genetics

Welcome to Dr. Kate Brilakis' Learning Portal

Here's a sex-linked cross to try for practice...
The ability to see color is due to a gene found on the X chromosome. The dominant allele permits color sight while the recessive allele does not. A carrier woman mates with a colorblind male. What is the chance they produce a daughter who is colorblind?

Pssst....the answer is: 1/2 of their daughters will be colorblind. The only way for a couple to produce a colorblind female is when the father is colorblind...

male X female

Epistasis
occurs when genes found at two different loci both affect the expression of a single trait. A gene can either modify or mask the phenotype.

Romanov's
and Hemophilia

To solve a dihybrid cross...
select a letter to represent each of the two genes. Assign the upper/lower case to the dominant and recessive allele for each gene. Write out the cross using the letters. For example: in the problem above, I will select F for fur color and E for eye color. F = Brown fur and f = white fur
E = black eyes and e = red eyes
Use these letters to write out the cross:

A heterozygote = FfEe is crossed with a white, red eyes mouse = ffee so the cross is FfEe X ffee
Now, separate this dihybrid cross into two mono crosses crossing the alleles for fur color with each other and the alleles for eye color with each other... Ff x ff and Ee x ee
Complete these two punnet squares (refer to this pic from above :
This time, the results of the punnet squares are:
Cross 1: fur color: 1/2 brown fur (Ff) 1/2 white fur (ff)
Cross 2: eye color: 1/2 black eyes (Ee) 1/2 red eyes (ee)
To answer the question,
Multiply fur x eye color (foil) 1. 1/2 brown fur x 1/2 red eyes eyes = 1/4 of the F1 will have brown fur and red eyes.
2. Do this for all of the possible offspring. The phenotypic ratio for the cross FfEe x ffee is: 1/4:1/4:1/4:1/4 or 1:1:1:1

Genetics:
the study of how characteristics are inherited

Quick review of mitosis and meiosis...

In humans, the allele that produces brown eyes is dominant over the one that results in blue eyes. A heterozygous man mates with a heterozygous woman. What is the chance they have a brown eyed child? What is the genotypic ratio of the F1 from this cross?

CFTR Gene: Cystic Fibrosis

The "Father" of genetics explained what would become
Mendelian Genetics

Terrific Genetics Info

what's incomplete dominance??

vs single gene autosomal mutations that cause disease

what is a repeat gene?

the answer is
X-inactivation

Here's a blood type cross to try for practice...
A blood type A+ woman has a father who is O- .
(hint: telling you her dad's blood type allows you to determine her genotype.)
She mates with a man who is blood type AB-. What is the chance they have a child who is blood type B+?
Solve this as you would any dihybrid cross.
Two genes: the ABO gene and the Rh gene

Rh + = Rh antigen present
Rh - = Rh antigen absent

how are these
blood types inherited?

Rh
factor

and cystic fibrosis

Mendel's observations

D = Brown Hair
d= blond hair
T = curly hair
t = straight hair

parent cross =
DdTt X DdTt

What are the genotypic and
phenotypic ratios of the F1?

Please refer to the Genetics outline for
instructions on how to solve a basic genetics problem.
Here's a dihybrid cross to try for practice...(try it first and then see the answer below)
In mice, white fur is recessive to brown fur and red eyes are recessive to black eyes. A heterozygote for both traits is crossed with a white, red eyed mouse.
1. What is the chance that their litter contains brown mice with red eyes?
2. What is the phenotypic ratio of the F1 from this cross?

inheritance pattern of Huntington's disease =

autosomal dominant

Genetics Movies

phenotype genotype(s)
Type A AA, AO
Type B BB, BO
Type AB AB
Type O OO

due 11/29th hard copy
1. Consider two genes in horses, one for color and one for movement.
Color: black color (B) dominates chestnut color (b).
Movement: trotting gait (T) dominates pacing gait (t).
A horse that is homozygous recessive for both color and movement is crossed with
a horse that is heterozygous for both genes.
List the genotypic and phenotypic ratios of all possible offspring.

2. In humans, a type of hemophilia is due to the recessive allele of a gene found on the X chromosome. A carrier female mates with a male who has blood that clots normally.
List the genotypic and phenotypic rations of their offspring.

3. What blood types are not possible for children with parents who have blood types AB and O?

4. Freckles occur in some humans after sun exposure. Freckles are due to a dominant allele. A couple, both of whom have freckles, have a child who does not freckle.
What are the genotypes of the parents and their child?

5. Cystic fibrosis is a homozygous recessive disorder. A female with cystic fibrosis mates with a man who is heterozygous for this gene. What fraction of their offspring might develop this disease?

6. In chickens, feathered legs are dominant over bald legs and a pea comb is dominant over a single comb. Two roosters, Tim and Jerry and two hens, Dolores and Peggy, all
exhibit feathered legs and pea-combs.
Tim, when bred to either hen, produces chicks that are all feathered and pea-combed. Jerry, when bred to Dolores, produces chicks both feathered and clean legged that all have pea combs.
When Jerry is bred to Peggy, all of their chicks are feathered. Some of these chicks exhibit pea combs and some single-combs.
What are the genotypes of Tim, Jerry, Dolores and Peggy?

7. In cattle, there is an allele called the dwarf allele which, in a heterozygote, produces calves with legs which are shorter than normal. Homozygous dwarf calves spontaneously abort early or are stillborn, a condition called homozygous lethality. If a dwarf bull is mated to 400 dwarf cows, what phenotypic ratio would you expect among the living offspring?

8. In cats, there is a coat color gene located on the X chromosome. This gene has two alleles: orange and black. A female who has an orange allele on one of her X chromosomes and a black allele on the other X chromosome winds up expressing a tortoiseshell color = a splotchy mix of both orange and black).
Predict the genotypic and phenotypic ratios of the F1 of the following crosses.
a. Black female X Orange male
b. Orange female X Black male
c. Tortoiseshell female X Orange male

9. A woman who has Type A / Rh negative blood mates with a man with Type AB / Rh positive blood.. This couple has four children: A-, B+, AB-, A+. What are the genotypes of the people in this family?

10. A woman with blood type O is given blood type AB in a transfusion. Her blood clots.
Why? (be specific)

...so A+ means
the A antigen is present
and
the Rh antigen present

What about my buckskin??...
Enter the Cream Allele (CCr)

The cream "gene" is a mutant allele of the SLC45A2 gene, which encodes membrane-associated transporter protein (MATP).
Variations in this gene produce:
buckskin (dilute bay)
palomino (dilute chestnut)
smoky black (dilute black)

and of course there's the horse...

and cystic fibrosis

we have 23 pairs of homologous chromosomes in each somatic cell.
for each of our homologous pairs of chromosomes,
one chromosome of the pair was inherited from your mom
and one was inherited from your dad.

Mendel used the term factors to describe inheritance. Today we know these factors are genes with variations called alleles.

X-linked cross
also called sex-linked cross

SO MANY GENES!

In vertebrates, one of the functions of Runx-2 is to regulate the differentiation of osteoblasts, which are the cells responsible for forming new bone. Runx-2 contains two repeats, one coding for 18–20 glutamines (the poly-Q region) and another coding for 12–17 alanines (the poly-A region). A statistical comparison of the total length of both of these repeats (poly-Q + poly-A) with various parameters of canine skull size revealed a correlation with the dogs' midface length, as well as with a property called clinorhynchy, or dorsoventral nose bend. To better understand what clinorhynchy looks like, think about the distinctive, long nose of bull terriers, which features a downward droop. Bull terriers tend to have a short pair of tandem repeats in Runx-2, and they have long midfaces and pronounced downturn of the snout. The breed has been intentionally selected for this trait, and museum specimens over the past 70 years show increased prominence of this feature .

what are the antigens and antibodies and why does it matter?

solving dihybrid cross using
16 box punnet square

one X chromosome in each cell is transcriptionally silenced via histone modification...H4 is hypoacetylated on the inactive X chromosome

is the CF allele maintained in the human population because it confers resistance?

more genetics crosses to try for practice...

1. In dogs, a long coat is caused by a recessive genetic mutation in the FGF5 gene. Because it is a recessive mutation, a dog must have two copies of the recessive long-hair allele (l) to express that gene and cause the dog to have long hair. A dog that has short hair could have one or two copies of the dominant short-hair allele (L) to have the short-hair phenotype. If a heterozygote mates with another heterozygote, what % of their pups should have long hair?

2. Hemophilia is due to the presence of a recessive allele/absence of the dominant allele for the clotting factor VIII gene which is located on the X chromosome.
a. Why is a boy more likely than a girl to exhibit hemophilia?
b. If a female who is a carrier for hemophilia mates with a hemophiliac male, what percent of their daughters should statistically have hemophilia?

3. A person who is type AB mates with a person who is type O.
a. What percent/fraction of their offspring should be type A?
b. If the parents were also both heterozygous for the Rh gene, what percent of their offspring would be Rh-?
c. Combining both of these crosses, what percent of their offspring would be type A-?

4. Why would a blood clot form if a person who is blood type A is given a transfusion of type B blood?
  Please be specific re the antigen/antibody reaction. I will provide a different transfusion scenario on
  the assessment...this is just an example.

examples of X inactivation in humans exhibited in phenotype

The images below are Punnet Squares. We use Punnet squares to determine the likelihood that offspring will exhibit certain genotypes/phenotypes. The genotypes of both parents are "crossed". The four boxes in the center represent the genotypes of potential offspring. Here's one to try for practice...
In cattle, the allele that produces a black coat is dominant to the allele that produces a red coat. A heterozygous bull mates with a heterozygous cow. What is the chance that they produce a calf who is red? What is the genotypic ratio of the F1 from this cross?

3 alleles are A, B, and O
A and B are co-dominant
O is recessive to both A and B

Mendel's laws

The Rh alleles are written

as R = Rh antigen present (dominant)
r - Rh antigen absent (recessive)
or you can write them as + and -

blood type inheritance

Chance of getting DDTT: 1/4 X 1/4 = 1/16
What is the genotypic ratio? DDTT = 1/16 DDTt =1/8 (2/16) DDtt = 1/16
DdTT 1/8 (2/16) DdTt 1/4 (4/16) Ddtt 1/8 (2/16)
ddTT 1/16 ddTt 1/8 (2/16) ddtt 1/16
ratio = 1:1:1:2:4:2:1:2:1
What is the phenotypic ratio of the F1?

more blood type inheritance

so females have two X chromosomes and males have only 1...how does that work?