Monohybrid Cross Practice Problems Worksheet

# Monohybrid Cross Practice Problems Worksheet

Punnett squares
Original parents in any given set of crosses are called the parent generation or parentals, while the two subsequent generations are denoted with the symbols F1 and F2 (a cross of two F1 individuals).Punnett Squares are one method for visually demonstrating the probability of offspring genotypes and offspring phenotypes.

Example 1: (Monohybrid Cross) For humans, brown eyes are dominant (B) over blue eyes (b).A heterozygous brown-eyed man marries a heterozygous brown-eyed female.What are the possible genotypes and phenotypes of the offspring?

Parents: Male = Bb; Female = Bb
Record the probabilities for genotypes and phenotypes of the offspring (F2 generation) as percents and ratios.Use the following format to write genotypic ratios: homozygous dominant: heterozygous: homozygous recessive.

Use the following format to write phenotypic ratios:

dominant phenotype: recessive phenotype.
Monohybrid Practice Problems 1.Cystic fibrosis is a recessive genetic disorder.Ron is homozygous dominant (FF) and Nancy is a carrier (Ff) of cystic fibrosis.Use a Punnett square to predict the probability that one of their children will have cystic fibrosis? Show all work and box your final answer.

2.

Patty is homozygous dominant for freckles (SS), while Charlie is homozygous for no freckles (ss).Draw a Punnett square predicting the probability if their children will have Percents

Phenotypic Percents

Genotypic Percents

Phenotypic Percents

B
b Conv2.Dominant allele (upper case) written befo-Linked Worksheet 2 3.Eddie has brown eyes, while Cybil has blue.If brown eyes are known to be dominant, and blue eyes are recessive, use a Punnett square to predict their offspring.Assume Eddie doesnt carry a recessive allele.
4.Larry and Lola Little have achondroplasia, a form of dwarfism.

Both are heterozygotes.

Their son, Big Bob Little, is 71.Use a Punnett square to show how Big Bob got his genotype.
5.Woody Guthrie, who wrote This Land is Your Land, was heterozygous for Huntingtons disease (Hh).His wife was homozygous recessive and perfectly normal (hh).Huntingtons disease is caused by a latent dominant gene, meaning that it is not phenotypically (physically) expressed until later in life.

Dominant disease genes are expressed in homozygous dominant and heterozygous people (HH or Hh).Draw a Punnett square for Woody ad his wife.

The normal female condition is a result of the chromosomal pairing XX, while the normal male condition is XY.Certain genes located on the X chromosome, not associated with female sex characteristics, cause sex-linked recessive traits.As a result, females must receive two recessive alleles to exhibit any particular characteristic associated with one of these genes, while males need only receive one allele.The reason for the male anomaly is that the Y chromosome does not carry versions of the same genes as the X chromosome.

Consequently, only females can be true heterozygotes (one dominant allele and one recessive allele).

! Hemophilia is a rare heredity human disease of the blood.

The blood of individuals with this condition does not clot properly.Without the capacity for blood clotting, even a small cut can be lethal
.In ahemophiliac allele and (h) for the hemophilia alle-Linked Worksheet 3 Sex-Linked Practice Problems 1.In humans colorblindness (b) is an example of a sex-linked recessive trait.In this problem, a male with colorblindness marries a female who is not colorblind but carries the (b) allele.Using a Punnett square, determine the genotypic and phenotypic probabilities for their potential offspring.

2.

In fruit flies red eye color (R) is dominant to white eyes (r).In a cross between two flies, 50% of the male and 50% of the female offspring had red eyes.The other half of the males and females had white eyes.

What are the phenotype, and all possible genotypes, of the o

Dihybrid crosses use Punnett squares to distribute parental alleles from two genes into gametes (eggs and sperm or pollen and ovum) as would be predicted by meiosis.

! In garden peas, tallness (T) is dominant to shortness (t) and axillary flowers (A) are dominant to terminal flowers (a).

What are the expected ratios for the genotypes and phenotypes of the offspring if a heterozygous tall, heterozygous axillary plant is crossed wi-Linked Worksheet 4 Dihybrid Practice Problems 1.

In horses, the coat color black is dominant (B) over chestnut (b).

The trotting gait is dominant (T) over the pacing gait (t).

If a homozygous black pacer is mated to a homozygous chestnut, heterozygous trotter, what will be the ratios for genotype a-Sachs.

The female is homozygous recessi.
Genetics worksheet 2
3: Set up a Punnett Square for your Step 4: Multiply the decimal percentage short-haired and black? short-haired and brown?long-haired and black?long haired and brown?Incomplete dominance and co-dominanceSome pairs of alleles arent fully dominant or recessive.In these special cases, the heterozygotes seem to have both inherited traits or a blend of both traits.This is because both alleles are expressed equally.

Incomplete dominance is where the traits appear to blend.An example is the Four-oclock flower, where white-flowered plants crossed with red-flowered plants can produce pink-flowered plants.

An example of co-dominance is the color roan in horses and other animals, where a red-coated horse crossed with a white-coated horse can produce a foal with a roan coat (both white and red hairs).

The underlying genetics in both cases are the same.Instead of A or a to express dominant or recessive alleles, Is it possible for these two people.

Genetics wkst
Ratios tell you

there is an even chance of having offspring with black eyes as there is for havingoffspring with red eyes.That would be the same as a 50% probability of having red eyes, or a50% probability of having black eyes..**On the following pages are several problems.With each new problem, one sample is illustrated, makesure you look over the sample.

In your spiral notebook you must copy the first problem from eachsection along with its solution.

This means you should have a total of eight problems written outBrown eyesPart 3b.Bb or heterozygousc.There were too few offspring to form a conclusionPart 6.
1.a.Should have a filled in Punnett Squareb.Round Yellow : Round Green : Wrinkled Yellow : Wrinkled Green
c.3131
2 hY x hhPart 101.a.1.ff

2.

?

3.Ff

4.Ff

c.

Nod.12 is the grandson of 22.1.bb

2.?

3.Bb

4.Bb.