Mendelian Genetics Problems: (20 points)

Answers are in blue.

Directions- Answer the following questions. Feel free to work in groups to find and understand the answers, but please indicate whom you worked with above. This assignment is due Thursday if you do not complete in class.

1. Use a Punnett square to predict the ratio of offspring genotypes and phenotypes for the following parents: Bb X bb, where B is black and b is white. (2)

Gametes:	B	b
b	Bb (black)	bb (white)
b	Bb (black)	bb (white)

1 Bb : 1 bb is the genotype ratio

1 Black : 1 white is the phenotype ratio

2. List all the possible gametes that the following genotype can produce from independent assortment assuming that each gene is on separate chromosomes: AABbCcdd (1).

ABCd, ABcd, AbCd, Abcd (no other possible combinations can occur)

3. What are linked genes? What is genetic recombination? (2)

Linked genes are those genes found on the same chromosome, thus where ever that chromosome goes so do all those genes’ alleles (for example recall the blonde-blue hair example)

Genetic recombination occurs when linked alleles become separate as a result of crossing over during meiosis. (that’s when you could get a brunett with blue eyes in my make believe example)

4. If the genes for A and B alleles are linked what would the possible gamete genotypes be for the an individual with the following genotype if crossing over does not occur?(1) AaBb [It might help to draw chromosomes with loci for each gene]

Assume dominant alleles are on the same chromosome and recessive are on the other homologous chromosome. Thus you can only get two possible combinations of gametes:

AB and ab (you cannot get Ab or aB unless a crossing-over event occurs so that the different linked alleles are exchanged).

5. Most harmful genetic diseases are recessive traits. Why do you think that is? (1)

Well imaging if you have a dominant deadly disease. If you had just one allele you would have the disease and thus you would die. As a result you would not pass your genes to the next generation so that allele would die with you! Dominant deadly diseases selected themselves out of the population through evolution! The individuals that would survive to pass there genes on to the next generation would be homozygous recessives thus the next generation would only contain the healthy recessive gene. An alternative answer is to address how deadly mutations in genes typically manifest. A deadly genetic disease is typically the result of a broken gene that does not produce the functional protein it is suppose to. If you are carrying at least one working gene than that gene will be dominant over the “broken” gene.

6. Cross the following parents and give the expected genotype and phenotype ratios of the offspring: AaBb X Aabb (3) A

Gametes	AB	Ab	aB	ab
Ab	AABb	AAbb	AaBb	Aabb
ab	AaBb	Aabb	aaBb	aabb

Notice that for the second parent only two possible gamete combinations are possible so you only need two rows in the punnett square.( Recall that A is brown hair, a is blond, B in brown eyes, b is blue eyes)

Here are the possible genotypes: 1 (AABb) : 2 AaBb: 1 AAbb : 2 Aabb : 1 aaBb : 1 aabb

notice that the sum of the ratio totals the number of cells in the punnett square.

The possible phenotype are: 3 brown hair, brown eyes (2 AABb, and 1 AaBb) : 3 Brown hair, blue eyes (1 AAbb, 2 Aabb) : 1 Blond brown-eyes (aaBb) : 1 blond, blue eyes (aabb).

7. Imagine that you bred two fruit flies: One that is red-eyed and the other is white eyed. You find that you get all red-eyed offspring in the F1 generation.

a. Which allele for eye-color is dominant? How do you know? (1)

Red must be dominate since the white parents trait is suddenly “hidden” in all offspring.

b. What are the likely genotypes of the parents? P 1__AA__ P 2___aa__ (1)

The white parent is easy since aa is the only genotype that would be white. choosing between AA and Aa for the first parent involves compare the possible offspring a cross with the other parent would produce. If it were Aa then half the offspring would be white… this was not the case so the parent must be AA.

c. What would be the phenotypic ratio of the F2 generation? (1)

Recall that an F2 cross is a cross between the offspring of an F1 cross. So, to answer this question you must first determine what the offspring look like. A cross between the parents would produce all heterozygous (Aa) offspring. A cross between two of these individuals would produce 1 AA : 2 Aa : 1 aa or a 3:1 ratio of red to white-eyed offspring (do the punnett square to convince yourself)

8. In snapdragon plants flower color can exhibit incomplete dominance. If you cross fertilize a heterozygous pink snapdragon with a homozygous recessive white snapdragon, what percentage of the offspring will have pink flowers? (2)

To do this cross you must first determine what the genotypes of the parents are. In this case, this is not difficult since all three possible genotype produce unique phenotypes (red AA, pink Aa or white aa). Thus the parent in this example are Aa (pink) and aa White. A punnett square will show you that half the offspring will be Aa (pink) and half will be aa (white). So half or 50% will be pink.

9. Why do we expect half of our children to be male and half to be female? Why doesn’t this always happen in families with even numbers of children? (2)

Sex is determined by whether you have 2 X chromosomes (female condition in humans) or one X and one Y chromosome (males in humans). All eggs (female gametes) will have X chromosomes, but half the sperm will carry the X and half will carry the Y chromosome. Thus there is a 50:50 chance depending on which sperm penetrates the egg first of which sex the resulting embryo will be. The reason why we don’t see this perfect 50:50 ratio in families or even larger populations for that matter, is the same reason why tossing a coin will not always give you 50% heads and 50% tails… there is chance and each event is independent of the other. If you have 6 girls the chances that your next child will also be a girl is still 50%! Just like by pure luck you toss a coin six times and get heads all 6 times. The chance of getting heads on your 7^th toss is still 50%.

10. Explain how a recessive trait can be expressed in a male when he carries only one such recessive allele? (hint: what chromosome must the gene be on? (1)

For a recessive trait to be express even though only one copy exists, it must exist on the sex chromosomes. In fact it like is found on the X chromosome since the Y chromosome carries very few genes. Since males only have one X chromosome, and allele regardless of dominance will be expresses since it is the only copy the male individual has. This is, in part any way, why male pattern baldness exists- you inherit male baldness from your mother not your father.

11. Is an organism the sum of all its expressed genes? Why or why not? (1)

No. They are the sum of all the genes plus the interactions between those genes and between those genes and the environment.

12. How many possible combinations of chromosomes can a human couple produce in their offspring? (1) The book states that there are more than 64 trillion possible combinations of each chromosome. Here how they figured this: Each parent individual donates 23 chromosomes. For just a single individual there are 2²³ possible combinations of chromosomes in the gametes, since there are 2 varieties of each chromosome. That equates to more than 8 million combinations. For two individuals, that would be 8 million multiplied by itself, which is 64 trillion.

Bonus: A breeding experiment was performed to attempt to determine the genotype of two adult fruit flies. The two adult flies were crossed and the total number of all their offspring along with the parental phenotypes are below (4 points):

Sex Wings Body

Parents: male wrinkled brown

X

female wrinkled yellow

Offspring: 143 male wrinkled brown

148 female wrinkled brown

58 male normal brown

51 female normal brown

Total: 400

Answer the following questions:

a. Based on these results are the traits autosomal or sex-linked or both? How do you know? (1) To answer this you have to pay attention to the sexes and the phenotypes. If a trait was linked to a sex chromosome males and females would have different phenotypic ratios since males would in reality by haploid for those traits. In the above case males and females express the traits in about the same proportions. So, there is no indication that these are sex-linked traits. Thus they are autosomal.

b. What trait for each character is dominant? Explain how you determined this. (1)

Look at each trait separately. We crossed a brown body with a yellow body, but all offspring were brown bodies. That tells us the brown is dominant to yellow since yellow was hidden in the offspring population. For the wings, we crossed two wrinkled wing flies and got wrinkled wing and normal winged individuals. For normal wings individuals to appear in the offspring the parent must be carriers of that trait and it is hidden by the dominant wrinkle phenotype. Thus wrinkle wings and brown bodies are dominant.

c. What should the genotypes be for each parent (Use the symbol ‘A/a’ for wing traits and ‘B/b’ for body color). (1) To get all brown bodies offspring the parents must be homozygous for each allele. The ratio of wrinkled to normal wings is about a 3:1 ratio. This is what you would expect if both parents were heterozygous for this trait. Thus the genotypes must be those given below:

Male: AaBB Female: Aabb__

d. What would the expected phenotypic ratio be for this cross? (1)

The expected phenotype ratio can be determined by completing a punnett square for a cross between these two genotypes:

	AB	aB
Ab	AABb (wrinkled- brown)	AaBb (wrinkled- brown)
ab	AaBb (wrinkle- brown)	aaBb (normal- brown)

3 wrinkled brown: 1 normal-brown. Notice the true numbers resemble that ratio (150 is three time 50).