B. If the mutant allele in this cross is a dominant. autosomal allele, what phenotype(s) should be present in the F2 of your cross, and what fraction of the F2 should each phe- notype represent? (Include "male" and "female" as part of your description for a given phenotype, only if it is necessary to do so.) C. If the mutant allele in this cross is a recessive, autosomal allele, what phenotype(s) should be present in the F2 of your cross, and what fraction of the F2 should each phe- notype represent? (Include "male" and "female" as part of your description for a given phenotype, only if it is necessary to do so.)

B. If the mutant allele in this cross is a dominant. autosomal allele, what phenotype(s) should be present in the F2 of your cross, and what fraction of the F2 should each phe- notype represent? (Include "male" and "female" as part of your description for a given phenotype, only if it is necessary to do so.) C. If the mutant allele in this cross is a recessive, autosomal allele, what phenotype(s) should be present in the F2 of your cross, and what fraction of the F2 should each phe- notype represent? (Include "male" and "female" as part of your description for a given phenotype, only if it is necessary to do so.)

Name: Answer the questions as if you were crossing flies with a “big head” m
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: Answer the questions as if you were crossing flies with a “big head” mutant trait with flies that have normal sized heads Q6 Answer this question for your particular cross. Include separate information for the tworeciprocal crosses. Your instructor m

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 5QP: Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of...

See similar textbooks

Similar questions

Show Comments Track Changet O Markup OptionS Track Changes wious Next Genetics Lab Exercise (by Dr. Lapik) Question 2. 田 BB = black Analyze the following dihybrid cross: Bb = black Parents: bbLI bb = white LL = short hair U = short hair Gametes: U= long hair Punnett Square: How many of the offspring are: Black, Short Black, Long White, Short White, Long Autupay Credentials Username ENNIN Citrix
GI Gi| GI Gi gI gi 5 2 | 3 23 gI 1 gi [X 46 zoom in G = green g=yellow I = inflated i = constricted
tion 8: below is the pedigree of inheritance of phenylketonuria (PKU). We will designate the letter Caven for the dominant allele and "p" for the recessive allele. 4 The pedigree shows that the pattern of inheritance for the allele for phenylk ylketonuria is: I. II. 1 III. IV. Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive b. The parents in generation I have how many children: I. 3 Boys II. 3 Girls III. IV. 3 Boys and 1 Girl 3 Girls and 1 Boy c. What is the genotype of individual 1 in generation III: I. PP II. pp III. Pp " O 1 III. 50% E III 1 ▬ 2 2 IV. 25% 1 3 IV. Can be PP or Pp ii. Suppose that a man having type AB blood marries a woman having type O blood. What is the probability that their child will have type A blood? I. 100% II. 75% 2 4 3
My Question is what is the probability their first child will have hemophilia and drawn pedigrees for family members with genotypes. My explantion so far: A man has both X and Y chromosomes as sex chromosomes in his body. Here, though the brother of the man is hemophiliac, a man can’t be a carrier of hemophilia. So, it can be said that his chromosome is “XnY”.Here, the “n” stands for “normal”.Though the paternal uncle is hemophiliac, a man cannot be a carrier of hemophilia, his niece will not be a career. So it can be said that the woman is also not a carrier and has the “XnX” chromosome.So, as the mother is not a carrier, their first child does not have a chance of having hemophilia. This can be determined as it is known that there is no hidden carrier of hemophilia in the family.
/d/1n5NtidRwTwUzcDkDPi5Z9P_SHPZ91A-XH-pfftLbhNc/edit (1) O pols Add-ons Help Last edit was seconds ago BIUA ミ: 12 + ext Calibri I|1 6 I 2 Section 5: Trihybrid cross and Laws of probability For a trihybrid cross, in which inheritance of alleles for three genes is tracked, drawing a Punnett square that combines all three genes may not be practical. Instead the laws of probability may be used. The product law of probabilities says that when alleles for separate genes segregate independently, we can figure out the probability of a particular combined genotype by multiplying the probability of the alleles for each gene. 13. We cross a homozygous tall pea plant with yellow, round seeds to a homozygous dwarf pea plant with green, wrinkled seeds. All the F1 offspring are all tall plants with yellow, round seeds. a. What are the expected F2 ratios (use fractions) of tall and dwarf plants? b. What are the expected F2 ratios (use fractions) of yellow and green seeds? C. What are the expected F2…
PEDIGREES: Problem (continued) This pedigree shows the inheritance of cystic fibrosis in this family. I • QUESTIONS ••. 5. What is the genotype of individual II-3? Use the letter "f" to 1 2 represent the disease allele. II 1 2 3 6. Individuals II-I and II-2 are sisters. Explain how it is possible for one sister to have cystic fibrosis but NOT the other. III 1 2 3
Let us practice it again! Analyze the pedigree below to answer the questions that follow. Huntington's disease a disorder in which nerve cells waste away, or disintegrate, is passed down through families. certain parts of the brain Huntington's diseate llustration ereated in htps://pregenygenetion.com/ 1. What members of the family above are affected with the Huntington's disease? 2. Tnere are no carriers ior Huntungton's disease you either have it or you do not. Is Huntington's disease caused-by a dominant or recessive trait? 3. Identify the genotypes of the following individuals using the pedigree above. (homozygous dominant, homozygous recessive, heterozygous). I- 1 II -1: II -3: III - 4 : 4. How many children did individuals I-1 and I-2 have? 5. How many girls did II-1 and II-2 have? How many have Huntington's Disease? 6. How are individuals III-2 and II-4 related? I-2 and III-5?
Drosohpila Punnet Square of Crosses. I need results of F1 & F2 generation using Punnett Squares for: Make Punnet Squares of the following crosses •Drosophila Female wildtype cross Male White-eye •Drosophila Male wildtype cross Female White-eye •Drosophila Female Wild Type cross Male Scarlet Eye •Drosophila Male Wild Type cross Female Scarlet Eye Also, Which allele is heterozygous and which is homozygous, & which is dominant and which is recessive?
ABO Blood TypeThe following pedigree shows the incidence of ABO blood types in a family. dentify the genotypes of the following individuals: Individual Genotype II-1 II-2 II-4 II-5 III-2 III-3
Please answer all of them, they are all connected. PEDIGREE ANALYSIS and SYMBOLOGY: Examine the pedigree which has X linked Dominant inheritance of disorder. Use letter X* (asterisk denotes disorder) as genotype of the individuals which can be XX, XY, X*X*, X*X and X*Y. a. What is the genotype of IV-6? b. What is the genotype of III-6? c. What is the genotype of II-3? d. What is the genotype of III-8? e. If couple I-1 and I-2 will have a son, what is the probability of having the disorder? f. If couple III-8 and III-9 will have another child, what is the probability of having the disorder? g. Theoretically, if individual IV-3 and individual IV-5 will marry and will have a child, what is the probability of having a child without the X-linked disorder?
Bb Take Test: Mod 9 Lab - Genetic X + ← ] C G QUESTION 3 In the pedigree below, all shaded individuals express the gene in question. For example, Arlene "has" the trait, she displays the phenotype in question. For example, if we were following the inheritance pattern of a widow's peak, Arlene has a widow's peak (that is NOT the trait here, just an example). Unshaded individuals (blank circles and squares) do not manifest the trait in question, but their specific genotype is unknown - they could be heterozygous, homozygous dominant, or homozygous recessive. What is the mechanism of inheritance of this trait? Sandra Daniel recessive dominant Tom 990 Alan QUESTION 4 George Sam Tina Christopher Arlene Wilma Ann O Carla R Update Michael In the pedigree below, all shaded individuals express the gene in question. For example, Arlene "has" the trait, she displays the phenotype in question. For example, if we were following the inheritance pattern of a widow's peak, Arlen a widow's peak (that…
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?