Answered: Suppose a man and a woman are each… | bartleby

Related questions

Question

Suppose a man and a woman are each heterozygous carriers of a mutation causing a fatal hereditary disease not on
the RUSP list. Prenatal genetic testing can identify the genotype of a fetus with regard to this disease and can identify
fetuses with the disease. What do you think are the three or four most important factors this couple should consider in
their decision making about having children?

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

You are to discover the chromosomal location (autosomal or X linked) and mode of inheritance for two new fly mutations coughrotato (cp) and happyhour, (bb) The homozygous couchpotato (cpcp) phenotype is (sedentary) an extreme lack of activity The homozygous happyhour (hhhh) can drink any other fly under the table so phenotype (drinker) PARENTS Cross Male double homozygous mutants cp hh with Female wild wild CP HH are to be crossed BUT FIRST you don't know if either gene is on the X chromosome Parental diploid genotypes could be: male çpcphbbh female CPCPHHHH if Both genes autosomal or Male XcpYhbhh female XCPXCPHHHH if cp on the X chromosome and bh autosomal or Male cpcpXhhY female CPCPXHHXHH if hh on the X chromosome and cp autosomal What are the female and male PARENTAL GAMETE GENOTYPES for the above parental crosses if: a) Both genes were autosomal ((REMEMBER BOTH GENDER HAVE ONE GAMETE GENO) b) cp gene on the X chromosome and hh autosomal (REMEMBER MALES WITH HAVE 2 GAMETE GENO) c)…
Hemophilia, disease in which the blood lacks a clotting factor, is caused by an X linked recessive gene. Joe doe not have hemophilia and Lucille is heterozygous for the condition. What is the chance that their MALE child will have hemophilia? (Note: you are calculating the probability for their MALE children only, in other words if they have 1 male child, what is the probability that he will be born with the disease?) 0% 1/4 or 25% 2/4 or 50% 3/4 or 75% 4/4 or 100%
please dont copy other answers and give your own answer. i will dislike if you dont follow what i said
Examine the pedigree and answer the following questions; shaded individuals show the trait; genotypes are all unknown. The individual marked with the question mark is of an unknown genotype but does have the trait. A. What mode of inheritance is the most likely for this trait, autosomal recessive or autosomal dominant? State your rationale for full credit. B. What is the genotype of the individual marked with the question mark? (Heterozygous, homozygous, or unknown)
Fill out the blanks of the pedigree (AA, Aa, or aa) How many children did the original couple have? How many grandchildren do they have?.
The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominant trait (polydactyly - P; wildtype - p). Cystic fibrosis (CF) is an autosomal recessive trait (cystic fibrosis - f; wildtype - F). A PD woman, otherwise normal in phenotype, marries a healthy normal man. Their 4 children are: 1) normal, 2) PD, 3) CF, 4) CF + PD. You will walk through a series of steps to answer this question: What is the probability that their 5th child will have at least one of these conditions? Here is the first step: 1. What is the cross? (Hint: You can use the 4 existing children to determine the genotypes of the parents.) O PpFF (female) x ppFF (male) O PoFf (female) x ppft (male) O pof female) x PPFI (male) O PPFF Ifemale) x ppFf (male) 2. What is/are the target genotypes? O pof OP.F. O pott O P.M 3. What is the probability the child will have PD AND cystic fibrosis? Answer to two decimal places (eg. 0.88). 4. What is the probability that their 5th child will have at…
Frank and Susan both have freckles but they're firstborn son Lawrence does not demonstrate the characteristics. What are the most likely genotype of Frank Susan and Lawrence?
Genetic Inheritance Patterms Retinitis pigmentosa (RP) can be autosomal recessive, autosomal dominant, or x-linked. Apparently the dominant forms are often less severe. Eric's form of RP is Usher's Syndrome, which is an autosomal recessive inheritance (i.e., you must get a copy of the defective gene from your Mom and one from your Dad). Autosomal means it is not carried on one of the chromosomes that determines sex. Usher's Type II is recessive, so for Eric this means that both his Mom and Dad are carriers of this condition. His brother, Dirk, does not have any symptoms of RP. Draw a pedigree of Eric's family showing possible genotypes and chances of having RP. Your pedigree must include his parents, Eric, his brother Dirk, and a potential daughter of Eric's. Draws a pedigree that shows the correct genotypes and chances of having retinitis pigmentosa for Eric and his family. D.Focus (United States) a S W 37 F 八口