BIOL 3600 - CH.3 HW

Imagine that 2 unlinked autosomal genes with simple dominance code in goats for size where T is tall and t is short. For color, B is brown and b is white. If a short/white male goat mates with a tall/brown female goat of an unknown genotype what's the probability they would make short/white offpsring?

0.0625 probability

In mice an allele for apricot eyes (a) is recessive to an allele for brown eyes a+. At an independently assorting locus an allele for tan t coat color is recessive to an allele for black t+ coat color. A mouse that's homozygous for brown eyes/black coat color is crossed with a mouse with apricot eyes/tan coat. The F1 are intercrossed to make the F2. In a litter of 8 F2 mice what's the probability that exactly 2 will have apricot eyes/tan coats?

0.074 probability

Suppose that goats have 1 gene that codes for color where A is brown and a is white. The goats have another gene that codes for height where B is tall and b is short. If these 2 genes are unlinked, what's the probability that a cross between AaBb x Aabb parents will make 2/6 offspring that're white and tall?

0.137 probability

Suppose 2 parents are both heterozygous for sickle cell anemia, an autosomal recessive disease. They have 8 children. Use the binomial therorem to determine the probability that 3 of the children have sickle cell anemia and 5 of the children are healthy, round to the nearest tenth.

0.2 probability

Pink eye and albinism are 2 recessive traits found in the deer mouse, Peromyscus maniculatus. In mice with pink eye, the eye is devoid of color and appears pink from the blood vessels within it. Albino mice are completely lacking color both in their fur and in their eyes. Clark crossed pink‑eyed mice with albino mice the F1 had normal coloration in their fur/eyes. Then crossed F1 mice with mice that were pink‑eyed and albino and obtained the mice shown in the table below. It is very hard to distinguish between mice that are albino and mice that are both pink‑eyed and albino, so he combined these two phenotypic classes. Match the expected numbers of progeny with each phenotype if the genes for pink‑eye and albinism assort independently. Phenotype-Observed-Expected Wild/wild-12-(1) Wild/pink eyes-62-(2) Albino/wild/or pink eyes-78-(3) Total-152-152 Use a chi-square test to determine if the observed numbers of progeny fit the number exected with independent assortment. Does this chi-square value fit the number expected with independent assortment?

1. 38 2. 38 3. 76 33.05=X2 No

Phenylketonuria (PKU) is a disease that results from a recessive gene. 2 normal parents make a child with PKU. What's the probability that a sperm from the father will have the PKU allele? What's the probability an egg from the mother will have the PKU allele? What's the probability their next child will have PKU? What's the probability their next child will be heterozygous for the PKU gene?

1/2, 1/2, 1/4, and 1/2

In cats curled ears are from an allele Cu that's dominant over cu for normal ears. Black color is from an independently assorting allele G that's dominant over an allele for gray g. A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black with curled ears. What phenotypes and proportions are expected from the 2 crosses?

2 of the F1 cats mate: 9/16 black cats/curled ears, 3/16 black cats/normal ears, 3/16 gray cats/curled ears, 1/16 gray cats/normal ears F1 mates with gray/normal ear stray: 1/4 black cats/curled ears, 1/4 black cats/normal ears, 1/4 gray cats/curled ears, 1/4 gray cats/normal ears

Suppose parents in the interactive made 208 peas. Determine the expected number of yellow and wrinkled offspring. Suppose 1 of the pea plants is homozygous recessive for both genes resulting in YyRr x yyrr. Calculate the percentage of green and wrinkled offpsring. Suppose 1 of the pea plants is homozygous recessive for seed color in the cross YyRr x rrRr. Select the correct phenotypic ratio for the offpsring

39 expected yellow and wrinkled offspring 25% green and wrinkled offspring 6 yellow/round: 2 yellow/wrinkled: 6 green/round: 2 green/wrinkled

In pea plants, plant height is controled by a single autosomal dominant gene. Tall plants (H) are dominant to short plants (h). In a cross of 2 tall heterozygous plants, which phenotype ratio is expected from the resulting offspring?

3:1

Alleles A and a are located on a pair of metacentric chromosomes. Alleles B and b are located on a pair of acrocentric chromosomes. A cross is made between individuals having the genotypes AaBb and aabb. Label the gametes of the aabb parent with the chromosomes carrying the correct alleles. Which gametes can the AaBb parent generate? For example progeny resulting from this cross, match the chromosomes as they would appear in a cell at G1, G2, and metaphase of mitosis.

4 ab gametes AB, ab, aB, and Ab G1: AaBb and Aabb G2: AAaaBBbb and AAaabbbb Metaphase of mitosis: AAaaBBbb and AAaabbbb

Huntington's disease is an inherited autosomal dominant disorder that can affect men and women. Imagine a couple has had 7 children and later in life the husband develops Huntington's disease. He's tested and it's discovered he's heterozygous for the disease allele Hh. The wife is also genetically tested for the disease allele and her test results show she is unaffected hh. What's the percent probability that the first child of this couple will have Huntington's disease? What's the percent probability that 6 of the 7 children will have Huntington's diease?

50% and 5.5%

In pea plants the allele for red flower color F is completely dominant to the allele for white flower color f. Complete the Punnett square showing the genotypes possible among the offspring when 2 heterozygous individuals, FfxFf, are crossed. In this cross between 2 pea plants, what are the chances that an offspring with red flowers will be made?

75%

In watermelons, bitter fruit B is dominant over sweet fruit b and yellow spots S are dominant over no spots s. The genes for these 2 characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to make the F2. What will be the phenotypic ratio in the F2? If an F1 plant is backcrossed with the bitter/yellow spotted parent what phenotypes and proportions are expected in the offspring? If an F1 plant is backcrossed with the sweet/nonspotted parent what phenotypes and proportions are expected in the offspring?

9/16 bitter fruit/yellow spots, 3/16 bitter fruit/no spots, 3/16 sweet fruit/yellow spots, 1/16 sweet fruit/no spots All bitter fruit with yellow spots 1/4 bitter fruit/yellow spots, 1/4 bitter fruit/no spots, 1/4 sweet fruit/yellow spots, 1/4 sweet fruit/no spots

Suppose there's a vial with a single generation of flies from a cross. There's an interesting phenotype where many individuals have abnormally long hairlike bristles and sensory organs extending from the dorsal thorax as opposed to short wirelike wild type bristles among the other siblings. References state this mutant has a dominant mutation called Suave (Su) and the phenotype of flies that're heterozygous or homozygous for Su appear phenotypically identical. Which fly should be crossed to a Suave male from this vial to generate progeny that help determine the male's genotype?

A wild type female sibling

In humans, oculocutaneous (OCA) albinism is a collection of autosomal recessive disorders characterized by an absence of the pigment melanin in skin, hair, and eyes. So, normal pigmentation A is dominant over albino characteristics a. Assume the phenotype is determined by a single gene with 2 alleles. If both parents have normal pigmentation, what are all of the possible genotypes that may be seen in their offspring?

AA, Aa, or aa

At the end of your biology class, your professor asks you to develop a project to determine the genotype of a plant with red flowers. Red petal color R is dominant to pink flower color r. To accomplish this task, you cross the plant with the unknown genotype with heterozygous red flowered plants. What ratios are valid predictions of lower colors in the offspring when the unknown is paired with an Rr plant?

All red flowers 3 red:1 pink flowers

The figure represents a pair of homologous chromosomes. Assign the appropriate term to each copmponent.

BB-homozygous dominant locus dd-homozygous recessive locus Nn-heterozygous locus

How did Mendel use self-pollination and cross-pollination techniques in his experiments with flower color to observe the basic patterns of inheritance?

By cross-pollinating a parental generation of plants with different colored flowers and allowing the F1 generation to self-pollinate, Mendel saw the basic patterns of inheritance in the F2 generation.

Horse coat color is controlled by many genes, including extension and cream. The extensions gene E determines pigment color. E−, representing either Ee or EE, produces a bay coat, and ee produces a red coat. The cream gene Cr is incomplete dominant and functions by decreasing the intensity of the E color phenotype. The crcr genotype does not alter the red/bay phenotype. The Crcr genotype changes red into palomino and bay into buckskin (single dilute). The CrCr genotype alters red into cremello and bay into perlino (double dilute). Suppose two buckskin horses, each with the Ee Crcr genotype, are crossed. Use the forked‑line (branched diagram) method to determine the proportion of each phenotype and each genotype in the next generation.

EeCrcr x EeCrcr 3/4 E-bay -> 1/4 crcr, 1/2 Crcr, 1/4 CrCr -> 3/16 bay, 3/8 buckskin, 3/16 perlino 1/4 ee red -> 1/4 crcr, 1/2 Crcr, 1/4 CrCr -> 1/16 red, 1/8 palomino, 1/16 cremello

Suppose a father with genotype AaBbCcDdee and mother aaBbCCDdEe want to have children. Assume each locus follows Mendelian inheritance patterns for dominance. What proportion of the offpsring will have each of the specified characteristics?

Father's genotype: 0.03 Mother's genotype: 0.03 Father's phenotype: 0.14 Mother's phenotype: 0.14 Neither parent's phenotype: 0.72

In fruit flies, gray bodies G are dominant over black bodies g and brown pigments N are dominant over yellow pigments n. Each individual has 2 alleles for each trait. If a fly that's homozygous dominant for both traits is crossed with a fly that's homozygous recessive for both traits, what's the predicted genotype of the offpsring?

GgNn

What can be concluded from Gregor Mendel's experiments with pea plants?

Only 1 allele determines the phenotype in heterozygous individuals and allelic combinations for different genes may differ between parents and their offspring

Use the image to observe the results of a cross between a tall pea plant and a short one. What phenotypes and proportions will be made for the 2 crosses?

Tall F1 progeny backcrossed to the short parent: 1/2 tall and 1/2 short Tall F1 progeny backcrossed to the tall parent: 3/4 tall and 1/4 short

A geneticist discovers an obese mouse in his laboratory colony. He breeds this obese mouse with a normal mouse. All the F1 mice from this cross are normal in size. When he interbreeds 2 F1 mice, 8 are normal in size and 2 obese. The geneticist then intercrosses two of his obese mice, and he finds that all of the progeny from this cross are obese. These results lead the geneticist to conclude that obesity in mice results from a recessive allele. A second geneticist at a different university also discovers an obese mouse in her laboratory colony. She carries out the same crosses as the first geneticist and obtains the same results. She also concludes that obesity in mice results from a recessive allele. 1 day, the 2 geneticists meet at a genetics conference, learn of each others experiments, and decide to exchange mice. They both find that, when they cross two obese mice from the different laboratories, all the offspring are normal. However, when they cross two obese mice from the same laboratory, all the offspring are obese. What option best explains their results?

The 2 obesity alleles are recessive to the wild type alleles but are located at different loci

White w coat color in guinea pigs is recessive to black W. In 1909 Castle and Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offpsring from the mating were black in color. What explains the results of this cross? What's the genotype of the offpsring from this cross? What are the conclusions indicated by this experiment regarding the validity of the pangenesis and the germplasm theories in this chapter?

The color of the offpspring was determined by genes in the transplanted ovary, not the genes of the female who gave birth. Ww The production of black guinea pig offpsring suggests that the allele for black coat color was passed along to the offpsring from the transplanted ovary, supporting the germplasm theory. And because no white guinea pigs were made, no white coat alleles traveled to the ovary and into the gametes of the white female, indicating pangenesis did not happen.

Imagine that a scientist studies 2 traits in peas. The scientists noticed that round is dominant over wrinkled with pea shape. Yellow is dominant over green with pea color. To determine if these traits are linked, 2 individuals that are heterozygous for both traits were crossed. That data in the table represents the number of offpsring made by this dybrid cross. Phenotypic ratios represents the predicted proportion of offspring with each set of traits what would be made if the traits independently assort.. Traits-Phenotypic Ratio-Observed-Expected Round/yellow-9/16-491-498 Round/green-3/16-179-166 Wrinkled/yellow-3/16-165-166 Wrinkled/green-1/16-50-55 What can be determined about these traits based on chiaquare analysis?

These traits assort independently

How is a true breeding purple flowered pea plant different from a hybrid purple flowered pea plant?

They have the same phenotype but different genotypes

Suppose a man is heterozygous for heterochromia (an autosomal dominant disorder that causes 2 different colored eyes in an individual) makes 25 offspring with his normal eyed wife. Of their children, 15 were heterochromatic and 10 were normal. Calculate the chi-square value for this observation. Identify the statement that best interprets the results of the X2 analysis.

X2=1 It's not unusual that a heterozygous man made 15/25 offspring with heterochromia