Biology 211: Genetics Problem Set 1

In summer squashes, white fruit is dominant to yellow fruit. The allele that codes for white fruit is represented by W and the allele that codes for yellow fruit by w. What are the genotypes of all possible gametes produced by plants with the following genotypes?

Do Punnett squares for the following crosses and give the fruit color for the offspring.

Hemophilia in humans is due to a sex-linked recessive gene, h. A normal woman with no history of hemophilia in her family marries a normal man who had a hemophilic father. What percentage of their children will have hemophilia?

Use the same information in question #3 but change the man to having a hemophilic mother.

Normal vision (C) in humans is dominant to color blindness © and is sex-linked. A normal-visioned man, whose father was colorblind, marries a colorblind woman. What are the chances that a son will be colorblind? What are the chances that a daughter will be colorblind?

The determiner for brown eyes (B) is dominant to blue eyes (b) and is not sex linked. A colorblind man with brown eyes, whose mother was blue-eyed, marries a normal-visioned, blue-eyed woman, whose father was colorblind. Show the expected phenotype ratio of their children involving eye color and color blindness.

Blood types in humans are caused by a combination of two of three possible alleles I^A, I^B and i. The i allele is recessive and when an individual is homozygous for this allele, they have type O blood. Blood type A may be due to a homozygous condition (I^A, I^A) or the heterozygous genotype (I^A, i). Blood type AB is caused by having a copy of both the I^A and I^B alleles. Rh factor is another marker on human red blood cells and is either positive (dominant R) or negative (recessive r). Give all possible genotypes for each blood type regarding I^A, I^B and i alleles.

A brown-eyed man marries a blue-eyed woman and they have eight children, all brown eyed. Accepting the hypothesis that a single gene codes for eye color, what would be the genotypes of each individual in this family?

A blue-eyed man, whose parents were brown-eyed, marries a brown-eyed woman whose father was brown-eyed and whose mother was blue-eyed. They have one child, who is blue-eyed. Assuming these traits are controlled by a single pair of alleles, B and b as in #6, what are the genotypes of all the individuals?

In incomplete dominance the heterozygous condition results in some intermediate trait. In humans, curly hair is homozygous dominant and straight hair is homozygous recessive. The heterozygous individual will have wavy hair. Using the alleles C and © give the genotypes for each of the above three phenotypes. If a father has wavy hair and the mother has wavy hair give the percentage of the offspring with each genotype and phenotype.

WW : W
homozygous dominant parent: since both alleles of the parent are W, all gametes will have the W allele
Ww : W & w
heterozygous parent: since W & w alleles are present in the parent, both W & w gametes can be produced
ww : w
homozygous recessive parent: since both alleles of the parent are w, all gametes will have the w allele

Ww x ww
WW x Ww
WW x ww

Ww x ww
WW x Ww
WW x ww

: Sex-linked genes are generally carried on the X chromosome. Males ordinarily are more likely to inherit X-linked disorders since they only have one X chromosome (& one Y chromosome), whereas females have 2 X chromosomes. The term "carrier" is generally used to describe a woman with one normal X chromosome & one X chromosome carrying an allele for an X-linked disorder. In most cases, a carrier will exhibit no signs or very slight signs of the disorder, but can pass on the disorder to offspring. The designation for an X-linked disorder is generally added as a superscript to the X chromosome...
So for hemophilia, the normal allele is X^H and the hemophilic allele is X^h.; Since there's no history of hemophilia in the woman's family, we'll assume she has 2 normal alleles (X^HX^H). The fact that the man's father had hemophilia has no bearing on his genotype, since the gene for hemophilia is carried on the X chromosome & ordinarily a man inherits only a Y chromosome from his father & the X chromosome from his mother. Regardless, since the man is normal, his copy of the X chromosome must be normal (X^H) & his genotype X^HY.

: Ignoring all other factors, if the man's mother had hemophilia (a very rare disorder in women, since it requires a hemophilic father & the mother at least be a carrier), he must now have hemophilia (X^hY). Since males inherit their X chromosome from the mother, & the mother must have had 2 hemophilic alleles (X^hX^h) to have hemophilia, the man must have inherited a hemophilic allele (X^h) from his mother. The woman we'll still assume to be normal (X^HX^H).

: Again, the fact that the man's father was colorblind has no bearing on his genotype, since colorblindness is an X-linked recessive disorder... so the gene for colorblindness is carried on the X chromosome, which the man inherited from his mother. If the man has normal vision, his genotype must be X^CY. The woman is colorblind, so her genotype must be X^©X^©.

: Here we're looking at two traits for each individual, so things get a little tricky. Each individual will have 2 alleles for each trait, & they'll get placed together for a total of 4 alleles for each individual. We're told that eye color is controlled by one gene & is not sex-linked, while we know now colorblindness is sex-linked. The colorblind man will have the genotype X^©Y for colorblindness, but there's also eye color to deal with. He's said to have brown eyes, & since we know his mother had blue eyes (which would be bb), it follows that his genotype for eye color is Bb (must have at least one B for brown eyes & must have received one b from his mother with blue eyes). His wife has blue eyes, which again must be bb, & normal vision, which could be X^CX^C or X^CX^©. Since her father was colorblind (X^©Y like her husband), she must have received a X^© from her father, so her genotype is X^CX^©. Putting the genotypes together, the man's genotype is X^©YBb & the wife's genotype is X^CX^©bb.

This type of cross, looking at two different traits at once, is called a dihybrid cross. The punnett square is below... it normally has 16 rather than 4 squares, as there are 4 possible gametes for each parent. Each gamete has one allele for each trait. To determine the gametes for each parent, each allele for one trait is paired with each allele for the other trait. This is commonly done using the "FOIL" method (First, Outer, Inner, Last), pairing the first allele for each trait, the two outer alleles, the two inner alleles, & the last allele for each trait.; Cross: X^©YBb x X^CX^©bb

A man that has blood type A negative, whose father was B positive and whose mother was AB positive, married a woman that was B positive. The woman's mother was A negative and her father was AB positive. Give the genotypes of all the above mentioned individuals.
Give the genotypes and phenotypes of the possible offspring from this marriage.

O-:	iirr
O+:	iiRR, iiRr
A-:	I^AI^Arr, I^Airr
A+:	I^AI^ARR, I^AI^ARr, I^AiRR, I^AiRr
B-:	I^BI^Brr, I^Birr
B+:	I^BI^BRR, I^BI^BRr, I^BiRR, I^BiRr
AB-:	I^AI^Brr
AB+:	I^AI^BRR, I^AI^BRr

A man that has blood type A negative, whose father was B positive and whose mother was AB positive, married a woman that was B positive. The woman's mother was A negative and her father was AB positive. Give the genotypes of all the above mentioned individuals.

Man: I^Airr
  Man's father: I^BiRr
  Man's mother: I^AI^BRr Woman: I^BiRr
  Woman's mother: I^Airr
  Woman's father: I^AI^BRR  or  I^AI^BRr
Give the genotypes and phenotypes of the possible offspring from this marriage.

: We'll call the dominant allele B for Brown eyes... the recessive allele will be b. Since the mother has blue eyes (bb), each child will receive a b allele from the mother. Although we aren't given the genotype for the father (with brown eyes, the father can have either genotype BB or Bb), if all 8 children are to have brown eyes, it is statistically most likely the father's genotype is BB. In this case, all of the children would be heterozygous for eye color (Bb) with brown eyes.; If the father's genotype was Bb (heterozygous for brown eye color), statistically only 50% of the children would have brown eyes (the brown-eyed children would be heterozygous with genotype Bb).

: If this is indeed a case of simple dominance, with B as the dominant allele specifying brown eyes & b specifying blue eyes, an individual with the recessive phenotype (blue eyes) will have the genotype bb (it doesn't matter if his or her mother, father or pet gopher has brown eyes, it'll still be bb), while those with brown eyes could have the genotype BB or Bb, pending additional information.
Here, it's stated that both of the man's parents had brown eyes... in order for the man to have blue eyes, both parents must have been heterozygous (Bb), since each had to contribute a b to the man's genotype. The wife has brown eyes also, but noting that their son has blue eyes, her genotype must also be Bb, since she contributed a b to her son's genotype. Rounding things up, the wife's mother with blue eyes had the genotype bb & her father could have had the genotype BB or Bb (we can't know for sure without additional information about his parents or the wife's siblings).

Man: bb
  Man's father: Bb
  Man's mother: Bb

Wife: Bb
  Wife's mother: bb
  Wife's father: BB or Bb

Son: bb

: Genotype for curly hair (homozygous dominant):  CC
Genotype for straight hair (homozygous recessive):  ©©
Genotype for wavy hair (heterozygous):  C©; Both the father & mother are heterozygous (C©) with wavy hair...