BnBn

    BnBs

    Bs

    BnBs

    BsBs

    genotypic ratio = 1 BnBn: 2 BnBs : 1 BsBs

    phenotypic ratio = 1 normal : 2 mild anemia: 1 sickle cell anemia

  3. a. For a gene with two alleles, one dominant and one recessive, the phenotype of a heterozygous individual is indistinguishable from that of a homozygous dominant. Why is this so?

the recessive allele is not expressed in the heterozygote

  1. A farmer says that the way to determine if an individual with dominant phenotype is homozygous or heterozygous is to cross it with a homozygous recessive individual. Is the farmer correct or not? Give sample crosses to prove your answer.

you are testing to see if the genotype is Aa or AA

the farmer says to cross this individual with aa

monohybrid cross

if individual is AA:

parents:

AA

aa

gametes:

A

a

 

A

a

Aa

genotypic ratio: all Aa

phenotypic ratio: all dominant phenotype

if individual is Aa

parents:

Aa

aa

gametes:

A, a

a

 

a

A

Aa

a

aa

genotypic ratio: 1 Aa : 1 aa

phenotypic ratio: 1 dominant : 1 recessive

so the farmer is correct!

  1. In peach trees, one gene determines whether or not there will be glands at the base of the leaves (presence of glands is dominant). Another gene determines whether the peaches will have smooth (recessive) or fuzzy (dominant) skins.

dihybrid cross

  1. What kinds of gametes can be produced by plants with the following genotypes: FfGg, FFGg, Ffgg, ffgg.
  2. FfGg – FG, Fg, fG, fg

    FFGg – FG, Fg

    Ffgg – Fg, fg

    ffgg – fg

  3. What are the possible genotypes of a tree which has glands and fuzzy-skinned fruit? Glands but smooth-skinned fruit?
  4. glands and fuzz: FFGG, FfGG, FFGg, FfGg

    glands and smooth: ffGG, ffGg

  5. One of the glandular, fuzzy-fruited plants in part b. above is a particularly good specimen. The fruit farmer decides to use it as a parent to produce similar offspring, but wishes to determine if it is heterozygous or homozygous for the gland and skin genes. Explain how he could do it, showing the possible outcomes.

tree: F_G_

to determine if heterozygous or homozygous, cross with ffgg

if FFGG:

parents:

FFGG

ffgg

gametes:

FG

fg

 

FG

fg

FfGg

all have fuzz and glands

if FfGG:

parents:

FfGG

ffgg

gametes:

FG, fG

fg

 

fg

FG

FfGg

fG

ffGg

½ offspring have fuzz and glands, the other half have smooth with glands

if FFGg:

parents:

FFGg

ffgg

gametes:

FG, Fg

fg

 

fg

FG

FfGg

Fg

Ffgg

½ offspring have fuzz and glands, the other half have fuzz with no glands

if FfGf:

parents:

FfGg

ffgg

gametes:

FG, Fg, fG, fg

fg

 

fg

FG

FfGg

Fg

Ffgg

fG

ffGg

fg

ffgg

¼ will have fuzz and glands, ¼ will have fuzz and no glands

¼ will be smooth with glands, ¼ will be smooth and have no glands

  1. A couple living together has a child with B blood type. The man has blood type AB and the woman has type O blood.

multiple alleles

  1. What is the genotype of the child?
  2. Woman must be genotype: ii

    Man must be genotype: IAIB

    parents:

    IAIB

    ii

    gametes:

    IA, IB

    i

     

    i

    IA

    IAi

    IB

    IBi

    The child must be genotype IBi

  3. A question of paternity arises when the mother seeks child support payments after the couple splits up. The father claims that the woman had been unfaithful and the child was fathered by an acquaintance whose blood type is A. Is this possible? Explain.

Acquaintance must be either IAIA or IAi. Either way he could not give his child the IB allele which would be necessary for the kid to have B blood type.

  1. What are the possible genotypes and phenotypes for children of a couple that have blood types A and B if both are heterozygous for the blood type gene? If both are homozygous for the blood type gene?
  2. multiple alleles

    if homozygous for the traits:

    parents:

    IAIA

    IBIB

    gametes:

    IA

    IB

     

    IB

    IA

    IAIB

    all kids are genotype IAIB and blood type AB

    if heterozygous for the trait

    parents:

    IAi

    IBi

    gametes:

    IA, i

    IB , i

     

    IB

    i

    IA

    IAIB

    IAi

    i

    IBi

    ii

    genotypic ratio = 1 IAIB : 1 IAi : 1 IBi : 1 ii

    phenotypic ratio = 1 type AB : 1 type A : 1 type B : 1 type O

  3. Determine which of the following are true and which are false:
  1. A type O child may have two type A parents. true (IAi x IAi à ii)
  2. A type O child may have one type AB parent. false IAIB cannot give allele "i"
  3. A type A child may have two type AB parents. true IAIB x IAIB à IAIA
  4. A type AB child may have one type O parent. false IAIB child cannot get an allele from an ii parent

 

  1. Hemophilia (bleeder’s disease) results from the expression of a sex-linked recessive gene h. Women who are heterozygous for the H/h gene have normal blood clotting, but are called carriers because they can pass the recessive allele on to their offspring.

sex-linked XH = normal Xh = hemophilia

  1. A normal man marries a woman who is a carrier. If they have four boys, what is the chance that one will be a hemophiliac? If they had four girls, what is the chance that one will be a hemophiliac?
  2. XHY x XHXh

    parents:

    XHY

    XHXh

    gametes:

    XH, Y

    XH, Xh

     

    XH

    Y

    XH

    XHXH

    XHY

    Xh

    XHXh

    XhY

    if they have 4 boys, 1/2 are expected to be hemophiliacs

    if they have 4 girls, none will be hemophiliacs

  3. Is it more common for male or for female offspring of a carrier to be hemophiliacs? Why? male offspring of carriers are more likely to be hemophiliacs because their only X chromosome is from their mother. (It is also unlikely that males that are hemophiliacs will mate… therefore they will not pass on their X chromsome to their daughters)
  4. A king with hemophilia married a normal woman who is not a carrier. They have two princes and two princesses. What proportion of each sex are expected to have hemophilia? To be carriers?

parents:

XhY

XHXH

gametes:

Xh, Y

XH

 

Xh

Y

XH

XHXh

XHY

none of the children will have hemophilia (but all of the daughters will be carriers of the disease)

 

  1. The ability to taste PTC (phenylthiocarbamide), an antithyroid drug with slows thyroid function, is dominant over non-tasting. Also free earlobes in people is dominant over attached earlobes.

dihybrid cross T = taster, t = non-taster, F = free earlobes, f = attached earlobes

  1. A male taster with free earlobes marries a female non-taster with attached earlobes. All their offspring are tasters with free earlobes. What are the genotypes of the father, mother and children?
  2. male is either FFTT or FFTt or FfTt or FfTT

    female is fftt

    all kids get ft from their mom

    because they are all tasters with free earlobes, they must have gotten FT from their dad (and therefore are genotype FfTt)

    any of the proposed genotypes for dad is possible. Depending on the number of offspring, it is likely that dad is FFTT… but we can't know for sure!

  3. A man and woman heterozygous for both genes have children. What are the phenotypes of the parents? What are the possible genotypes and phenotypes of the children?

man and woman both have genotype FfTt

parents:

FfTt

FfTt

gametes:

FT, Ft, fT, ft

FT, Ft, fT, ft

 

FT

Ft

fT

ft

FT

FFTT

FFTt

FfTT

FfTt

Ft

FFTt

FFtt

FfTt

Fftt

fT

FfTT

FfTt

ffTT

ffTt

ft

FfTt

Fftt

ffTt

fftt

the possible genotypes of the children are shown in the above table (Punnett square)

the phenotypic ratio is:

9 tasters with free lobes : 3 tasters with attached lobes :

3 non-tasters with free lobes : 1 non-taster with attached lobes

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